<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] 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.
</s>