Krdytkyu

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 .

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.