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

A: As suggested by Chen Wang, this integral is related to an integral of the form
$$
\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }
$$
which appears in this forum's
Question 879089
by the same author.
Here's a more direct route than given by Kirill, without
conversions between definite integrals and hypergeometric functions.
A differential $dx \left/ \left(\sqrt[3]{A+Bx} \, (1-x^2)^{2/3}\right)\right.$
has four poles of fractional order, two at $x = \pm 1$ of order $2/3$,
and two at $x = \infty$ and $x = -B/A$ of order $1/3$.
Hence the differential is invariant under an involution
in the form of a fractional linear transformation that switches
$-1 \leftrightarrow +1$ and $\infty \leftrightarrow -B/A$.
To exploit this symmetry it is convenient to use a coordinate
$u = (cx+1)/(x+c)$: for any $c$ we have $u=\pm 1$ at $x = \pm 1$
and $u = c$ at $x = \infty$, and we choose $c>1$ so that $u=-c$ at $x=-B/A$.
Explicitly, this change of variable gives
$$
{\large\int}_{-1}^1
  \frac{dx}{\sqrt[3]{c^2+1+2cx}\ \left(1-x^2\right)^{\small2/3}}
=
{\large\int}_{-1}^1
  \frac{du}{\sqrt[3]{c^2-u^2}\ \left(1-u^2\right)^{\small2/3}} .
$$
By symmetry the second $\int_{-1}^1$ is $2\int_0^1$ of the same integrand,
and then the change of variable $u^2 = t$ gives
$$
{\large\int}_0^1 \frac{dt}{\sqrt{t} \, (c^2-t)^{\small1/3} (1-t)^{\small2/3}}.
$$
Now integrating with respect to $X=1-t$ instead of $t$ gives
$$
{\large\int}_0^1 \frac{dX}{\sqrt{1-X} \; X^{2/3}
  \left(c^2-1+X\right)^{\small1/3}},
$$
and we have reached an integral of the desired form.
In the present case we take $c = \frac12 \! \sqrt 5$.
Then $c^2+1 + 2cx = (9 + 4\sqrt{5}x)/4$, so our integral is
$4^{1/3}$ times the one with $\sqrt[3]{9 + 4\sqrt{5}x}$ in the denominator.
Hence $c^2-1 = 1/4$, so we gain another factor of $4^{1/3}$ and we've reached 
the integral in the first display with $z=-4$.  Vladimir Reshetnikov
already noted in a comment to his question that the integral seemed to
have an exact value at $z = -4$, and the analysis I gave in my answer there
(or Kirill's answer here) leads to the period of a CM elliptic curve
with $j$-invariant $146329141248 \sqrt{5} - 327201914880$; this curve is
related by $5$-isogeny with a curve of $j$-invariant zero, and thus
has a period proportional to a Beta integral.
