Prove ${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$ I discovered the following conjectured identity numerically (it holds with at least $1000$ digits of precision). How can I prove it?
$${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$$

Update: It looks like this hypergeometric function assumes algebraic values at algebraic points (it's only a guess because I have only approximations to those algebraic numbers). Looking at those values, I was able to further conjecture that the hypergeometric function for $x<0$ is actually the following elementary function:
$$
{_2F_1}\left(\frac16,\frac12;\frac13;x\right)\stackrel{\color{#808080}?}=
\\
\frac1{\sqrt[4]2\sqrt3}\cdot\sqrt{\frac{\alpha}{1-x}+\frac{1}{\alpha}
\sqrt{\frac{4\left(\alpha\sqrt{2}+2\right)+x\left(\sqrt[3]{4\beta}-2\left(\alpha\sqrt{2}+4\right)\right)+2\sqrt[3]{2\beta^2}}{1-x}}}~,
$$
where
$$\alpha=\sqrt{2-2x+\sqrt[3]{2\beta^2}}~,\qquad\beta=x(x-1)~.$$
 A: Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector
$$\vec{y}(z)=\left(\begin{array}{c}
y_1 \\ y_2
\end{array}\right)=\left(\begin{array}{c}
_2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z)
\end{array}\right).\tag{1}$$
This is a single-valued vector function on $\mathbb{C}\backslash\{(-\infty,0]\cup[1,\infty)\}$. Its analytic continuation along a closed loop $\gamma$ gives rise to monodromy representation of $\pi_1(\mathbb{C}\backslash\{0,1\})$:
$$ \gamma\mapsto M_{[\gamma]},\qquad y(\gamma z)=M_{[\gamma]}y(z).$$
The monodromy group $G\subset GL(2,\mathbb{C})$ of the hypergeometric equation is generated by two matrices corresponding to simple loops around $0$ and $1$. In the case we are interested in these matrices are explicitly given by
$$M_0=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{-2\pi i /3}\end{array}\right),\qquad
M_1=C\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{2\pi i /3}\end{array}\right)C^{-1},\tag{2}$$
where the connection matrix $C=\left(\begin{array}{cc} 1 & 2^{\frac43} \\ -2^{\frac83} & 8\end{array}\right)$. If $G$ is finite, then $\vec{y}(z)$ has a finite number of branches, and moreover (Schwarz 1872), is algebraic. 
It is not difficult to check that the monodromy group $G$ generated by $M_0$, $M_1$ from (2) is indeed finite. In particular, note that
$$M_0^3=M_1^3=I,\qquad M_1^2=-M_0M_1M_0, $$ $$M_1M_0M_1=-M_0^2,\qquad M_1M_0^2M_1=M_0^2M_1M_0^2.$$ 
It turns out that $G$ has order $24$ and is isomorphic to the binary tetrahedral group:
$$G\cong 2T=\langle s,t\,|\,(st)^2=s^3=t^3\rangle, $$
where the generators can be identified as $s=M_0M_1M_0M_1M_0$, $t=M_1M_0M_1M_0M_1$.
Corollary: The hypergeometric functions in (1) are algebraic.

Algebraic solutions of the hypergeometric equations are classified by the so-called Schwarz table, and have been studied by many mathematicians, see e.g. the bibliography in this paper by R.Vidunas. Their explicit construction is somewhat involved but relatively straightforward - at least when the corresponding algebraic curve has genus $0$ (the genus can be determined independently from the Riemann-Hurwitz formula). 
In our case the task simplifies even more as our parameter values can be obtained from the genus $0$ tetrahedral formula (2.4) of the above mentioned paper by a combination of a linear trasformation (sending $\frac56$ to $\frac43-\frac56=\frac12$) and differentiation (transforming $\frac43$ into $\frac13$). The result is
$$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}.$$
Corollary: The antiderivative $\displaystyle\int \mathcal{R}\left(x,y(x)\right)dx$, where $y(x)={}_2F_1\left(\frac16,\frac12;\frac13;-x\right)$ and $\mathcal{R}(x,y)$ is rational in both arguments, can be expressed in terms of elementary functions.

Example:
The transformation $r\mapsto x(r)=\frac{r(r+2)^3}{(r+1)(1-r)^3}$ bijectively maps $(0,1)$ to $(0,\infty)$, and therefore the initial integral becomes
\begin{align}\mathcal{I}&=\int_0^1 \left(\frac{\sqrt{1-r^2}}{2r+1}\right)^{12}\left(\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)'dr=\\&=
2\int_0^1 \frac{(1+r)^4(1-r)^2(r+2)^2}{(2r+1)^{10}} dr=\\&=\frac{80\,663}{153\,090}.
\end{align}
