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.
A fast and simple solution idea by Cornel Ioan Valean
We will use the power of the ideas and strategies from the books (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
Okay, let's start!
From the book, More (Almost) Impossible Integrals, Sums, and Series (2023), Sect. $4.5$, pages $396$-$398$, we have that
$$\displaystyle \color{blue}{-\frac{\log(1-x^2)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n} \frac{1}{4^n}\binom{2n}{n}(2 H_{2n}-H_n),\ |x|<1} ,$$
and exploiting this fact and turning the left-hand side into a double integral, $\displaystyle \int _0^{\pi/2}\left(\int _0^{\sin (x)}\frac{\log \left(1-y^2\right)}{y\sqrt{1-y^2}}\textrm{d}y\right)\textrm{d}x$, immediately reveals (after changing the integration order) that
$$\sum _{n=1}^{\infty } \binom{2 n}{n}^2 \frac{2 H_{2 n}-H_n}{n 2^{4 n}}$$
$$ =\frac{\pi^2}{2}+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1-x)}{x} \textrm{d}x+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x)}{x} \textrm{d}x$$
$$-\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x^2)}{x} \textrm{d}x. \tag1$$
On the other hand, exploiting that
$$\int_0^{\pi/2} \log(\sin(x)) \sin^{2n}(x) \textrm{d}x=\frac{\pi}{2}\frac{1}{2^{2n}}\binom{2n}{n}\left(H_{2n}-H_n-\log(2)\right),$$
which is also found in the sequel, page $191$, multiplying both sides by $\displaystyle \frac{1}{n 2^{2n}}\binom{2n}{n}$, making the summation from $n=1$ to $\infty$, and rearranging, we get that
$$\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{H_{2n}-H_n-\log(2)}{n 2^{4n}}$$
$$\small =-\frac{8 }{\pi }\int_0^1 \frac{\log ^2(1+x^2)}{1+x^2} \textrm{d}x+\frac{8 }{\pi } \log (2)\int_0^1 \frac{ \log(1+x^2)}{1+x^2}\textrm{d}x$$
$$+\frac{8 }{\pi }\int_0^1 \frac{\log(x) \log(1+x^2)}{1+x^2} \textrm{d}x.\tag2$$
The rest is known and trivial, and the separate series involving the numbers $H_{2n}$ and $H_n$ are extracted by a system of relation consisting of $(1)$ and $(2)$.
End of story
Thank you so much Cornel for your work and your life-changing books, (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
(Very) important: Exploiting such techniques involving the series in blue for building systems of relations, one can derive extremely difficult series. Some (simple) paper(s) showing such examples will be written soon.
Update 1: Indeed, the series taken apart also contain a Trilogarithm with a complex argument. The forms of the two key identities (that is, using integrals) avoid that appearance, and to get directly the desired value without touching the Trilogarithm with a complex argument, all we need is to turn the integrals from the identity in $(2)$ into arctan-log integrals. Then we need to exploit the arctan-log integral identities given in the sequel, Sects. $1.36$-$1.38$, pages $48$-$51$. That's all (and done)!
Update 2: A full solution with all the details of this problem will be found soon in a separate paper (the main series plus the two separated series).
Update 3: Binoharmonic Series with the Squared Central Binomial Coefficient And Their Integral Transformation Using Elliptic Integrals by Cornel Ioan Valean
Best Answer
In the linked answer (Integrals of elliptic integrals.) we proved:
$$K(s)E(s)=\frac{\pi^2}{8}\frac{2-s^2}{\sqrt{1-s^2}}\sum_{n=0}^{\infty}\frac{(2n+1)(2n)!^3}{2^{8n}n!^6}{\left(\frac{s^4}{ s^2-1}\right)}^n.\tag{1}$$ Which is a consequence of Clausen's formula: $$K^{2}(s)=\frac{\pi^2}{4}\frac{1}{\sqrt{1-s^2}}\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{8n}n!^6}\left(\frac{s^4}{s^2-1} \right)^{n}.\tag{2}$$
To find $p,q,s$ such $pK(s)E(s)+qK^2(s)$ is a known constant, you will need the elliptic singular moduli. If you consider $s=\frac{1}{\sqrt{2}}$ then :
$$K^2(s)=\frac{\sqrt{2}\pi^2}{4}\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3}{2^{9n}n!^6}=\frac{\Gamma{(1/4)}^4}{16\pi},\tag{3}$$ and $$E(s)K(s)=\frac{3\pi^2}{8\sqrt{2}}\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3(2n+1)}{2^{9n}n!^6}=\frac{\pi}{4}+\frac{\Gamma(\frac{1}{4})^4}{32\pi}.\tag{4}$$ So if $p=1$, $q=-\frac{1}{2}$ and $s=\frac{1}{\sqrt{2}}$ then: $$pK(s)E(s)+qK^2(s)=\frac{\pi}{4}.\tag{5}$$ If you consider Landen's transformation $K(s)(1+s)=K(\frac{2\sqrt{s}}{1+s})$ applied to $(1)$ and $(2)$ and $s=\sqrt{2}-1$ you will get Bauer's series and you will find another relation of the type of $(5)$.