Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$ Here is one more conjecture I discovered numerically:
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right)$$
How can we prove it?
Note that $\sqrt[3]{9+4\sqrt5}=\phi^2$.
Mathematica can evaluate this integral, but gives a large expression in terms of Gauss and Appel hypergeometric functions of irrational arguments.

I've found a generalization of your conjecture:

If $k>0$ real number, then
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4\sqrt5k}{9}\,x}\
 \left(1-x^2\right)^{\small2/3}}\stackrel{?}{=}
 \frac{1}{\sqrt[3]{k}}\cdot\frac{3^{\small13/6}}{2^{\small4/3}5^{\small5/6}\pi}\Gamma^{3}\left(\tfrac{1}{3}\right).$$

For $k=9$ with $+$ sign it gives back your integral.
Another family of this type:

If $k>0$ real number, then
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4k}{5}\,x}\
 \left(1-x^2\right)^{\small2/3}}\stackrel{?}{=} \frac{1}{\sqrt[3]{k}}\cdot\frac{\sqrt[3]{5}}{\sqrt[3]{2}\sqrt{3}\pi}\Gamma^3\left(\tfrac{1}{3}\right).
$$

