I'm trying to evaluate $$\int_0^1 z^p(1-z)^q \hspace{3pt}{}_2 F_1(a,a;2a,z) \hspace{3pt}{}_2 F_1(b,b;2b,z) dz$$ where $a,b\in \mathbb{R}$
Using the series definition of hypergeometric function ${}_2 F_1(a,a;2a,z)=\sum_n \frac{(a)_n(a)_n}{(2a)_n}\frac{z^n}{n!}$ in above integral I get $$\sum_{n,m}\frac{1}{n!m!}\frac{(a)_n(a)_n}{(2a)_n}\frac{(b)_m(b)_m}{(2b)_m}\int_0^1z^{p+n+m}(1-z)^{q}dz$$ $$=\sum_{n,m}\frac{1}{n!m!}\frac{(a)_n(a)_n}{(2a)_n}\frac{(b)_m(b)_m}{(2b)_m}B(p+n+m+1,q+1)$$ $$=\sum_{n,m}\frac{1}{n!m!}\frac{(a)_n(a)_n}{(2a)_n}\frac{(b)_m(b)_m}{(2b)_m}\frac{\Gamma(p+n+m+1)\Gamma(q+1)}{\Gamma(p+q+n+m+2)}$$ Since the final expression involve summation over two variable it seemed like it can be written in compact form as Appell series at $x=y=1$ something like $$=\frac{\Gamma(p+1)\Gamma(q+1)}{\Gamma(p+q+2)}\sum_{n,m}\frac{(a)_n(a)_n}{(2a)_n}\frac{(b)_m(b)_m}{(2b)_m}\frac{(p+1)_{n+m}}{(p+q+2)_{n+m}}\frac{1}{n!m!}$$ but the factor $\frac{(a)_n}{(2a)_n}\frac{(b)_m}{(2b)_m}$ spoils identifying above expression to $F_1(q+1,a,b;p+q+2;1,1)$. Another strategy is to write ${}_2F_1$ as Barnes integral and then close the contour appropriately but it also leads to above series expression. Can this integral be written in some closed form either by some other integration method or giving the last series some closed form?