Evaluate the double integral

The substitutions $1-x=u^2$ , $y=v^2$ give $I \; = \; \int_0^1 \int_0^1 \frac{\sin^{-1}(\sqrt{1-x}\sqrt{y})}{\sqrt{1-y}\sqrt{1-y+xy}}\,dx\,dy \; = \; 4\int_0^1 \int_0^1 \frac{uv\sin^{-1}(uv)} {\sqrt{1-v^2}\sqrt{1-u^2v^2}}\,du\,dv$ and so $\begin{array}{c}\ I & = \displaystyle 4\int_0^1 \frac{dv}{\sqrt{1-v^2}} \int_0^1 \frac{uv \sin^{-1}(uv)}{\sqrt{1-u^2v^2}}\,du \; = \; 4\int_0^1 \frac{dv}{v\sqrt{1-v^2}}\int_0^v \frac{u\sin^{-1}u}{\sqrt{1-u^2}}\,du \\ & = \displaystyle 4\int_0^1 \frac{dv}{v\sqrt{1-v^2}}\left\{ \Big[-\sin^{-1}u \sqrt{1-u^2}\Big]_{u=0}^v + \int_0^v\,du\right\} \; = \; 4\int_0^1 \frac{dv}{v\sqrt{1-v^2}}\left\{ v - \sqrt{1-v^2}\sin^{-1}v\right\} \\ & = \displaystyle 4\int_0^1 \frac{dv}{\sqrt{1-v^2}} - 4\int_0^1 \frac{\sin^{-1}v}{v}\,dv \; =\; 4\Big[\sin^{-1}v\Big]_0^1 - 4\Big[\ln v \, \sin^{-1}v\Big]_0^1 + 4\int_0^1 \frac{\ln v}{\sqrt{1-v^2}}\,dv \\ & = \displaystyle 2\pi + \int_0^1 w^{-\frac12}(1-w)^{-\frac12} \ln w\,dw \; = \; 2\pi + \frac{\partial B}{\partial \alpha}(\alpha,\beta)\Big|_{\alpha=\beta=\frac12} \\ & = \displaystyle 2\pi + B(\alpha,\beta)\big[\psi(\alpha) - \psi(\alpha+\beta)\big]\Big|_{\alpha=\beta=\frac12} \\ &= \displaystyle 2\pi + \pi\big(\psi(\tfrac12) - \psi(1)\big) \; = \; 2\pi + \pi(-\gamma - \ln4 + \gamma) \\ & = \displaystyle 2\pi(1 - \ln2) \end{array}$ making the answer $2+1+2=\boxed{5}$ .