<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] 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.
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