Let $\gamma>1$ be fixed. For $1\leq x\leq y\leq \gamma$, I would like to evaluate the following integral : $$F(x,y)=\frac{1}{\pi}\int_1^x\frac{dt}{t\sqrt{t-1}}\arcsin \sqrt{\frac{y-t}{\gamma-t}}.$$ What I know about this are the boundary values :
1) trivially, $F(1,y)=0$;
2) For $y=\gamma$, we have $$F(x,\gamma)=\frac{1}{\pi}\int_1^x\frac{dt}{t\sqrt{t-1}}=\arcsin(1)-\arcsin\sqrt{\frac{1}{x}};$$
3) On the diagonal $y=x$, we have $$F(x,x)=\arcsin\sqrt{\frac{x}{\gamma}}-\arcsin\sqrt{\frac{1}{\gamma}},$$ as explained in the following post : evaluation of some integral
My question is: For $1<x<y<\gamma$, can $F(x,y)$ be expressed in terms of elementary functions, or is there some simple reason why it cannot?
If it helps, I also know the partial derivatives : $$\partial_x F(x,y)=\frac{1}{\pi x\sqrt{x-1}}\arcsin\sqrt{\frac{y-x}{\gamma-x}},$$ and $$\partial_y F(x,y)=\frac{1}{\pi \sqrt{y}\sqrt{\gamma-y}}\arcsin\sqrt{\frac{y(x-1)}{x(y-1)}}.$$ The first one is trivialy obtained (so it won't help), but the second one requires the evaluation of the integral $$\int_1^x\frac{dt}{t\sqrt{t-1}\sqrt{y-t}}.$$ This can be done by using Euler substitution method.
