Given real parameters $\left(\alpha,\beta,\gamma,\delta\right)\in\mathbb{R}^{4}$ such that $0<\alpha<\delta$ and real arguments $\left(x,y\right)\in\left(-\infty,1\right)^{2}$, we can express the Appell $F_{1}$ function via the integral representation
$$\begin{align}
F_{1}{\left(\alpha;\beta,\gamma;\delta;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(\alpha,\delta-\alpha\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\alpha-1}\left(1-t\right)^{\delta-\alpha-1}}{\left(1-xt\right)^{\beta}\left(1-yt\right)^{\gamma}}.\\
\end{align}$$
Starting from the integral representation of the $F_{1}$ function for the particular set of parameters that we're interested in, we obtain an integral of a simple algebraic function with elementary antiderivative: for any fixed but arbitrary $\left(x,y\right)\in\left(-\infty,1\right)^{2}$,
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(1,1\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{1}^{0}\mathrm{d}u\,\frac{\left(-1\right)\left(1-x\right)}{\left(1-xu\right)^{2}}\cdot\frac{\left(1-xu\right)\sqrt{1-xu}}{\left(1-x\right)\sqrt{\left(1-y\right)-\left(x-y\right)u}};~~~\small{\left[t=\frac{1-u}{1-xu}\right]}\\
&=\frac{1}{\sqrt{1-y}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-y}{1-y}\right)u\right]}}.\\
\end{align}$$
Suppose $a\in\left(0,1\right)$ and $x<1\land x\neq0$. Setting $y=ax$, we then find
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,ax\right)}
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-ax}{1-ax}\right)u\right]}}\\
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{1-a}{1-ax}\right)xu\right]}}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{x}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v\right)\left[1-\left(\frac{1-a}{1-ax}\right)v\right]}};~~~\small{\left[u=\frac{v}{x}\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[1-\left(\frac{1-a}{1-ax}\right)\left(1-w\right)\right]}};~~~\small{\left[v=1-w\right]}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[\left(1-ax\right)-\left(1-a\right)\left(1-w\right)\right]}}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1-x}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[\left(1-ax\right)-\left(1-a\right)t\right]}};~~~\small{\left[w=1-t\right]}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left[a+\left(1-a\right)t\right]}};~~~\small{\left[w=\left(1-x\right)t\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[1-\left(\frac{1-a}{1-ax}\right)t\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-u\right)\left[1-\left(1-a\right)u\right]}};~~~\small{\left[t=1-u\right]}\\
&=\frac{2}{x\sqrt{1-ax}}\,{_2F_1}{\left(\frac12,1;\frac32;\frac{1-a}{1-ax}\right)}-\frac{2}{x}\,{_2F_1}{\left(\frac12,1;\frac32;1-a\right)},\\
\end{align}$$
where in the last line above we've used the Euler integral representation formula to express the remaining integrals in terms of the ${_2F_1}$ function:
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)};~~~\small{z<1\land0<\beta<\gamma}.$$
$$\tag*{$\blacksquare$}$$
Let $\mathcal{S}$ denote the sum of the following (convergent) infinite series:
$$\mathcal{S}:=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2},\tag{1}$$
where here $n!!$ denotes the so-called double factorial of a number $n$.
(Note: My definition of $\mathcal{S}$ has an additional scalar factor of $4$ so as to simplify its expression in terms of the generalized hypergeometric function $_4F_3$.)
