Integral involving hypergeometric function $\int_0^1[{}_2F_1(\frac13,\frac23;1;x^3)]^2dx$ 
Question: How to prove $$I=\int_0^1\bigg[{}_2F_1\left(\frac13,\frac23;1;x^3\right)\bigg]^2dx=\frac{\sqrt3}{32\pi^5}\Gamma\left(\frac13\right)^9?$$

Source: An integral competition post of my country. 
Attempt
Recall the series definition of hypergeometric function $$_2F_1(a,b,c,x)=\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_nn!}x^n,$$
we can transform $I$ into the series form
$$I=\sum_{n,m=0}^\infty\frac{(a)_n(a)_m(b)_n(b)_m}{(c)_n(c)_mn!m!(3n+3m+1)}$$
But I can not handle this series.
I also thought of using complex method. $$I=\int_0^1\bigg[{}_2F_1\left(\frac13,\frac23;1;x\right)\bigg]^2\frac{x^{-2/3}}3dx$$ then let $f(z)=\bigg[{}_2F_1\left(\frac13,\frac23;1;z\right)\bigg]^2(-z)^{-2/3}$ and use keyhole contour, where $(\cdot)^{-2/3}$ is the principal branch of the multi-valued function. But the nature of the branch of the integrand in the inteval $[1,\infty)$ is too complex for me to handle. It involves another definite integral which is similar to $I$.
 A: Not an answer, but an extended comment for now.
This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8.
The formulas 15.8.25 and 15.8.26 both apply here.
However, the most interesting one is so called Ramanujan’s Cubic Transformation (15.8.33):
$${_2 F_1} \left( \frac13, \frac23;1;x^3 \right)= \frac{1}{1+2 x} {_2 F_1} \left( \frac13, \frac23;1;1- \frac{(1-x)^3}{(1+2x)^3}\right)$$
Update:
The iteration:
$$x_{n+1}=\left( 1- \frac{(1-x_n)^3}{(1+2x_n)^3}\right)^{1/3}$$
Converges to $x_{\infty}=1$ for any $x \in (0,1]$. Not sure how to use this, because ${_2 F_1} \left( \frac13, \frac23;1;1 \right)= \infty$.

This transformation is related to the cubic analogue of the arithmetic-geometric mean. See the references at DLMF and also these questions:
Integral identity related with cubic analogue of arithmetic-geometric mean
Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$
Some formulas from the question above (and Nemo's answer) might be useful here, for example:
$$\int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right)= \\ =\frac{2 \pi}{3 \sqrt{3}p^{1/3}}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right)$$
This is just an application of already linked transformations, and can be applied backwards in this case.
