Indefinite integral of $frac{tan{(1+x^2)}}{1+x^2}$












6












$begingroup$


Let $f(x) = frac{tan{(1+x^2)}}{1+x^2}$ , find $int f(x)dx$ . I've tried many substitutions (including trigonometric substitutions like $x=tan theta$ ) and also integration by parts but didn't work . We can apply power series but it doesn't solve problem .






real-analysis calculus integration indefinite-integrals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I am afraid that, even using special functions, closed form solutions could be difficult to get.
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 8:20






  • 1




    $begingroup$
    Hmm.. this guy is fighting back pretty hard..
    $endgroup$
    – InertialObserver
    Dec 22 '18 at 8:21










  • $begingroup$
    Where did you find this beast?
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 22 '18 at 8:26










  • $begingroup$
    @MohammadZuhairKhan My friend asked that beast!
    $endgroup$
    – S.H.W
    Dec 22 '18 at 8:31






  • 3




    $begingroup$
    Are you sure that he/she is a friend ?
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 9:00
















6












$begingroup$


Let $f(x) = frac{tan{(1+x^2)}}{1+x^2}$ , find $int f(x)dx$ . I've tried many substitutions (including trigonometric substitutions like $x=tan theta$ ) and also integration by parts but didn't work . We can apply power series but it doesn't solve problem .






real-analysis calculus integration indefinite-integrals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I am afraid that, even using special functions, closed form solutions could be difficult to get.
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 8:20






  • 1




    $begingroup$
    Hmm.. this guy is fighting back pretty hard..
    $endgroup$
    – InertialObserver
    Dec 22 '18 at 8:21










  • $begingroup$
    Where did you find this beast?
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 22 '18 at 8:26










  • $begingroup$
    @MohammadZuhairKhan My friend asked that beast!
    $endgroup$
    – S.H.W
    Dec 22 '18 at 8:31






  • 3




    $begingroup$
    Are you sure that he/she is a friend ?
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 9:00














6












6








6


3



$begingroup$


Let $f(x) = frac{tan{(1+x^2)}}{1+x^2}$ , find $int f(x)dx$ . I've tried many substitutions (including trigonometric substitutions like $x=tan theta$ ) and also integration by parts but didn't work . We can apply power series but it doesn't solve problem .






real-analysis calculus integration indefinite-integrals







share|cite|improve this question









$endgroup$




Let $f(x) = frac{tan{(1+x^2)}}{1+x^2}$ , find $int f(x)dx$ . I've tried many substitutions (including trigonometric substitutions like $x=tan theta$ ) and also integration by parts but didn't work . We can apply power series but it doesn't solve problem .






real-analysis calculus integration indefinite-integrals




real-analysis calculus integration indefinite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 '18 at 8:08









S.H.WS.H.W

1,2031923




1,2031923








  • 2




    $begingroup$
    I am afraid that, even using special functions, closed form solutions could be difficult to get.
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 8:20






  • 1




    $begingroup$
    Hmm.. this guy is fighting back pretty hard..
    $endgroup$
    – InertialObserver
    Dec 22 '18 at 8:21










  • $begingroup$
    Where did you find this beast?
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 22 '18 at 8:26










  • $begingroup$
    @MohammadZuhairKhan My friend asked that beast!
    $endgroup$
    – S.H.W
    Dec 22 '18 at 8:31






  • 3




    $begingroup$
    Are you sure that he/she is a friend ?
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 9:00














  • 2




    $begingroup$
    I am afraid that, even using special functions, closed form solutions could be difficult to get.
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 8:20






  • 1




    $begingroup$
    Hmm.. this guy is fighting back pretty hard..
    $endgroup$
    – InertialObserver
    Dec 22 '18 at 8:21










  • $begingroup$
    Where did you find this beast?
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 22 '18 at 8:26










  • $begingroup$
    @MohammadZuhairKhan My friend asked that beast!
    $endgroup$
    – S.H.W
    Dec 22 '18 at 8:31






  • 3




    $begingroup$
    Are you sure that he/she is a friend ?
    $endgroup$
    – Claude Leibovici
    Dec 22 '18 at 9:00








2




2




$begingroup$
I am afraid that, even using special functions, closed form solutions could be difficult to get.
$endgroup$
– Claude Leibovici
Dec 22 '18 at 8:20




