Evaluating $\int_0^{\infty}\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3 dx$ I’m looking for a closed form of this integral $$I_3=\int_0^{\infty}\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3 dx$$ I’ve managed to evaluate $$I_1=\int_0^{\infty} \frac{\ln(1+x^2)}{1+x^2}dx=\pi \ln 2$$ with the change of variable $x =\tan \theta $, $dx=(1+(\tan \theta)^2) d\theta $ giving $$I_1=-2\int_0^{\frac{\pi}{2}} \ln(\cos \theta ) d\theta$$ which is classic. But I’m stuck with the power $(\ln (\cos \theta))^3$ in $I_3$. Thanks for your help.
 A: We have

$$\int_0^{+\infty}\!\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3\!{\rm d}x=\frac{9\pi}{4}  \zeta(3)+\frac{3 \pi}{2} \ln^3 2-\frac{21\pi}{8}\ln^2 2-\frac{33\pi}{16} \ln 2-\frac{7 \pi ^3}{32}-\frac{57 \pi }{64}.$$

Proof. Observe that
$$
\left(\frac{\ln(1+x^2)}{1+x^2}\right)^3=-\partial_s^3 \left.\left(\frac{1}{(1+x^2)^s}\right)\right|_{s=3} \tag1
$$
Then you may write
$$\begin{align}
\int_0^{+\infty}\frac{1}{(1+x^2)^s}{\rm d}x
&=\frac12\int_0^1u^s\left(\frac1u-1\right)^{-1/2}\frac{{\rm d}u}{u^2},\quad \color{purple}{u=\frac{1}{1+x^2},\, x=\left(\frac1u-1\right)^{1/2}}\\\\
&=\frac12\int_0^1u^{s-3/2}\left(1-u\right)^{-1/2}{\rm d}u\\\\
&=\frac12\frac{\Gamma\left(s-\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(s)}\\\\
&=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(s-\frac12\right)}{\Gamma(s)} \qquad \qquad (2)
\end{align}
$$ where, in the last step, we have used the classic Euler beta function integral representation. 
Now, performing $\displaystyle -\partial_s^3 $ to $(2)$ and using $(1)$, you easily get a closed form expression for your integral with the help of Wolfram|Alpha for the computation of
$$
-\partial_s^3 \left.\left(\frac{\Gamma\left(s-\frac12\right)}{\Gamma(s)}\right)\right|_{s=3}
$$ 
A: Try this 
$$
I_3 =\lim_{\beta\rightarrow 0}\frac{\partial^3}{\partial \beta^3}\int \left(1+x^2\right)^{\beta-3}dx
$$
