<s>[INST] Evaluate $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$ I am trying to evaluate this integral:
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$$
I coverted the region by using polar coordinate system for xy-plane and it became:
$$\displaystyle\int_{0}^{1}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz+\displaystyle\int_{0}^{1}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz$$
After IBP
$$\int r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}dr=\frac{1}{3}(r^2+z^2)^{\frac{3}{2}}\arctan{\sqrt{r^2+z^2}}-\frac{1}{6}(r^2+z^2)+\frac{1}{6}\ln(r^2+z^2+1)+C $$ and replace upper and lower limits, it is scary-looking.
My experience in solving this tyle of integral is still weak, I think there is a way to get rid of one variable from the beginning integral, but i can't figure it out. Hope everyone can help me, thanks.
 [/INST] Very similar steps from (Almost) Impossible Integrals, Sums, and Series, pages 261-261, can be applied where there we have that
$$\int_0^1\left( \int_0^1 \sqrt{x^2+y^2}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\textrm{d}x\right)\textrm{d}y$$ $$=\int_0^1 \left(\int_0^1 \left(\int_0^1 \frac{x^2+y^2}{x^2+y^2+z^2}\textrm{d}z\right)\textrm{d}x\right)\textrm{d}y.$$
Similarly, we have that
$$\int_0^1  \left( \int_0^1 \left(\int_0^1 \sqrt{x^2+y^2+z^2}\arctan\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)\textrm{d}x\right)\textrm{d}y\right)\textrm{dz}$$
$$=\int_0^1 \left(\int_0^1 \left(\int_0^1\left(\int_0^1 \frac{x^2+y^2+z^2}{x^2+y^2+z^2+w^2}\textrm{d}w\right)\textrm{d}z\right)\textrm{d}x\right)\textrm{d}y,$$
where since the last quadruple integral is immediately done by symmetry, the rest of the work to do is straightforward in various ways.
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