We'll make use of the following well-known integration formula for a subclass of Wallis' integrals (for proof see [wiki][1]):
$$\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}=\frac{(2n)!!}{(2n+1)!!};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$
Recall the definition of the [polylogarithm][2] as an infinite series. Given $s\in\mathbb{C}\land z\in\mathbb{C}\land|z|<1$, the polylogarithm $\operatorname{Li}_{s}{\left(z\right)}$ of order $s$ and argument $z$ is given by the (absolutely convergent) power series
$$\operatorname{Li}_{s}{\left(z\right)}=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}}.$$
For positive integer order, the polylogarithm can be defined iteratively by
$$\operatorname{Li}_{1}{\left(z\right)}:=-\ln{\left(1-z\right)};~~~\small{z\in\left(-\infty,1\right)},$$
$$\operatorname{Li}_{n+1}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}t\,\frac{\operatorname{Li}_{n}{\left(t\right)}}{t};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
Another useful integral representation for $\operatorname{Li}_{n+1}{\left(z\right)}$, which can be obtained from the previous one by repeated integration by parts, is
$$\operatorname{Li}_{n+1}{\left(z\right)}=\frac{(-1)^{n}}{n!}\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{n}{\left(t\right)}}{1-zt};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
An important auxiliary function pertaining to the polylogarithm is the so-called Nielsen generalized polylogarithm, defined for positive integer parameters via the integral representation
$$S_{n,p}{\left(z\right)}:=\frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{n-1}{\left(t\right)}\ln^{p}{\left(1-zt\right)}}{t};~~~\small{\left(n,p\right)\in\mathbb{N}^{2}\land z\in\left(-\infty,1\right]}.$$
The following integration formula will be useful to have on hand later:
$$\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t}=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)};~~~\small{z\in\left(-\infty,1\right]}.$$
Proof:
$$\begin{align}
\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t}
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}+\ln^{2}{\left(1-zt\right)}-\left[\ln{\left(1-t\right)}-\ln{\left(1-zt\right)}\right]^{2}}{2t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(\frac{1-t}{1-zt}\right)}}{2t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)}{\left(1-zu\right)^{2}}\cdot\frac{\ln^{2}{\left(u\right)}}{2\left(\frac{1-u}{1-zu}\right)};~~~\small{\left[t=\frac{1-u}{1-zu}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)\ln^{2}{\left(u\right)}}{2\left(1-u\right)\left(1-zu\right)}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\ln^{2}{\left(u\right)}}{2\left(1-u\right)}+\int_{0}^{1}\mathrm{d}u\,\frac{z\ln^{2}{\left(u\right)}}{2\left(1-zu\right)}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{2}{\left(t\right)}}{1-zt}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{t}\\
&=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)}.\\
\end{align}$$
Using the technique of interchanging the order of summation and integration, we obtain an expression for the power series $\mathcal{S}$ as a definite integral.
$$\begin{align}
\mathcal{S}
&=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2}\\
&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\cdot\frac{(2n)!!}{(2n+1)!!}\\
&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}\\
&=\sum_{n=0}^{\infty}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=0}^{\infty}\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=1}^{\infty}\frac{\cos^{2n-1}{\left(\varphi\right)}}{n^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\cos{\left(\varphi\right)}}\sum_{n=1}^{\infty}\frac{\left[\cos^{2}{\left(\varphi\right)}\right]^{n}}{n^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sec{\left(\varphi\right)}\operatorname{Li}_{2}{\left(\cos^{2}{\left(\varphi\right)}\right)}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos{\left(\varphi\right)}\operatorname{Li}_{2}{\left(1-\sin^{2}{\left(\varphi\right)}\right)}}{1-\sin^{2}{\left(\varphi\right)}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}};~~~\small{\left[\varphi=\arcsin{\left(x\right)}\right]}.\\
\end{align}$$
Then,
$$\begin{align}
\mathcal{S}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}}\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+x}{1-x}\right)}}{2}\cdot\frac{2x\ln{\left(x^{2}\right)}}{1-x^{2}};~~~\small{I.B.P.s}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(\frac{1-x}{1+x}\right)}}{1-x^{2}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left((1+x)^2\right)}}{1-x^{2}}\\
&=\frac12\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{4x\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x^{2}}\\
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(y\right)}\ln{\left(1-y\right)}}{1-y};~~~\small{\left[x^{2}=y\right]}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1+x}-\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t};~~~\small{\left[y=1-t\right]}\\
&~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(1+x\right)}}{x};~~~\small{I.B.P.s}\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2-t\right)}}{t};~~~\small{\left[x=1-t\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1+t\right)}}{t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2\right)}}{t}-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(1-\frac12t\right)}}{t}\\
&=\frac12\,S_{2,1}{\left(1\right)}-2S_{1,2}{\left(-1\right)}\\
&~~~~~+2\ln{\left(2\right)}\operatorname{Li}_{2}{\left(1\right)}-2\left[\operatorname{Li}_{3}{\left(\frac12\right)}+S_{1,2}{\left(\frac12\right)}\right]\\
&=3\ln{\left(2\right)}\,\zeta{\left(2\right)}-\frac74\,\zeta{\left(3\right)}.\blacksquare\\
\end{align}$$
Best Answer
Fortunately, there's nothing terribly difficult about this problem. Once you've committed a handful of the most important series and integral representations to memory, the result practically falls out of the definitions.
(Note: I've gone ahead and absorbed that factor of $4$ into $z$ since makes the integral less cumbersome and there's no good reason not to.)
