TITLE: Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$
QUESTION [11 upvotes]: 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!

REPLY [6 votes]: Here's the shortest closed-form:

$$\,_3F_2\left(\frac12,\frac12,\frac12;\frac32,\frac32;x\right) = -\,\frac{{\rm Li}_2(1-\beta^2)}{2\sqrt{-x}}-\frac{\ln^2 \beta}{2\sqrt{-x}}$$

where,
$$\beta = \sqrt{1-x}+\sqrt{-x}$$
Note: This was derived using WA's solution and the 4th identity here.