TITLE: Closed form for $\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$
QUESTION [26 upvotes]: Please help me to find a closed form for this integral:
$$\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$$

REPLY [24 votes]: The integral can be trasformed into a "computable" form. Indeed,
\begin{align}
I&=\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx=\\
&=2\int_0^{\infty}\frac{\arctan y^2\ln(1+y^4)}{1+y^4}y^2dy=\\
&=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+y^4}y^2dy=\\
&=\frac{1}{4}\int_{-\infty}^{\infty}\left(\frac{1}{1+iy^2}-\frac{1}{1-iy^2}\right)\Bigl(\ln^2(1+iy^2)-\ln^2(1-iy^2)\Bigr)dy=\\
&=\frac12\mathrm{Re}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+iy^2}dy,
\end{align}
where the logarithms are defined on their main sheets. The remaining integrals can be evaluated by suitably deforming the contours (or using Mathematica). The final result is
$$I=\frac{\pi}{6\sqrt{2}}\left(12G+9\ln^22+3\pi\ln2+\pi^2\right)\approx 11.7433,$$
where $G$ denotes the Catalan's constant.