Double Integral

The substitution $p = \sqrt{(1-x)y}$ gives $\begin{aligned} \int_0^1 \frac{\sin^{-1}\sqrt{(1-x)y}}{\sqrt{1-x}\sqrt{xy-y+1}\sqrt{y}}\,dy & = \frac{2}{1-x}\int_0^{\sqrt{1-x}}\frac{\sin^{-1}p}{\sqrt{1-p^2}}\,dp \\ & = \frac{1}{1-x}\Big[\big(\sin^{-1}p\big)^2\Big]_0^{\sqrt{1-x}} \; = \; \frac{\big(\sin^{-1}\sqrt{1-x}\big)^2}{1-x} \end{aligned}$ and so the integral $\begin{aligned} I \; = \; \int_0^1 \int_0^1 \frac{\sin^{-1}\sqrt{(1-x)y}}{\sqrt{1-x}\sqrt{xy-y+1}\sqrt{y}}\,dx\,dy & = \int_0^1 \frac{\big(\sin^{-1}\sqrt{1-x}\big)^2}{1-x}\,dx \; = \; \int_0^1 \frac{\big(\sin^{-1}\sqrt{x}\big)^2}{x}\,dx \\ & = 2\int_0^{\frac12\pi} q^2 \cot q\,dq \; = \; -4\int_0^{\frac12\pi} q\ln(\sin q)\,dq \end{aligned}$ using the substitution $q = \sin^{-1}\sqrt{x}$ and integration by parts. But then $I \; = \; -4\Big(\tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\ln2\Big) \; = \; \tfrac12\pi^2\ln2 - \tfrac74\zeta(3)$ using a result of Euler to evaluate this last integral. This makes the answer $7+4+1+2= \boxed{14}$ .