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,

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

**Questions:**

- What are the closed-forms of: $$\int_0^1\big(K_n(k)\big)^m dk=\,?$$ for power $m=2$ or $m=3$?
- 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.)