Conjectured closed form for $\int_0^1x^{2\,q-1}\,K(x)^2dx$ where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind I am interested in a general closed-form formula for integrals of the following form:
$$\mathcal{J}_q=\int_0^1x^{2\,q-1}\,K(x)^2dx,\tag0$$
where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind:
$$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\frac\pi2\\=\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-x^2\sin^2\phi}}=\int_0^1\frac{dz}{\sqrt{1-x^2z^2\vphantom{|^2}}\sqrt{1-z^2\vphantom{|^2}}}.\tag1$$
I could not find a suitable integral in DLMF or Gradshteyn—Ryzhik or Prudnikov tables of integrals. Mathematica was not able to evaluate it either, even when the parameter $q$ was fixed to an integer (Note: In Mathematica you would write EllipticK[x^2] for $K(x)$ because of a different convention for the modulus parameter).
But, numerical integration and lookups in WolframAlpha and ISC+ suggest that there are closed forms for (at least some) postitive integer values of the parameter $q$. For example,
$$\mathcal{J}_1\stackrel?=\frac{7\,\zeta(3)}4,\ \mathcal{J}_2\stackrel?=\frac{7\,\zeta(3)}8+\frac14,\\\mathcal{J}_3\stackrel?=\frac{77\,\zeta(3)}{128}+\frac{17}{64},\ \mathcal{J}_4\stackrel?=\frac{119\,\zeta(3)}{256}+\frac{881}{3456},\tag2$$
where $\zeta(3)$ is the Apéry constant.
It looks that the general formula for postitive integer values of $q$ is
$$\mathcal{J}_q\stackrel?=a_q\,\zeta(3)+b_q,\tag3$$
where $a_q,\,b_q$ are some rational coefficients. I tried to find general formulae for these coefficients, and actually discovered plausible candidates for both. 
Conjecturally, $a_q$ can be expressed in terms of a generalized hypergeometric function:
$$a_q\stackrel?=\frac{7\,\pi}4\cdot\frac{{_4F_3}\left(\begin{array}{c}\frac12,\,\frac12,\,1-q,\,1-q\\1,\,\frac32-q,\,\frac32-q\end{array}\middle|\ 1\right)}{\Gamma\left(\frac32-q\right)^2\ \Gamma(q)^2}.\tag4$$
Somewhat suprisingly, this scarish expression seems to evaluate only to rationals for all $q\in\mathbb{Z}^+$. 
For $b_q$ I found only a conjectural recurrence relation:
$$b_1\stackrel?=0,\ b_2\stackrel?=\frac14,\ b_q\stackrel?=\frac{2 \,(2\,q-3)\,\left(2\,q^2-6\,q+5\right)\cdot b_{q-1}-4\,(q-2)^3\cdot b_{q-2}+1}{4\,(q-1)^3},\tag5$$
but I could not find a general term formula for it.

  
*
  
*Can we prove that the closed forms shown in $(2)$ are correct?
  
*Can we prove that the formula $(3)$ holds for all $q\in\mathbb{Z}^+$ with some rational coefficients $a_q,\,b_q$?
  
*Are my conjectured formulae for these coefficients $(4),\,(5)$ correct?
  
*Can the formula $(4)$ be expanded in terms of simpler functions?
  
*Is there a general term formula for $(5)$?
  
*Can we find (or at least conjecture) a general closed-form formula for $\mathcal{J}_q$ where $q$ is not necessarily an integer?
  

 A: As pointed out in a comment, these and many related results are derived in the excellent paper "Moments of Elliptic Integrals", by James Wan (http://arxiv.org/abs/1101.1132).  Specifically, the author shows that
$$
K_{1}=\int_{0}^{1}xK(x)^2dx =\frac{7}{4}\zeta(3),
$$
that
$$
K_{3}=\int_{0}^{1}x^3 K^2(x)dx= \frac{1}{4}+\frac{7}{8}\zeta(3),
$$
and finally that the recurrence
$$
(n+1)^3K_{n+2} - 2n(n^2+1)K_{n}+(n-1)^3 K_{n-2}=2
$$
holds in general.  In terms of your integrals, where ${\cal J}_q=K_{2q-1}$, we find that
$$
{\cal J}_{q}=K_{2q-1}=\frac{2+2(2q-3)((2q-3)^2+1)K_{2q-3}-(2q-4)^3 K_{2q-5}}{(2q-2)^3}
\\
=\frac{1}{4(q-1)^3}+\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}{\cal J}_{q-1}-\frac{(q-2)^3}{(q-1)^3}{\cal J}_{q-2}.
$$
Then indeed ${\cal J}_q=a_q\zeta(3)+b_q$, where $a_1=7/4$, $a_2=7/8$, $b_1=0$, $b_2=1/4$; and $a_q$ and $b_q$ satisfy these recurrence relations:
$$
a_{q}=\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}a_{q-1}-\frac{(q-2)^3}{(q-1)^3}a_{q-2}
$$
and
$$
b_{q}=\frac{1}{4(q-1)^3}+\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}b_{q-1}-\frac{(q-2)^3}{(q-1)^3}b_{q-2}
$$
(where the latter matches your conjecture).
