I am trying to compute the indefinite integral $$ \int_0^u {}_2F_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv $$ for $0<u<1$. Using Clausen's formula for the square of the hypergeometric function, this can be written as $$ \int_0^u {}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;1-v^2\right)dv = \frac{1}{2}\int_{1-u^2}^{1} \frac{1}{\sqrt{1-x}}{}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;x\right)dx, $$ where I have performed a change of variables $x=1-v^2$. From Brudnikov & Brychkov, Volume 3, I know the following indefinite integral for generalized hypergeometric functions ${}_pF_q$: $$ \int x^{\alpha-1}{}_pF_q\left((a_p);(b_q);x\right)=\frac{x^\alpha}{\alpha}\,{}_{p+1}F_{q+1}\left((a_p),\alpha;(b_q),\alpha+1;x\right). $$ One ansatz is to expand the factor of $1/\sqrt{1-x}$ in a Taylor series, to integrate each summand and then to sum the series. Doing this, one obtains the expression $$ \left.\frac{1}{2}\sum_{n=0}^\infty\frac{\left(\frac{1}{2}\right)_{n}}{(n+1)!}x^{n+1}{}_4F_3\left(\frac{1}{2},\frac{3}{2},\frac{5}{2},n+1;2,3,n+2;x\right)\right|_{x=1-u^2}^{x=1}, $$ where $(\alpha)_n=\alpha(\alpha+1)\ldots(\alpha+n-1)$ is the Pochhammer symbol. Unfortunately, I do not know how to sum this.

Another ansatz is to use the identity $$ {}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;1-v^2\right)=\frac{64 \left((1+v) K\left(\frac{1-v}{2}\right)-2vE\left(\frac{1-v}{2}\right)\right)^2}{9 \pi^2 (1-v^2)^2}, $$ where $K$ and $E$ are the complete elliptic integrals of the first and second kind. However, I also do not know how to integrate these terms with squared complete elliptic integrals.