Prove that $\int_{0}^{1}x^2K^\prime(x)^3\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{640\pi^2} -\frac{\pi^4}{40}$

Question

I need to prove the following result $$ \int_{0}^{1}x^2K^\prime(x)^3\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{640\pi^2} -\frac{\pi^4}{40}, $$ where $K^\prime(x)=K\left(\sqrt{1-x^2}\right)$and $K(x)$ is a complete elliptic integral of the first kind, with the definition $K(x)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$ for $|x|<1$. $x$ is the elliptic modulus.
Note that the following results are known $$ \int_{0}^{1}K^\prime(x)^3\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{128\pi^2}, \int_{0}^{1}xK^\prime(x)^3\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{640\pi^2}. $$ If the integral is true, we can derive many closed-forms for certain integrals. For example, $$ \int_{0}^{1} \frac{K(x)^3}{1+x}\text{d}x =\frac{\,\Gamma\left ( \frac14 \right )^8}{640\pi^2} -\frac{\pi^4}{320}, $$ $$ \int_{0}^{1} \frac{K(x)K^\prime(x)^2}{1+x}\text{d}x =\frac{11\,\Gamma\left ( \frac14 \right )^8}{3840\pi^2}+\frac{\pi^4}{160}. $$ I am grateful for all your help.

Answer 1

A possible evaluation is based on the following. $$\displaystyle \int_{0}^{1} \left ( \frac{K^\prime}{K} \right )^{s-1}K(x)^3\left(1-5x^2\right){\text{d}x} =2^{s-2}\pi^{4-s}\Gamma(s)\lambda(s)\beta(s-4) \qquad[s\in\mathbb{C}] $$ Where $\beta(s)$(resp. $\lambda(s)$) denotes Dirichlet's Beta(resp. Lambda) function. Setting $s=4$ gives $$ \int_0^1K^\prime(x)^3\left(1-5x^2\right)\text{d}x =\frac{\pi^4}{8} $$ and proves the claim.
In the same manner, we list four known values for $x^2K^{s-1}K^\prime{}^{4-s}$. $$ \int_{0}^{1}x^2K^\prime(x)^3\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{640\pi^2} -\frac{\pi^4}{40}\qquad \int_{0}^{1}x^2K(x)^2K^\prime(x)\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{1920\pi^2}+\frac{\pi^4}{80} $$ $$ \int_{0}^{1}x^2K(x)K^\prime(x)^2\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{1280\pi^2} \qquad \int_{0}^{1}x^2K(x)^3\text{d}x =\frac{3\,\Gamma\left ( \frac14 \right )^8 }{6400\pi^2} +\frac{12}{5}\beta(4) $$