I am trying to evaluate this integral:$$I=\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\sqrt{x^2+y^2}\arctan{\sqrt{x^2+y^2}}dxdy$$
But I am getting stuck to exploit the symmetry. First try, convert this integral to polar coordinate system by: $x=r\cos{\phi}$ and $y=r\sin{\phi}$ then $I$ become:
$$I=2\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos{\phi}}}r^6(\cos^4{\phi}+\sin^2{\phi})\arctan{r}drd\phi=2\int_{0}^{\frac{\pi}{4}}(\cos^4{\phi}+\sin^2{\phi})d\phi\int_{0}^{\frac{1}{\cos{\phi}}}r^6\arctan{r}dr $$ and IBP for integral with variable $r$, and it became a mess.
Second try, rewrite: $$I=\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\sqrt{x^2+y^2}\arctan{\sqrt{x^2+y^2}}dxdy=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\frac{x^2+y^2}{1+z^2(x^2+y^2)}dzdxdy$$
But it seems more difficult, my experience told me that has a way to exploit the symmetry, but i can't figure it out yet.
I will be thankful if everyone can give me some hints. Thank you for reading and good day.