Evaluate $\int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2 K\left ( \sin\theta \right )\text{d}\theta$ Let us define $K(x)$ as complete elliptic integral of the first kind, where $x$ is elliptic modulus. A possible closed-form is ($G$ denotes Catalan's constant.)
$$
\int_{0}^{\pi/2}
\ln\left[ \tan\left (  \frac{\theta}{2}\right)  \right ]^2
K\left ( \sin\theta \right )\text{d}\theta
=\frac{\Gamma\left ( \frac14 \right )^4G }{8\pi}.
$$
It looks like a "product" of two solvable integrals (both are elementary):
$$\int_{0}^{\pi/2}
\ln\left[ \tan\left (  \frac{\theta}{2}\right)  \right ]^2\text{d}\theta
=\frac{\pi^3 }{8}$$
and
$$
\int_{0}^{\pi/2}
K\left ( \sin\theta \right )\text{d}\theta
=\frac{\Gamma\left ( \frac14 \right )^4 }{16\pi}.
$$

Question:
How can we evaluate the integral? I try to utilize the Fourier series $K(\sin\theta)
=\pi\sum_{n\ge0} \frac{\left ( \frac12 \right )_n^2 }{(n!)^2} 
\sin\left ( \left ( 4n+1 \right )\theta  \right )$ to prove, but seems not to go well. I appreciate for your help.

An Interesting Observation:
We find
$$
\int_{0}^{\pi/2}
\ln\left[ \tan\left (  \frac{\theta}{2}\right)  \right ]^4
K\left ( \sin\theta \right )\text{d}\theta
=\frac{3\,\Gamma\left ( \frac14 \right )^4}{4\pi}(G^2+\beta(4))
$$
where $\beta(.)$ is Dirichlet's $\beta$ function.
 A: We have
$$\int_0^{\pi/2} \log^{2n} (\tan \frac{x}{2}) K(\sin x) dx = \int_0^1 \frac{2 \log^{2n} t}{1+t^2} K(\frac{2t}{1+t^2}) dt$$
Their values can be extracted by differentiating $$\tag{*}\int_0^1 \frac{t^{4a} + t^{-4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt = \frac{\pi}{8}\cot(\pi(\frac{1}{4}-a))\frac{\Gamma \left(a+\frac{1}{4}\right)^2}{\Gamma \left(a+\frac{3}{4}\right)^2} \quad -1/4<\Re(a)<1/4$$
I give two different proofs of $(*)$, I come up with the longer proof first.

First proof of $(*)$: using quadratic transformation $(1+t^2)K(t^2) = K(2t/(1+t^2))$, we have
$$
\begin{aligned}\int_0^1 \frac{t^{4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt &= \frac{1}{2} \int_0^1 t^{2a-1/2} K(t) dt \\ &= \frac{1}{2} \frac{\pi}{1+4a}{_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1)
\end{aligned}$$
here we noted that $K(t)$ is a $_2F_1$. Using third formulas here gives
$${_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) = \frac{4 a+1}{4 a-1} {_3F_2(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} - a; 1, \frac{5}{4} - a; 1)} +\frac{\Gamma \left(\frac{1}{4}-a\right) \Gamma \left(a+\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}-a\right) \Gamma \left(a+\frac{3}{4}\right)}$$
we see the $a$ in $_3F_2$ becomes $-a$, giving $(*)$. Q.E.D.

Second proof of $(*)$: It's much longer. See edit history.
