Finding $p\cdot K\left(s\right)E\left(s\right)+q\cdot K\left(s\right)^{2}$ such that it is a known constant

Question

Recently I got interested in the Series of Form, $$\sum_{r=0}^{\infty}\frac{r^p}{A^{r}}\binom{2r}{r}^{b}$$
I was able to derive this, $$\left(\frac{\pi^{2}}{4}\right)\sum_{r=0}^{\infty}\frac{a+br}{x^{r}}\binom{2r}{r}^3=\left(\frac{b}{1-2\alpha}\right)K\left(\alpha\right)E\left(\alpha\right)+\left(a-\frac{16b}{32-x\alpha}\right)K\left(\alpha\right)^{2}$$ Where, $K(k)$ is the Complete Elliptical Integral of the First Kind and $E(k)$ is the Complete Elliptical Integral of the Second Kind with the following convention: $$K\left(k\right)=\int_{0}^{1}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-kt^{2}\right)}}dt$$ $$E\left(k\right)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k\sin^{2}t}dt$$ and, $$\alpha=\frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{64}{x}}$$

Now to make the series have an Elegant Form we need to eliminate the Elliptical Integrals from the Right Side.
What I mean by elegant form is something like, $$\sum_{r=0}^{\infty}\frac{a+br}{x^{r}}\binom{2r}{r}^3=\frac{C}{\pi^2}$$ Where $a, b$ are Integers and $C$ should be a known constant, preferably an Integer but other closed form expressions work too! $x$ would most likely be a power of $2$ but I am not sure on that.

Modifying the Equations a bit we can write it as, $$\left(\frac{\pi^{2}}{4}\right)\sum_{r=0}^{\infty}\left(\frac{\left[q+p\left(1-s\right)\right]+\left[p\left(1-2s\right)\right])r}{\left[\frac{16}{s\left(1-s\right)}\right]^{r}}\right)\binom{2r}{r}^3=p\cdot K\left(s\right)E\left(s\right)+q\cdot K\left(s\right)^{2}$$
Now the main problem is that I haven't been able to find $p, q, s$ such that $$p\cdot K\left(s\right)E\left(s\right)+q\cdot K\left(s\right)^{2}$$ is a known constant.

EDIT: I searched out some values for the Elliptical Integrals and was able to at least derive this: $$\sum_{r=0}^{\infty}\left(\frac{1+6r}{2^{8r}}\right)\binom{2r}{r}^3=\frac{4}{\pi}$$

Answer 1

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 $(5)$.