$begingroup$
I am afraid that, even using special functions, closed form solutions could be difficult to get.
$endgroup$
– Claude Leibovici
Dec 22 '18 at 8:20




1




1




$begingroup$
Hmm.. this guy is fighting back pretty hard..
$endgroup$
– InertialObserver
Dec 22 '18 at 8:21




$begingroup$
Hmm.. this guy is fighting back pretty hard..
$endgroup$
– InertialObserver
Dec 22 '18 at 8:21












$begingroup$
Where did you find this beast?
$endgroup$
– Mohammad Zuhair Khan
Dec 22 '18 at 8:26




$begingroup$
Where did you find this beast?
$endgroup$
– Mohammad Zuhair Khan
Dec 22 '18 at 8:26












$begingroup$
@MohammadZuhairKhan My friend asked that beast!
$endgroup$
– S.H.W
Dec 22 '18 at 8:31




$begingroup$
@MohammadZuhairKhan My friend asked that beast!
$endgroup$
– S.H.W
Dec 22 '18 at 8:31




3




3




$begingroup$
Are you sure that he/she is a friend ?
$endgroup$
– Claude Leibovici
Dec 22 '18 at 9:00




$begingroup$
Are you sure that he/she is a friend ?
$endgroup$
– Claude Leibovici
Dec 22 '18 at 9:00










1 Answer
1






active

oldest

votes


















3












$begingroup$

If the problem was
$$int_0^a frac{tan{(1+x^2)}}{1+x^2},dx qquad text{with} qquad 0 leq a lt sqrt{frac{pi -2}{2}}approx 0.7555$$ it could be possible to have an approximation of it using a Padé approximant built at $x=0$.



Using $k=tan(1)$, we should have
$$ frac{tan{(1+x^2)}}{1+x^2}=frac {3 k left(-2 k^2+2 k+1right)+3 left(-k^2+k+1right)x^2 +(2 k-3) left(k^2+1right)x^4} {3(-2 k^2+2 k+1)+ 6 (k-1)^2 k x^2+(6 k^3-7 k^2-4)x^4}$$ which can be integrated using partial fraction decomposition (leading to a nasty expression. Numerically, this would be "almost" (making the coefficient rational)
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{184 }{789}x+frac{1099} {3526}tan ^{-1}left(frac{529
}{801}xright)+frac{1409 }{1667}tanh ^{-1}left(frac{692
}{523}xright)$$ For a few values of $a$, some results (the so called "exact" being obtained by numerical integration)
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57609 & 0.57610 \
0.40 & 0.67171 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01775 & 1.01775 \
0.60 & 1.17192 & 1.17195 \
0.65 & 1.37127 & 1.37136 \
0.70 & 1.67673 & 1.67722 \
0.75 & 2.66909 & 2.68502
end{array}
right)$$



Edit



We can make the approximation better using the fact that
$$f(x)=left(x^2-frac{pi }{2}+1right)frac{ tan left(x^2+1right)}{x^2+1}$$ is a quite nice function. Using a $[4,2]$ Padé approximant, we get
$$f(x) simeqfrac {a_0+a_1 x^2+a_2 x^4 }{b_0+b_1 x^2 }$$ where
$$a_0=3 (pi -2) k left(pi (k ((k-1) k+2)-1)-2 left(k^3+kright)right)$$
$$a_1=6 pi -left(k^2+1right) left(3 (pi -2) pi k^2+((8-5 pi ) pi -8) k+3 pi ^2right)$$
$$a_2=left(k^2+1right) left(2 (pi -2) (1+pi ) k^2-2 pi (1+pi ) k-(pi -10) pi -4right)$$
$$b_0=6 (k (k (pi -(pi -2) k)-2 pi +2)+pi )$$
$$b_1=2 left(k left(k left(3 (pi -2) k^2-3 pi k+7 pi -8right)-6 pi right)+4 pi -2right)$$ leading to
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{a_2 }{b_1}x-frac{2 left(b_1 (a_0 b_1-a_1 b_0)+a_2
b_0^2right)}{sqrt{b_0} b_1^{3/2} (2 b_0+(pi -2) b_1)}tan ^{-1}left(frac{sqrt{b_1} }{sqrt{b_0}}xright) -frac{(4 a_0+(pi -2) (2 a_1+(pi -2)
a_2))}{sqrt{2 (pi -2)} (2 b_0+(pi -2) b_1)}tanh
^{-1}left(sqrt{frac{2}{pi -2}} xright) $$ Making the same table as before
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57610 & 0.57610 \
0.40 & 0.67172 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01776 & 1.01775 \
0.60 & 1.17197 & 1.17195 \
0.65 & 1.37142 & 1.37136 \
0.70 & 1.67740 & 1.67722 \
0.75 & 2.68589 & 2.68502
end{array}
right)$$ which looks quite better.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks , I keep this question open for more answers .
    $endgroup$
    – S.H.W
    Dec 22 '18 at 18:27










