What is $\int_0^1 \left(\tfrac{\pi}2\,_2F_1\big(\tfrac13,\tfrac23,1,\,k^2\big)\right)^3 dk$?

Question

As in this post, define the ff:

$$K_2(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)}$$

$$K_3(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac13,\tfrac23,1,\,k^2\right)}$$

$$K_4(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac14,\tfrac34,1,\,k^2\right)}$$

$$K_6(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac16,\tfrac56,1,\,k^2\right)}$$

We find that,

$$\int_0^1 K_2(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;1,\tfrac32;1\right)}=2G$$

$$\int_0^1 K_3(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac13,\tfrac23;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}2\, \ln2$$

$$\int_0^1 K_4(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac14,\tfrac34;1,\tfrac32;1\right)}=2\ln(1+\sqrt2)$$

$$\int_0^1 K_6(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac16,\tfrac56;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}4\, \ln(2+\sqrt{3})$$

where $G$ is Catalan's constant. However, this post gives a third power,

$$\int_0^1\big(K_2(k)\big)^3 dk=\frac35 \bigg(\tfrac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12,1,\tfrac12\big)\bigg)^4=\frac35 \big(K(k_1)\big)^4 = \frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2} \approx 7.0902$$

where $K(k_d)$ is an elliptic integral singular value while YuriyS in his comment below gives,

$$\int_0^1\big(K_3(k)\big)^3 dk \approx 6.53686311168760876289835638374 $$

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Questions:

1. What are the closed-forms of: $$\int_0^1\big(K_n(k)\big)^m dk=\,?$$ for power $m=2$ or $m=3$?
2. Or at least their numerical evaluation up to 20 digits?

P.S. My old version of Mathematica can't evaluate it with sufficient precision, nor does WolframAlpha. (Enough digits may make it amenable to the Inverse Symbolic Calculator.)

Answer 1

This is not an answer, but it might outline a viable approach, whose credits go to Yajun Zhou.

As outlined here, $K_3(k)$ is related to the Legendre function $P_{-1/3}$.

Legendre functions and products of Legendre functions with suitable indices have known Tricomi transforms,

$$ (\mathfrak{T} f)(x)=\text{PV}\int_{-1}^{1}\frac{f(\xi)}{\pi(x-\xi)}\,d\xi$$

and the Tricomi transform is a self-adjoint operator with respect to the dot product given by $\langle f,g\rangle = \int_{0}^{1}f(x)g(x)\,dx$.

This allowed Zhou to show that

$$ \int_{0}^{1}\mathbf{K}^3\left(\sqrt{1-k^2}\right)\,dk = 3\int_{0}^{1}\mathbf{K}^2(k)\mathbf{K}\left(\sqrt{1-k^2}\right)\,dk =\frac{\Gamma^8\left(\tfrac{1}{4}\right)}{128\pi^2}\tag{1}$$

and it is possible that something similar can be done also by replacing $\mathbf{K}$ with $K_3$.

On the other hand, there is a major difficulty to overcome: Zhou's proof, besides relying on generalized Clebsch-Gordan integrals and Beltrami/Ramanujan rotations, heavily relies on Landen's identity

$$ \mathbf{K}\left(\sqrt{1-k^2}\right) = \frac{2}{1+k}\,\mathbf{K}\left(\frac{1-k}{1+k}\right), $$ which can be seen as a consequence of the relation between $\mathbf{K}$ and the arithmetic-geometric mean.

$K_3$ is related to the cubic analogue of the $\text{AGM}$, so it also fulfills transformation formulas, but they are considerably more complex than Landen's identity for $\mathbf{K}$, hence mimicking Zhou in proving the analogue of $(1)$ might be pretty hard (see equation $(64)$ of Zhou's paper).

Rogers, Wan and Zucker follow a different approach, which is to consider suitable modular forms and perform an evaluation of their Eisenstein series for suitable exponents. That also might be viable for $K_3$.