In particular, we'll need the integral representations for the Gauss hypergeometric function ${_2F_1}$ and the generalized hypergeometric function ${_pF_q}$:
$$\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}=\int_{0}^{1}\mathrm{d}t\,t^{b-1}\left(1-t\right)^{c-b-1}\left(1-zt\right)^{-a};~~~\small{\Re{(c)}>\Re{(b)}>0},$$
$$\operatorname{B}{\left(p,q-p\right)}\,{_3F_2}{\left(a,b,p;c,q;z\right)}=\int_{0}^{1}\mathrm{d}t\,t^{p-1}\left(1-t\right)^{q-p-1}{_2F_1}{\left(a,b;c;zt\right)};~~~\small{\Re{(q)}>\Re{(p)}>0},$$
$$\operatorname{B}{\left(r,s-r\right)}\,{_4F_3}{\left(a,b,p,r;c,q,s;z\right)}=\int_{0}^{1}\mathrm{d}t\,t^{r-1}\left(1-t\right)^{s-r-1}{_3F_2}{\left(a,b,p;c,q;zt\right)};~~~\small{\Re{(s)}>\Re{(r)}>0},$$
where of course the beta function itself is given by
$$\operatorname{B}{\left(x,y\right)}:=\int_{0}^{1}\mathrm{d}t\,t^{x-1}\left(1-t\right)^{y-1};~~~\small{\Re{(x)}>0\land\Re{(y)}>0}.$$
There are a couple of trigonometric symmetries we can exploit right away to simplify the problem: for any continuous function $f:[-1,1]\rightarrow\mathbb{R}$, we have
$$\begin{align} \int_{0}^{\pi}\mathrm{d}\varphi\,f{\left(\cos{\left(\varphi\right)}\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\cos{\left(\varphi\right)}\right)}+\int_{\frac{\pi}{2}}^{\pi}\mathrm{d}\varphi\,f{\left(\cos{\left(\varphi\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\cos{\left(\varphi\right)}\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\cos{\left(\pi-\varphi\right)}\right)};~~~\small{\left[\varphi\mapsto\pi-\varphi\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\cos{\left(\varphi\right)}\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(-\cos{\left(\varphi\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\left[f{\left(\cos{\left(\varphi\right)}\right)}+f{\left(-\cos{\left(\varphi\right)}\right)}\right],\\ \end{align}$$
and
$$\int_{0}^{\pi}\mathrm{d}\varphi\,f{\left(\cos^{2}{\left(\varphi\right)}\right)}=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\cos^{2}{\left(\varphi\right)}\right)}=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,f{\left(\sin^{2}{\left(\varphi\right)}\right)}.$$
Then, for any $-1<z<1$ we obtain
$$\begin{align} \mathcal{I}{\left(z\right)} &=-\int_{0}^{\pi}\mathrm{d}\phi_{1}\int_{0}^{\pi}\mathrm{d}\phi_{2}\,\ln{\left(1-z\cos{\left(\phi_{1}\right)}\cos{\left(\phi_{2}\right)}\right)}\\ &=-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\phi_{1}\int_{0}^{\pi}\mathrm{d}\phi_{2}\,\left[\ln{\left(1-z\cos{\left(\phi_{1}\right)}\cos{\left(\phi_{2}\right)}\right)}-\ln{\left(1+z\cos{\left(\phi_{1}\right)}\cos{\left(\phi_{2}\right)}\right)}\right]\\ &=-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\phi_{1}\int_{0}^{\pi}\mathrm{d}\phi_{2}\,\ln{\left(1-z^{2}\cos^{2}{\left(\phi_{1}\right)}\cos^{2}{\left(\phi_{2}\right)}\right)}\\ &=-2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\phi_{1}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\phi_{2}\,\ln{\left(1-z^{2}\sin^{2}{\left(\phi_{1}\right)}\sin^{2}{\left(\phi_{2}\right)}\right)}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{2\ln{\left(1-z^{2}x^{2}y^{2}\right)}}{\sqrt{1-x^{2}}\sqrt{1-y^{2}}};~~~\small{\left[\phi_{1}=\arcsin{(x)},\,\phi_{2}=\arcsin{(y)}\right]}\\ &=-\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1-z^{2}tu\right)}}{2\sqrt{t}\sqrt{1-t}\sqrt{u}\sqrt{1-u}};~~~\small{\left[x=\sqrt{t},\,y=\sqrt{u}\right]}\\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{z^{2}\sqrt{t}\sqrt{u}}{2\sqrt{1-t}\sqrt{1-u}}\left[-\frac{\ln{\left(1-z^{2}tu\right)}}{z^{2}tu}\right]\\ &=\frac{z^{2}}{2}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\sqrt{1-t}}\int_{0}^{1}\mathrm{d}u\,\frac{\sqrt{u}}{\sqrt{1-u}}\,{_2F_1}{\left(1,1;2;z^{2}tu\right)}\\ &=\frac{z^{2}}{2}\cdot\frac{\pi}{2}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\sqrt{1-t}}\,{_3F_2}{\left(1,1,\frac32;2,2;z^{2}t\right)}\\ &=\frac{\pi^{2}z^{2}}{8}\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;z^{2}\right)}.\blacksquare\\ \end{align}$$