  • $begingroup$
    @S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 23 '18 at 2:34










  • $begingroup$
    @Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
    $endgroup$
    – Awe Kumar Jha
    Dec 24 '18 at 12:38











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

If the problem was
$$int_0^a frac{tan{(1+x^2)}}{1+x^2},dx qquad text{with} qquad 0 leq a lt sqrt{frac{pi -2}{2}}approx 0.7555$$ it could be possible to have an approximation of it using a Padé approximant built at $x=0$.



Using $k=tan(1)$, we should have
$$ frac{tan{(1+x^2)}}{1+x^2}=frac {3 k left(-2 k^2+2 k+1right)+3 left(-k^2+k+1right)x^2 +(2 k-3) left(k^2+1right)x^4} {3(-2 k^2+2 k+1)+ 6 (k-1)^2 k x^2+(6 k^3-7 k^2-4)x^4}$$ which can be integrated using partial fraction decomposition (leading to a nasty expression. Numerically, this would be "almost" (making the coefficient rational)
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{184 }{789}x+frac{1099} {3526}tan ^{-1}left(frac{529
}{801}xright)+frac{1409 }{1667}tanh ^{-1}left(frac{692
}{523}xright)$$ For a few values of $a$, some results (the so called "exact" being obtained by numerical integration)
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57609 & 0.57610 \
0.40 & 0.67171 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01775 & 1.01775 \
0.60 & 1.17192 & 1.17195 \
0.65 & 1.37127 & 1.37136 \
0.70 & 1.67673 & 1.67722 \
0.75 & 2.66909 & 2.68502
end{array}
right)$$



Edit



We can make the approximation better using the fact that
$$f(x)=left(x^2-frac{pi }{2}+1right)frac{ tan left(x^2+1right)}{x^2+1}$$ is a quite nice function. Using a $[4,2]$ Padé approximant, we get
$$f(x) simeqfrac {a_0+a_1 x^2+a_2 x^4 }{b_0+b_1 x^2 }$$ where
$$a_0=3 (pi -2) k left(pi (k ((k-1) k+2)-1)-2 left(k^3+kright)right)$$
$$a_1=6 pi -left(k^2+1right) left(3 (pi -2) pi k^2+((8-5 pi ) pi -8) k+3 pi ^2right)$$
$$a_2=left(k^2+1right) left(2 (pi -2) (1+pi ) k^2-2 pi (1+pi ) k-(pi -10) pi -4right)$$
$$b_0=6 (k (k (pi -(pi -2) k)-2 pi +2)+pi )$$
$$b_1=2 left(k left(k left(3 (pi -2) k^2-3 pi k+7 pi -8right)-6 pi right)+4 pi -2right)$$ leading to
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{a_2 }{b_1}x-frac{2 left(b_1 (a_0 b_1-a_1 b_0)+a_2
b_0^2right)}{sqrt{b_0} b_1^{3/2} (2 b_0+(pi -2) b_1)}tan ^{-1}left(frac{sqrt{b_1} }{sqrt{b_0}}xright) -frac{(4 a_0+(pi -2) (2 a_1+(pi -2)
a_2))}{sqrt{2 (pi -2)} (2 b_0+(pi -2) b_1)}tanh
^{-1}left(sqrt{frac{2}{pi -2}} xright) $$ Making the same table as before
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57610 & 0.57610 \
0.40 & 0.67172 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01776 & 1.01775 \
0.60 & 1.17197 & 1.17195 \
0.65 & 1.37142 & 1.37136 \
0.70 & 1.67740 & 1.67722 \
0.75 & 2.68589 & 2.68502
end{array}
right)$$ which looks quite better.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks , I keep this question open for more answers .
    $endgroup$
    – S.H.W
    Dec 22 '18 at 18:27










  • $begingroup$
    @S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 23 '18 at 2:34










  • $begingroup$
    @Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
    $endgroup$
    – Awe Kumar Jha
    Dec 24 '18 at 12:38
















3












$begingroup$

If the problem was
$$int_0^a frac{tan{(1+x^2)}}{1+x^2},dx qquad text{with} qquad 0 leq a lt sqrt{frac{pi -2}{2}}approx 0.7555$$ it could be possible to have an approximation of it using a Padé approximant built at $x=0$.



