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 ?.
 A: A hypergeometric solution: Modulo Beta function $I_0=\int_0^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx=\frac{2 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{7}{6}\right)}{\sqrt{\pi }}$ one may evaluate $I_1=\int_2^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx$ instead. Substitute $x\to\frac 1x$ and binomial expansion gives
$$I_1=\sqrt{2} \, _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{1}{8}\right)=\frac{2 \sqrt{\frac{\pi }{3}} \Gamma \left(\frac{7}{6}\right)}{\Gamma \left(\frac{2}{3}\right)}$$
Where the last step has invoked the following formula $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{4}{3}-2 a\right)}$$
Computing $I_0-I_1$ gives the desired result.

Update: Hypergeometric method can also establish @pisco's result (the case of $4$-torsion)

$$\int_0^\alpha  {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732$$

Since by binomial expansion again, it equals $$\left(\sqrt{3}-1\right) {_2F_1}\left(\frac{1}{3},\frac{1}{2},\frac{4}{3},10-6 \sqrt{3}\right)=\frac{\sqrt{\frac{1}{2} \left(6 \sqrt{3}-9\right) \pi } \Gamma \left(\frac{1}{3}\right)}{3\ 3^{3/4} \left(\sqrt{3}-1\right) \Gamma \left(\frac{5}{6}\right)}$$ due to certain transformation of hypergeometric series (see Special values of hypergeometric series by Akihito Ebisu). The rest are trivial.
A: 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
$$\int_0^\alpha  {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$
$$\int_0^\alpha  {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{2}{\dd x \over \root{1 + x^{3}}}}  =
2\int_{0}^{1}\pars{1 + 8x^{3}}^{-1/2}\,\dd x
\\[5mm] = &
{2 \over 3}\int_{0}^{1}x^{-2/3}\,\bracks{1 + 8x}^{-1/2}\,\dd x
\\[5mm] & =
{2 \over 3}\int_{0}^{1}x^{\color{red}{1/3}\ -\ 1}\,
\pars{1 - x}^{\color{#0f0}{4/3}\ -\ \color{red}{1/3}\ -\ 1}
\,\bracks{1 -\pars{\bf\color{red}{-8}}x}^{\,-{\bf 1/2}}\,\dd x
\\[5mm] & =
{2 \over 3}\,\mrm{B}\pars{\color{red}{1 \over 3},
\color{#0f0}{4 \over 3} - \color{red}{1 \over 3}}
\mbox{}_{2}\mrm{F}_{1}\pars{\bf{1 \over 2},
\color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8}
\label{1}\tag{1}
\\[5mm] & =
{2 \over 3}\,{\Gamma\pars{1/3}\Gamma\pars{1} \over \Gamma\pars{4/3}}
\,\mbox{}_{2}\mrm{F}_{1}\pars{\bf{1 \over 2},
\color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8}
\\[5mm] & =
\bbx{\large {\Large 2}\,\,\mbox{}_{2}\mrm{F}_{1}\!\!\!\pars{\bf{1 \over 2},
\color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8}}
\approx 1.4022 \\ &
\end{align}
Step (\ref{1}): See the
Euler Type Hypergeometric Function.
A: Consider
$$
y^2=4x^3+4,g_2=0,g_3=-4
$$
The real period of $\wp(z)$ is
$$
\omega_1=\int_{-1}^{\infty} \frac{1}{\sqrt{1+x^3}} \text{d}x
=\frac{\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{2\sqrt{\pi} }
$$
And
$$
\wp\left ( \frac{\omega_1}{2}  \right ) =-1
$$

We have the addition formula:
$$
\begin{aligned}
&\wp(3z) = \frac{1}{4}\left [  \frac{\wp'(z)-\wp'(2z)}{\wp(z)-\wp(2z)} \right ]^2
-\wp(z)-\wp(2z)\\
&\wp(2z)=\frac{1}{4} \left [ \frac{\wp''(z)}{\wp'(z)}  \right ] ^2-2\wp(z)\\
&\wp(u+v)
=\frac{1}{4} \left [ \frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)}  \right ]^2
-\wp(u) -\wp(v)
\end{aligned}
$$
where $3z,2z,z$ is not at the period lattices.
Let $x=\wp\left ( \frac{\omega_1}{6}  \right ) ,y=\wp\left ( \frac{\omega_1}{3}  \right )$.And they have two relations:
$$
\begin{aligned}
&\frac{1}{4} \left [ \frac{\sqrt{4x^3+4}-\sqrt{4y^3+4} }{x-y}  \right ]^2-x-y  = -1\\
&y=\frac{9x^4}{4x^3+4} -2x
\end{aligned}
$$
Solve it and gives,
$$
\wp\left ( \frac{\omega_1}{6}  \right )=2,
\wp\left ( \frac{\omega_1}{3}  \right )=0
$$
So,
$$
\begin{aligned}
I:&= \int_{0}^{2} \frac{1}{\sqrt{1+x^3} } \text{d}x\\
&=2\left(\frac{\omega_1}{3} -\frac{\omega_1}{6} \right)\\
&=\frac{\omega_1}{3} 
\end{aligned}
$$

We have $\wp\left ( \frac{\omega_1}{4}  \right ) =\sqrt{3}-1$,hence
$$
\int_{0}^{\sqrt{3} -1} \frac{1}{\sqrt{1+x^3} }\text{d}x
=\frac{\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{12\sqrt{\pi} }
$$

We have
$t=\wp\left ( \frac{\omega_1}{8}  \right ) =
-1+\sqrt{3} +\sqrt{3}\sqrt{2-\sqrt{3} }  
+\sqrt{3} \sqrt{2-\sqrt{3}+2\sqrt{2-\sqrt{3} }  }$,hence
$$
\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{5\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{24\sqrt{\pi} }
$$

We have
$t=\wp\left ( \frac{\omega_1}{12}  \right ) 
=2+\sqrt{3}+\sqrt{3} \sqrt{3+2\sqrt{3} }$,hence
$$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{4\sqrt{\pi} }$$

We have
$
t=\wp\left ( \frac{\omega_1}{9}  \right ) 
=2^{1/3}
\left ( 16+3\cdot2^{2/3}\sqrt[3]{37+i\sqrt{3} }+
6\cdot2^{-1/3}\sqrt[3]{37-i\sqrt{3} }
 \right )^{1/3}\approx4.5707391614928...
$,hence
$$
\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{2\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{9\sqrt{\pi} }
$$

If $t\ge0$,and
$$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=R\cdot\frac{\Gamma\left ( \frac{1}{3}  \right )\Gamma\left ( \frac{1}{6}  \right ) }{\sqrt{\pi} }$$
where $R$ is rational with $0\le R\le\frac{1}{3}$.$t$ must be an algebraic number.
