Evaluate hypergeometric $_6F_5\left(\{\frac12\}_6;1,\{\frac32\}_4;1\right)$ Background: I'm searching for $_pF_q$ representations for MZVs. In related article On the interplay between hypergeometric series, Fourier-Legendre expansions and Euler sums by M. Cantarini and J. D’Aurizio, the series $_5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$ is transformed to and Euler sum i.e. $\sum _{n=1}^{\infty } \frac{(-1)^n \left(\sum _{k=1}^n \frac{1}{2 k+1}\right){}^3}{2 n+1}$ by using FL expansion, which I invoke MZV values to give a closed-form successfully:

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*$ \pi  \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=64 \Im(\text{Li}_4(1+i))-\frac{2}{3}  \pi  \log ^3(2)-\pi ^3 \log (2)-\frac{1}{32} \left(\psi ^{(3)}\left(\frac{1}{4}\right)-\psi ^{(3)}\left(\frac{3}{4}\right)\right)$

Problem: I wonder if the higher-weight case can be evaluated by similar means:

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*$ \pi \sum _{n=0}^{\infty } \left(\frac{\binom{2 n}{n}}{4^n}\right)^2\frac{1}{(2 n+1)^4}=\pi \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$
FL expansion of $\frac{\log ^3(x)}{\sqrt{x}}$ is needed here, but I'm not able to compute it.

Update: Using Jack's formula one may deduce

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*$\small \pi  \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=-40 \Im(\text{CMZV}(4,\{4,1\},\{1,0\}))+\frac{152}{3} \Im(\text{CMZV}(4,\{4,1\},\{1,2\}))-256 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{16}{3} \beta(4) \log (2)+\frac{25 \pi ^5}{64}+\frac{1}{6} \pi  \log ^4(2)+\frac{3}{4} \pi ^3 \log ^2(2)$
See here for detailed expanlation.

All right, I am going to re-do my computations from scratch. This will probably take some time, so please do not downvote this answer in the meanwhile. For any $n\geq 3$ we have
$$ (-1)^{n+1}x^{n+1/2}\cdot \frac{d^n}{dx^n} \frac{\log^3(x)}{\sqrt{x}} = A_n+B_n\log(x)+C_n\log^2(x)+D_n\log^3(x)=S_n\tag{S}$$
with $A_n,B_n,C_n,D_n\in\mathbb{Q}$ related to each other by induction / recurrence relations. Once the explicit form of these constants is known we also have the explicit FL-expansion of $\frac{\log^3(x)}{\sqrt{x}}$ by Rodrigues formula, since
$$ \int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}P_n(2x-1)\,dx = \frac{1}{n!}\int_{0}^{1}x^n (1-x)^n \left[\frac{d^n}{dx^n}\frac{\log^3(x)}{\sqrt{x}}\right]\,dx \tag{B}$$
and the RHS is given by derivatives of the Beta function.
The coefficient which is the easiest to guess is $D_n$:
$$ D_n = -\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^n} = -\frac{(2n)!}{4^n n!}$$
then we may differentiate both sides of $(S)$ and write down the induced recurrence relations.
$$ \left(n+\tfrac{1}{2}\right)S_n - S_{n+1} = B_n + 2C_n \log(x)+3D_n\log^2(x)\tag{R}$$
Focusing on the coefficient of $\log^2(x)$ on both sides we get
$$ \left(n+\tfrac{1}{2}\right)C_n-C_{n+1} = 3D_n \tag{D_n}$$
and it is practical to introduce rescaled coefficients for simplifying the recursion. Letting $D_n=\frac{(2n)!}{4^n n!}d_n$ (and the same for $A_n,B_n,C_n$) we get
$$ \left(n+\tfrac{1}{2}\right)(c_n-c_{n+1}) = 3d_n = -3 $$
so
$$ c_{n+1} = c_n+\frac{6}{2n+1} $$
and
$$ C_n = \frac{(2n)!}{4^n n!}\cdot 6\sum_{k=0}^{n-1}\frac{1}{2k+1}.\tag{C_n}$$
[...] Continuing on this route, once we define $\mathscr{H}_n^{(k)}$ as $\sum_{h=0}^{n}\frac{1}{(2h+1)^k}$ we get
$$ \boxed{\small\frac{\log^3(x)}{\sqrt{x}}\stackrel{\mathcal{D}}{=}32\sum_{n\geq 0}(-1)^{n+1}P_n(2x-1)\left[4\mathscr{H}_n^3+2\mathscr{H}_n^{(3)}-\frac{6\mathscr{H}_n^2}{2n+1}+\frac{6\mathscr{H}_n}{(2n+1)^2}-\frac{3}{(2n+1)^3}\right]}.$$
Since $\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{(2n+1)^4}$ can be represented in terms of $\frac{2}{\pi}\int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}K(x)\,dx $, the simple FL-expansion of $K(x)$ gives that the first hypergeometric series can be computed in terms of five Euler sums with weight five. The simplest of them are
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5} = \frac{5\pi^5}{1536}$$
and
$$ \sum_{n\geq 0}\frac{(-1)^n\mathscr{H}_n}{(2n+1)^4}=-\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}\sum_{n\geq 0}(-1)^n\mathscr{H}_n x^n\,dx $$
which equals
$$ -\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)\arctan(x)}{x(1+x)}\,dx=\frac{5\pi^5}{24576}-\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)\arctan(x)}{x+1}\,dx. $$
The factor $\frac{dx}{x+1}$ is invariant with respect to the substitution $x\to\frac{1-x}{1+x}$, so the last integral can also be expressed in terms of $\int_{0}^{1}\text{arctanh}^3(x)\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{x+1}$, where
$$ \int_{0}^{1}\text{arctanh}^3(x)\frac{dx}{x+1}\,dx = \int_{0}^{+\infty}x^3(1-\tanh x)\,dx = \frac{7\pi^4}{960}$$
by the integral representations for the $\eta$ and $\zeta$ functions.
