Conjecture about integral $\int_0^1 K\left(\sqrt{\vphantom1x}\right)\,K\left(\sqrt{1-x}\right)\,x^ndx$

I'm interested in the following integral: $$\mathcal J(n)=\int_0^1 K\left(\sqrt{\vphantom1x}\right)\,K\left(\sqrt{1-x}\right)\,x^ndx,\tag1$$ where $K(z)$ is the complete elliptic integral of the 1ˢᵗ kind: $$K(z)={_2F_1}\left(\begin{array}c\tfrac12,\tfrac12\\1\end{array}\middle|\ z^2\right)\cdot\frac\pi2.\tag2$$ Using numerical integration and inverse symbolic calculation techniques, I found a conjectural formula for this integral, that only though seems to work for $n\in\mathbb N_0$: $$\mathcal J(n)\stackrel?=\frac{\pi^3}{8\cdot4^n}\left(\frac{(2n-1)!!}{n!}\right)^2{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,-n,-n\\1,\tfrac12-n,\tfrac12-n\end{array}\middle|\ 1\right).\tag3$$ This formula produces the following sequence of results for $n=0,1,2,...$ $$\frac{\pi^3}8,\frac{\pi^3}{16},\frac{11\,\pi^3}{256},\frac{17\,\pi^3}{512},\frac{1787 \,\pi^3}{65536},\frac{3047\,\pi^3}{131072},\frac{42631\,\pi^3}{2097152},\frac{75937\,\pi^3}{4194304},...\tag4$$

• Can we prove the conjecture $(3)$ is correct?
• Is it possible to find a generalization that works for non-integer values of $n$ as well?
• Zhou deals with similar integrals in arxiv.org/pdf/1301.1735. An effective technique is to exploit $$K(k) = 2\sum_{n\geq 0}\frac{P_n(2k-1)}{2n+1}$$ like I did here. Shifted Legendre polynomials give an orthogonal base of $L^2(0,1)$, so $\mathcal{J}(n)$ is easily related with the series $$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}.$$ – Jack D'Aurizio Aug 4 '15 at 17:39

\begin{align*}&\; \int^1_0x^nK\left(\sqrt{\vphantom1x}\right)K\left(\sqrt{1-x}\right)dx\\ &=\int^1_0\left(\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}x^{m+n}\right)K\left(\sqrt{1-x}\right)dx\\ &=\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}\int^1_0x^{m+n}K\left(\sqrt{1-x}\right)dx\\ &=\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}\frac{2^{4m+4n+1}((m+n)!)^4}{(2m+2n+1)!^2}\\ &=2^{4n}\pi\sum^{\infty}_{m=0}\frac{(2m)!^2}{(m!)^4}\frac{((m+n)!)^4}{(2m+2n+1)!^2}\\ &=\frac{2^{4n}\pi(n!)^4}{(2n+1)!^2}{}_4F_3\left(\frac12,\frac12,n+1,n+1;1,n+\frac32,n+\frac32\middle|\,1\right). \end{align*} This form checks numerically for non-integer $n$, and I think it should be related to OP's conjectured form via some $_4F_3$ transformation.
Zhou deals with similar integrals in arxiv.org/pdf/1301.1735. An effective technique is to exploit: $$K(k) = 2\sum_{n\geq 0}\frac{P_n(2k-1)}{2n+1}$$ like I did here.
The shifted Legendre polynomials give an orthogonal base of $L^2(0,1)$, so $\mathcal{J}(n)$, for any $n\in\mathbb{N}$, is easily related with the series: $$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$ that is a particular case of the series computed here.