Is there a simple expression for $\sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^2?$

Is there a simple expression for the power series $$\sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^2?$$ This question came up in a quantum mechanics problem. Mathematica only returns $$ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x^2\right).$$ We know that $$\sum_{n=0}^{\infty} (4x)^n \frac{(n!)^2}{(2n+1)!} = \frac{\arcsin \sqrt x}{\sqrt{x(1-x)}},$$ but it is not clear how to proceed for the other sum.

I will award a 50 point bounty to a correct closed-form expression if desired.

This isn't quite an answer, but I think it can lead to one: with \begin{align}z:&=_2\!\!\mathrm{F}_1(\tfrac12,\tfrac12;1;x)=\frac2\pi K(\sqrt{x})\\ y:&=\pi\frac{_2\mathrm{F}_1(\tfrac12,\tfrac12;1;1-x)}{_2\mathrm{F}_1(\tfrac12,\tfrac12;1;x)}=\pi\frac{z(1-x)}{z(x)}=\pi\frac{K'(\sqrt{x})}{K(\sqrt{x})},\tag1\\ \sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^2&={_3\mathrm{F}_2}\left(1,1,1;\frac{3}{2},\frac{3}{2};x^2\right)=\frac{4K(x)}{\pi x}\sum_{k\ge1}\frac1{(2k-1)^2 \cosh(k-1/2)y(x^2)}\tag2 \end{align}(from Ramanujan's Notebooks Part III; $(1)$, $(2)$, pgs. 101, 153 resp. [entry 6 in both]) and \begin{align}\int_0^1 u\cos(2k-1)\pi u\,du&=\left[\frac{u\sin(2k-1)\pi u}{(2k-1)\pi}+\frac{\cos(2k-1)\pi u}{(2k-1)^2\pi^2}\right]_0^1\\ &=\frac{-1-1}{(2k-1)^2\pi^2}=\frac{-2}{(2k-1)^2\pi^2}\\ \end{align} $$\frac{Kk}\pi\operatorname{cn}\left(\frac{2Ku}\pi,k\right)=\sum_{n\ge1}\frac{2\cos(2n-1)u}{e^{(n-1/2)y(k^2)}+e^{-(n-1/2)y(k^2)}}=\sum_{n\ge1}\frac{\cos(2n-1)u}{\cosh\pi(n-1/2)K'/K}$$(the last formula is from DLMF 22.11) thus\begin{align}\int_0^1\frac{Kku}\pi\operatorname{cn}\left(2uK,k\right)\, du&=\int_0^1\sum_{n\ge1}\frac{u\cos(2n-1)\pi u}{\cosh\left(\frac{\pi K'}{K}(n-\tfrac12)\right)}\,du\tag3\\ &=-\frac2{\pi^2}\sum_{k\ge1}\frac1{(2n-1)^2\cosh\pi(n-\tfrac12)\frac{K'}K} \end{align} I thought to add a new variable $v$ to the first integral in $(3)$ to make it$$\frac{Kk}\pi\int_0^1u\operatorname{cn}\left(2K(u+v),k\right)\,du\tag4$$I figured since $\operatorname{cn}(2Ku,k)$ has periods $2$ and $1+iK'/K$, so too will the integral (for $v$)...but I'm not completely sure because the simpler $\int_0^1\operatorname{cn}(u+v,k)\,du$ yields a multi-valued $k^{-1}[\sin^{-1}(k\operatorname{sn}(1+v,k))-\sin^{-1}(k\operatorname{sn}(v,k))]$. But if I'm right, then $(4)$ can be studied as an elliptic function in $v$ and so $(2)$ evaluates to a combination of theta and elliptic functions.