Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$ 
I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$

I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving $\mathrm{Li}_2$ for any value of $x$ that I've tried so far, so I can only assume that a general closed form exists.
I've started my attempts by noticing that 
$$f(x)=\frac12\int_0^1 \frac{_2F_1(\tfrac12,\tfrac12;\tfrac32;xt)}{\sqrt{t}}dt,$$
because
$$\frac12\int_0^1 \frac{(xt)^n}{\sqrt{t}}dt=\frac{x^n}{2n+1}$$
which would introduce another factor of $$\frac{n+1/2}{n+3/2}$$
when computing the ratio of the terms. Similarly,
$$_2F_1\left(\tfrac12,\tfrac12;\tfrac32;x\right)=\frac12\int_0^1 \frac{_1F_0(\tfrac12;;xt)}{\sqrt{t}}dt.$$
The last hypergeometric I was able to recognize as $$_1F_0\left(\tfrac12;;xt\right)=\frac1{\sqrt{1-xt}}.$$
So, all in all,
$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu,$$
which looks like the Beta function's evil cousin.
I do not know how to turn this integral into something containing $\mathrm{Li}_2$ and I need some help. Thanks!

Starting from the integral representation, we let $v = w/u$ and change the order of integration to find
\begin{align} 
f(x) &= \frac{1}{4} \int \limits_0^1 \int \limits_0^u \frac{1}{u \sqrt{w(1-xw)}} \, \mathrm{d} w \, \mathrm{d} u = \frac{1}{4} \int \limits_0^1 \frac{\mathrm{d} w}{\sqrt{w(1-xw)}} \int \limits_w^1 \frac{\mathrm{d} u}{u} \\
&= \frac{1}{4} \int \limits_0^1 \frac{-\log(w)}{\sqrt{w(1-xw)}} \, \mathrm{d} w \, .
\end{align}
The combination of the next few substitutions can be written as $w = \sin^2(t/2)/x$, which yields
\begin{align} f(x) &= \frac{1}{2 \sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(\frac{\sin\left(\frac{t}{2}\right)}{\sqrt{x}}\right) \, \mathrm{d} t \\
&= \frac{\arcsin(\sqrt{x})}{\sqrt{x}} \log(2 \sqrt{x}) + \frac{1}{2\sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(2 \sin\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\
&= \frac{1}{\sqrt{x}} \left[\arcsin(\sqrt{x}) \log(2\sqrt{x}) + \frac{1}{2} \operatorname{Cl}_2(2\arcsin(\sqrt{x}))\right]
\end{align}
for $x \in (0,1]$, while $f(0) = 1$. The Clausen function is of course related to the dilogarithm. Interesting special values include $f(1) = \frac{\pi}{2} \log(2)$, $f\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} \left[\frac{\pi}{4} \log(2) + \mathrm{G}\right]$ and $f\left(\frac{1}{4}\right) = \operatorname{Cl}_2\left(\frac{\pi}{3}\right)$  (see this question).
For $y > 0$ similar steps lead to
\begin{align}
f(-y) &= \frac{\operatorname{arsinh}(\sqrt{y})}{\sqrt{y}} \log(2 \sqrt{y}) + \frac{1}{2\sqrt{y}} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(2 \sinh\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{1}{2} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(1 - \mathrm{e}^{-t}\right) \, \mathrm{d} t \right] \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left(\mathrm{e}^{-2\operatorname{arsinh}(\sqrt{y})}\right) \right] \\
&= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left[\left(\sqrt{1+y} - \sqrt{y}\right)^2\right] \right] ,
\end{align}
which closely resembles Claude Leibovici's result (some dilogarithm identities should do the trick). Here $f\left(-\frac{1}{4}\right) = \frac{\pi^2}{10}$ looks rather nice, as does $f\left(-\frac{1}{8}\right) = \frac{\pi^2 - 3 \log^2(2)}{6 \sqrt{2}}$.