Using $k=tan(1)$, we should have
$$ frac{tan{(1+x^2)}}{1+x^2}=frac {3 k left(-2 k^2+2 k+1right)+3 left(-k^2+k+1right)x^2 +(2 k-3) left(k^2+1right)x^4} {3(-2 k^2+2 k+1)+ 6 (k-1)^2 k x^2+(6 k^3-7 k^2-4)x^4}$$ which can be integrated using partial fraction decomposition (leading to a nasty expression. Numerically, this would be "almost" (making the coefficient rational)
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{184 }{789}x+frac{1099} {3526}tan ^{-1}left(frac{529
}{801}xright)+frac{1409 }{1667}tanh ^{-1}left(frac{692
}{523}xright)$$ For a few values of $a$, some results (the so called "exact" being obtained by numerical integration)
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57609 & 0.57610 \
0.40 & 0.67171 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01775 & 1.01775 \
0.60 & 1.17192 & 1.17195 \
0.65 & 1.37127 & 1.37136 \
0.70 & 1.67673 & 1.67722 \
0.75 & 2.66909 & 2.68502
end{array}
right)$$



Edit



We can make the approximation better using the fact that
$$f(x)=left(x^2-frac{pi }{2}+1right)frac{ tan left(x^2+1right)}{x^2+1}$$ is a quite nice function. Using a $[4,2]$ Padé approximant, we get
$$f(x) simeqfrac {a_0+a_1 x^2+a_2 x^4 }{b_0+b_1 x^2 }$$ where
$$a_0=3 (pi -2) k left(pi (k ((k-1) k+2)-1)-2 left(k^3+kright)right)$$
$$a_1=6 pi -left(k^2+1right) left(3 (pi -2) pi k^2+((8-5 pi ) pi -8) k+3 pi ^2right)$$
$$a_2=left(k^2+1right) left(2 (pi -2) (1+pi ) k^2-2 pi (1+pi ) k-(pi -10) pi -4right)$$
$$b_0=6 (k (k (pi -(pi -2) k)-2 pi +2)+pi )$$
$$b_1=2 left(k left(k left(3 (pi -2) k^2-3 pi k+7 pi -8right)-6 pi right)+4 pi -2right)$$ leading to
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{a_2 }{b_1}x-frac{2 left(b_1 (a_0 b_1-a_1 b_0)+a_2
b_0^2right)}{sqrt{b_0} b_1^{3/2} (2 b_0+(pi -2) b_1)}tan ^{-1}left(frac{sqrt{b_1} }{sqrt{b_0}}xright) -frac{(4 a_0+(pi -2) (2 a_1+(pi -2)
a_2))}{sqrt{2 (pi -2)} (2 b_0+(pi -2) b_1)}tanh
^{-1}left(sqrt{frac{2}{pi -2}} xright) $$ Making the same table as before
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57610 & 0.57610 \
0.40 & 0.67172 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01776 & 1.01775 \
0.60 & 1.17197 & 1.17195 \
0.65 & 1.37142 & 1.37136 \
0.70 & 1.67740 & 1.67722 \
0.75 & 2.68589 & 2.68502
end{array}
right)$$ which looks quite better.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks , I keep this question open for more answers .
    $endgroup$
    – S.H.W
    Dec 22 '18 at 18:27










  • $begingroup$
    @S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 23 '18 at 2:34










  • $begingroup$
    @Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
    $endgroup$
    – Awe Kumar Jha
    Dec 24 '18 at 12:38














3












3








3





$begingroup$

If the problem was
$$int_0^a frac{tan{(1+x^2)}}{1+x^2},dx qquad text{with} qquad 0 leq a lt sqrt{frac{pi -2}{2}}approx 0.7555$$ it could be possible to have an approximation of it using a Padé approximant built at $x=0$.



