Is it true that $\int_0^1 \big(K(k^{1/2})\big)^2\,dk = \frac{7}2\zeta(3)$? Define the complete elliptic integral of the first kind as,
$$K(k) = \tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)$$
Part I. From the link above, we find some of the evaluations below,
$$\begin{aligned}
\int_0^1 K(k^{1/1})\,dk &= 2C\\ 
\int_0^1 K(k^{1/2})\,dk &= 2\\ 
\int_0^1 K(k^{1/3})\,dk &= \frac34(2C+1) \\ 
\int_0^1 K(k^{1/4})\,dk &= \frac{20}9 \\ 
\int_0^1 K(k^{1/5})\,dk &= \frac5{64}(18C+13)
\end{aligned}$$
and so on (?) where $C$ is Catalan's constant.
Part II. On a hunch, I decided to check 2nd powers. It turns out that,
$$\begin{aligned}
\int_0^1 \big(K(k^{1/2})\big)^2\,dk &= \frac{7}2\zeta(3)\\ 
\int_0^1 \big(K(k^{1/4})\big)^2\,dk &= \frac{7}2\zeta(3)+1\\ 
\int_0^1 \big(K(k^{1/6})\big)^2\,dk &= \frac{231}{64}\zeta(3)+\frac{51}{32}\\ 
 \int_0^1 \big(K(k^{1/8})\big)^2\,dk &= \frac{238}{64}\zeta(3)+\frac{881}{432}\\ 
\end{aligned}$$
and so on (?) where $\zeta(3)$ is Apery's constant.
Question: Does the pattern of Part I (involving Catalan's constant) and Part II (involving Apery's constant) really go on forever? What is the closed-form?
 A: In part I, the integrals can be written as
\begin{equation}
  I_n=n\int_0^1x^{n-1}K(x)\,dx
\end{equation}
The moments of the elliptic integrals seem to be well described in the literature (see
here for example)
Denoting $K_n$ and $E_n$ the moments of order $n$ of $K$ and $E$,
\begin{equation}
  K_n=\int_0^1x^nK(x)\,dx; \quad  E_n=\int_0^1x^nE(x)\,dx
\end{equation}
Then $I_n=nK_{n-1}$
the following recursions are derived:
\begin{equation}
  K_{n+2}=\frac{nK_n+E_n}{n+2};\quad E_n=\frac{K_n+1}{n+2}
\end{equation}
with
\begin{equation}
  K_0=2C;\quad E_0=C+\frac{1}{2};\quad K_1=1;\quad E_1=\frac{2}{3}
\end{equation}
which explains the observed pattern.
In part II, integrals can be written as
\begin{equation}
  J_{2p}=2p\int_0^1 x^{2p-1}K^2(x)\,dx
\end{equation}
Denoting the moment of order $n$ of $K^2$,
\begin{equation}
  {}_2K_n=\int_0^1x^nK^2(x)\,dx
\end{equation}
we have
\begin{equation}
  J_{2p}=2p\,_2K_{2p-1}
\end{equation}
In Moments of products of elliptic integrals
by J.G. Wan, Theorem 2 expresses that, when $p$ is odd, the p-th moments of $K'^2,
E'^2, K'E', K^2, E^2$ and $KE$ are expressible as $a+b\zeta(3)$. Moreover, the
moments of $K^2$ satisfy a recursion
\begin{equation}
  (n+1)^3 {}_2K_{n+2}-2n\left( n^2+1 \right) {}_2K_n+(n-1)^3 {}_2K_{n-2}=2
\end{equation}
and thus
\begin{equation}
 J_{2p+4}=\frac{p+2}{2\left( p+1 \right)^3}+\frac{\left( p+2 \right)\left( 2p+1 \right)\left( 2p^2+2p+1 \right)}{2\left( p+1 \right)^4}J_{2p+2}-\frac{p^2\left( p+2 \right)}{(p+1)^3}J_{2p}
\end{equation} 
In the linked paper, a method is given to obtain ${}_2K_1$ and ${}_2K_3$ using a theorem by Zudilin which express them in terms of a ${}_7F_6$ hypergeometric function. 
