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](https://en.wikipedia.org/wiki/Clausen%27s_formula "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](https://en.wikipedia.org/wiki/Elliptic_integral "Elliptic integral") of the first and second kind. However, I also do not know how to integrate these terms with squared complete elliptic integrals.



Indefinite integral of squared hypergeometric function
Using Maple to play with the equation ([details](http://dl.vwx.fr/8UAD7bnP6v0gtuD1/foo.pdf)) suggests that
$$
\int\_0^u {}\_2F\_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv
= -\frac{32}{\pi} + \frac83 u \cdot {}\_3F\_2\!\left(\frac12, \frac12, \frac52, 1, 2, 1-u^2\right).
$$
You can *almost* prove this equality by observing that both sides satisfy
$$
\left( u-1 \right) ^{2} \left( u+1 \right) ^{2}{\frac {{\rm d}^{4}}{
{\rm d}{u}^{4}}}y \left( u \right) +12\,u \left( u-1 \right) \left( u
+1 \right) {\frac {{\rm d}^{3}}{{\rm d}{u}^{3}}}y \left( u \right) +
 \left( 33\,{u}^{2}-9 \right) {\frac {{\rm d}^{2}}{{\rm d}{u}^{2}}}y
 \left( u \right) +15 u
\, {\frac {\rm d}{{\rm d}u}}y \left( u \right)= 0
$$
and their series expansions at $u=1$ agree. More precisely, $u=1$ is a regular singular point of the differential equation, with indicial polynomial $n(n+1)(n-1)$, so that solutions of this equation are characterized by the coefficients of $(u-1)^{-1}$, $1$, $\log(u-1)$ and $(u-1)$ in their generalized series expansions at $1$. The coefficients of $(u-1)^{-1}$, $\log(u-1)$, and $(u-1)$ are (respectively) $0$, $0$, and $1$ for both sides. Therefore, proving the above equality for all $u$ reduces to proving that it holds for $u=1$, i.e. that
$$\int\_0^1 {}\_2F\_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv = \frac83 - \frac{32}{9 \pi}.$$