Using $k=tan(1)$, we should have
$$ frac{tan{(1+x^2)}}{1+x^2}=frac {3 k left(-2 k^2+2 k+1right)+3 left(-k^2+k+1right)x^2 +(2 k-3) left(k^2+1right)x^4} {3(-2 k^2+2 k+1)+ 6 (k-1)^2 k x^2+(6 k^3-7 k^2-4)x^4}$$ which can be integrated using partial fraction decomposition (leading to a nasty expression. Numerically, this would be "almost" (making the coefficient rational)
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{184 }{789}x+frac{1099} {3526}tan ^{-1}left(frac{529
}{801}xright)+frac{1409 }{1667}tanh ^{-1}left(frac{692
}{523}xright)$$ For a few values of $a$, some results (the so called "exact" being obtained by numerical integration)
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57609 & 0.57610 \
0.40 & 0.67171 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01775 & 1.01775 \
0.60 & 1.17192 & 1.17195 \
0.65 & 1.37127 & 1.37136 \
0.70 & 1.67673 & 1.67722 \
0.75 & 2.66909 & 2.68502
end{array}
right)$$



Edit



We can make the approximation better using the fact that
$$f(x)=left(x^2-frac{pi }{2}+1right)frac{ tan left(x^2+1right)}{x^2+1}$$ is a quite nice function. Using a $[4,2]$ Padé approximant, we get
$$f(x) simeqfrac {a_0+a_1 x^2+a_2 x^4 }{b_0+b_1 x^2 }$$ where
$$a_0=3 (pi -2) k left(pi (k ((k-1) k+2)-1)-2 left(k^3+kright)right)$$
$$a_1=6 pi -left(k^2+1right) left(3 (pi -2) pi k^2+((8-5 pi ) pi -8) k+3 pi ^2right)$$
$$a_2=left(k^2+1right) left(2 (pi -2) (1+pi ) k^2-2 pi (1+pi ) k-(pi -10) pi -4right)$$
$$b_0=6 (k (k (pi -(pi -2) k)-2 pi +2)+pi )$$
$$b_1=2 left(k left(k left(3 (pi -2) k^2-3 pi k+7 pi -8right)-6 pi right)+4 pi -2right)$$ leading to
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{a_2 }{b_1}x-frac{2 left(b_1 (a_0 b_1-a_1 b_0)+a_2
b_0^2right)}{sqrt{b_0} b_1^{3/2} (2 b_0+(pi -2) b_1)}tan ^{-1}left(frac{sqrt{b_1} }{sqrt{b_0}}xright) -frac{(4 a_0+(pi -2) (2 a_1+(pi -2)
a_2))}{sqrt{2 (pi -2)} (2 b_0+(pi -2) b_1)}tanh
^{-1}left(sqrt{frac{2}{pi -2}} xright) $$ Making the same table as before
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57610 & 0.57610 \
0.40 & 0.67172 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01776 & 1.01775 \
0.60 & 1.17197 & 1.17195 \
0.65 & 1.37142 & 1.37136 \
0.70 & 1.67740 & 1.67722 \
0.75 & 2.68589 & 2.68502
end{array}
right)$$ which looks quite better.






share|cite|improve this answer











$endgroup$



If the problem was
$$int_0^a frac{tan{(1+x^2)}}{1+x^2},dx qquad text{with} qquad 0 leq a lt sqrt{frac{pi -2}{2}}approx 0.7555$$ it could be possible to have an approximation of it using a Padé approximant built at $x=0$.



Using $k=tan(1)$, we should have
$$ frac{tan{(1+x^2)}}{1+x^2}=frac {3 k left(-2 k^2+2 k+1right)+3 left(-k^2+k+1right)x^2 +(2 k-3) left(k^2+1right)x^4} {3(-2 k^2+2 k+1)+ 6 (k-1)^2 k x^2+(6 k^3-7 k^2-4)x^4}$$ which can be integrated using partial fraction decomposition (leading to a nasty expression. Numerically, this would be "almost" (making the coefficient rational)
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{184 }{789}x+frac{1099} {3526}tan ^{-1}left(frac{529
}{801}xright)+frac{1409 }{1667}tanh ^{-1}left(frac{692
}{523}xright)$$ For a few values of $a$, some results (the so called "exact" being obtained by numerical integration)
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57609 & 0.57610 \
0.40 & 0.67171 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01775 & 1.01775 \
0.60 & 1.17192 & 1.17195 \
0.65 & 1.37127 & 1.37136 \
0.70 & 1.67673 & 1.67722 \
0.75 & 2.66909 & 2.68502
end{array}
right)$$



