How this can be proved?
$$ E = {}_2F_1(-\frac{1}{2}, \frac{1}{3}, \frac{4}{3},-8) = \frac{6}{5} - \frac{\chi}{2} $$
where
$$ \chi = \frac{6\sqrt{\pi}}{5}\frac{\Gamma(\frac{1}{3})}{\Gamma(-\frac{1}{6})} $$
This is not resolved by WolframAlpha and is not in any of the lists in Wolfram Function n Site
I'm not too familiar with cases where the parameters are not integers and with different denominators, and also the $z$ values are not $\pm 1$
My attempts so far has been to search for a transformation to get an easier or solved case:
With the $z \rightarrow \frac{1}{1-z}$ transformation I got: $$ E = \frac{6}{5}\cdot {}_2F_1(-\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) - \chi $$ that I thought was a good progress since now $b=1$.
I also found that
$$ {}_2F_1(-\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) = \frac{5}{2} - \frac{4}{3} \cdot{}_2F_1(\frac{1}{2}, 1, \frac{1}{6},\frac{1}{9}) $$
and I was hoping to use the fact that $$ \big(-\frac{1}{2} \big)_n = \big(-\frac{1}{2} \big) \big(\frac{1}{2} \big) \\ _{n-1} $$$$ \big(-\frac{1}{2} \big)_n = \big(-\frac{1}{2} \big) \big(\frac{1}{2} \big)_{n-1} $$
But I couldn't find a way.