Closed form for $\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$

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.