<s>[INST] Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First obvious sub is $t = 1 + x^{3}$:
$$
\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t
$$
From here I tried many things like $\frac{1}{t}$, $t-1$, and more.  The trickiest part is the bounds! Reversing it from the answer the integral should be like
$$
\frac{1}{6}\int_{0}^{1}
x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t
$$
I'm not sure where the $1/2$ comes from and the $0$ to $1$ bounds.  Any idea or tip please ?.
 [/INST] An elementary solution: Consider the substitution
$$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$
$t$ is monotonic decreasing on $0<x<2$, and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$
this can be verified by explictly computing $(dt/dx)^2$ and compare it to $9(1+t^3)/(1+x^3)$. When $x=2, t=-1$, so
$$\int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}dx}  = \frac{1}{3}\int_{ - 1}^\infty  {\frac{1}{{\sqrt {1 + {t^3}} }}dt} $$
I believe you now have no difficulty to solve last integral via Beta function.

A conceptual solution: Consider the elliptic curve $E:y^2=x^3+1$, $P=(2,3),Q=(0,1)$ on $E$, $\omega = dx/y$ is the invariant differential on $E$. For the multiplication-by-$3$ isogeny $\phi:E\to E$, we have $3P=(-1,0), 3Q=O$. So $3\int_0^2 \omega \cong \int_{-1}^\infty \omega$ up to an element of $H_1(E,\mathbb{Z})$.
$t$ given above is the $x$-component of $\phi$ and $(1)$ is equivalent to $\phi^\ast \omega = 3\omega$.
The $P$ above is $6$-torsion, if we consider $4$ or $5$-torsion instead, we obtain
results like
$$\int_0^\alpha  {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$
$$\int_0^\alpha  {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$
</s>