Elliptic definite integral $\int_0^1 x \left(\left(7+x^2\right) E\left(x\right) - 4 \left(1-x^2\right) K(x)\right)^2 \mathrm{d}x$ Can this integral
$$\int_0^1 x \left(\left(7+x^2\right) E\left(x\right) - 4 \left(1-x^2\right) K(x)\right)^2 \mathrm{d}x$$
be found in a closed form?
As usual
$K(x) = \int_0^{\pi /2} \mathrm{d}\theta/\sqrt{1-x^2 \sin^2\theta}$ and $E(x) = \int_0^{\pi /2} \sqrt{1-x^2 \sin^2\theta} \,\,\mathrm{d}\theta$.
EDIT: ANSWER FOUND BY THE AUTHOR
$\frac{945 \zeta (3)}{1024}+\frac{10149}{512}$
 A: Heureka!: I was unsure whether it is possible to compute this, after series of calculations, I found the result myself. This integral is related to an expeted product of two sides of a random triangle formed by three independently uniformly chosen points from a unit disk. And since
$$\mathbb{E}\left[a b\right] = \frac{6766 + 315 \zeta (3)}{864 \pi ^2},$$
it is concluded that
$$\int_0^1 x \left((7+x^2)E(x)-4(1-x^2)K(x)\right)^2 \,\mathrm{d}x = \frac{945 \zeta (3)}{1024}+\frac{10149}{512} \approx 20.931585716294742...$$

EDIT:
Briefly, I will comment why this is related to the expectation given above. We first calculate the expectation $\mathbb{E}[a]$. Pick a random point $X$ in a unit disk, then we pick another point $Y$ and average out the distance $a$ fixing the first point, this expectation we call $g$. This procedure is done for all points $X'$. Due to symmetry, this must be a function of $t = |OX|$ only, so $g = g(t)$. We shift the coordinate system such that $X=O$, the disk moves wlog. to the right by $t$ with polar equation $r = t \cos\phi + \sqrt{1-t^2 \sin^2\phi}$, for the distance we have simply $|XY|=|OY|=r$ polar radius, and thus
$$g(t) = \mathbb{E}\left[|X Y|, X\text{ fixed}\right] = \frac{1}{\pi}\int_0^\pi \int_0^{t \cos\phi + \sqrt{1-t^2 \sin^2\phi}} r^2 \,\mathrm{d}r\mathrm{d}\phi = \ldots = \frac{4}{9\pi}\left[(7+t^2)E(t)-4(1-t^2)K(t)\right]$$
For the expectation $\mathbb{E}[a]$ we have, due to circular symmetry,
$$\mathbb{E}[a]= \mathbb{E}[|XY|]=\mathbb{E}[\mathbb{E}[|XY|, X\text{ fixed}]] = \mathbb{E}[g(t)] = 2\int_0^1 g(t) t \, \mathrm{d}t = \ldots =\frac{128}{45\pi}.$$
Similarly, if we pick three points $X,Y,Z$, then
$$\mathbb{E}[|XY||XZ|] = \mathbb{E}[\mathbb{E}[|XY||XZ|,X\text{ fixed}]] = 2\int_0^1 \mathbb{E}[|XY|]\mathbb{E}[|XZ|] \, t\,\mathrm{d}t = 2\int_0^1 t g(t)^2 \,\mathrm{d}t,$$
which is my original integral upto a scalar multiple.
