I would like to perform the integration ($u \in [0,1]$), \begin{equation} \int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u^2 \sqrt{-a^2 u^2-b^2+1} \tan ^{-1}\left(\frac{\left| a\right| }{\sqrt{-a^2-b^2+1}}\right) db da . \end{equation}

I also have a companion problem \begin{equation} \int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u \sqrt{-a^2-b^2+1} \tan ^{-1}\left(u \left| a\right| \sqrt{-\frac{1}{a^2 u^2+b^2-1}}\right) db da . \end{equation}

Solutions to these two problems would provide a resolution to the question "Compute a certain "separability probability" via a constrained 4D integration over $[-1,1]^4$", posed in https://mathematica.stackexchange.com/questions/193337/compute-a-certain-separability-probability-via-a-constrained-4d-integration-ov

Now, for the component functions of the first integrand, we have \begin{equation} \int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u^2 \sqrt{-a^2 u^2-b^2+1} db da = \frac{4}{3} u \left(\sqrt{1-u^2} u+\sin ^{-1}(u)\right), \end{equation} and \begin{equation} \int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} \tan ^{-1}\left(\frac{\left| a\right| }{\sqrt{-a^2-b^2+1}}\right) db da=\frac{1}{2} (\pi -2) \pi. \end{equation} Can some form of integration-by-parts--or other methodology--be performed? (I realize that more may certainly be needed in addition to the two above results.)

The much-viewed question https://math.stackexchange.com/questions/1167346/integration-by-parts-for-a-double-integral seems relevant here.