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.)