Edit



We can make the approximation better using the fact that
$$f(x)=left(x^2-frac{pi }{2}+1right)frac{ tan left(x^2+1right)}{x^2+1}$$ is a quite nice function. Using a $[4,2]$ Padé approximant, we get
$$f(x) simeqfrac {a_0+a_1 x^2+a_2 x^4 }{b_0+b_1 x^2 }$$ where
$$a_0=3 (pi -2) k left(pi (k ((k-1) k+2)-1)-2 left(k^3+kright)right)$$
$$a_1=6 pi -left(k^2+1right) left(3 (pi -2) pi k^2+((8-5 pi ) pi -8) k+3 pi ^2right)$$
$$a_2=left(k^2+1right) left(2 (pi -2) (1+pi ) k^2-2 pi (1+pi ) k-(pi -10) pi -4right)$$
$$b_0=6 (k (k (pi -(pi -2) k)-2 pi +2)+pi )$$
$$b_1=2 left(k left(k left(3 (pi -2) k^2-3 pi k+7 pi -8right)-6 pi right)+4 pi -2right)$$ leading to
$$int frac{tan{(1+x^2)}}{1+x^2},dx simeq frac{a_2 }{b_1}x-frac{2 left(b_1 (a_0 b_1-a_1 b_0)+a_2
b_0^2right)}{sqrt{b_0} b_1^{3/2} (2 b_0+(pi -2) b_1)}tan ^{-1}left(frac{sqrt{b_1} }{sqrt{b_0}}xright) -frac{(4 a_0+(pi -2) (2 a_1+(pi -2)
a_2))}{sqrt{2 (pi -2)} (2 b_0+(pi -2) b_1)}tanh
^{-1}left(sqrt{frac{2}{pi -2}} xright) $$ Making the same table as before
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0.05 & 0.07795 & 0.07795 \
0.10 & 0.15637 & 0.15637 \
0.15 & 0.23577 & 0.23577 \
0.20 & 0.31670 & 0.31670 \
0.25 & 0.39982 & 0.39982 \
0.30 & 0.48593 & 0.48593 \
0.35 & 0.57610 & 0.57610 \
0.40 & 0.67172 & 0.67172 \
0.45 & 0.77482 & 0.77482 \
0.50 & 0.88848 & 0.88848 \
0.55 & 1.01776 & 1.01775 \
0.60 & 1.17197 & 1.17195 \
0.65 & 1.37142 & 1.37136 \
0.70 & 1.67740 & 1.67722 \
0.75 & 2.68589 & 2.68502
end{array}
right)$$ which looks quite better.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 14:39

























answered Dec 22 '18 at 9:49









Claude LeiboviciClaude Leibovici

122k1157134




122k1157134












  • $begingroup$
    Thanks , I keep this question open for more answers .
    $endgroup$
    – S.H.W
    Dec 22 '18 at 18:27










  • $begingroup$
    @S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 23 '18 at 2:34










  • $begingroup$
    @Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
    $endgroup$
    – Awe Kumar Jha
    Dec 24 '18 at 12:38


















  • $begingroup$
    Thanks , I keep this question open for more answers .
    $endgroup$
    – S.H.W
    Dec 22 '18 at 18:27










  • $begingroup$
    @S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 23 '18 at 2:34










  • $begingroup$
    @Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
    $endgroup$
    – Awe Kumar Jha
    Dec 24 '18 at 12:38
















$begingroup$
Thanks , I keep this question open for more answers .
$endgroup$
– S.H.W
Dec 22 '18 at 18:27




$begingroup$
Thanks , I keep this question open for more answers .
$endgroup$
– S.H.W
Dec 22 '18 at 18:27












$begingroup$
@S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
$endgroup$
– Claude Leibovici
Dec 23 '18 at 2:34




$begingroup$
@S.H.W : For sure, keep it open. It is very interesting. I am still working on it. Cheers.
$endgroup$
– Claude Leibovici
Dec 23 '18 at 2:34












$begingroup$
@Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
$endgroup$
– Awe Kumar Jha
Dec 24 '18 at 12:38




$begingroup$
@Claude Leibovici, I think my Maxima made an error , soon I shall post a better answer.
$endgroup$
– Awe Kumar Jha
Dec 24 '18 at 12:38


















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