Indefinite integral of squared hypergeometric function

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.

• The squared hypergeometric function under the integral is in fact the Legendre function, may be it will help? – Sergei Aug 23 '16 at 20:03
• @Sergei Can you say how to write it as a Legendre function? Thank you – physicus Aug 25 '16 at 22:38

Using Maple to play with the equation (details) 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 $$\int_0^1 {}_2F_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv = \frac83 - \frac{32}{9 \pi}.$$