TITLE: How to integrate $\int\frac{1}{\sqrt{1+x^3}}\mathrm dx$?
QUESTION [24 upvotes]: In a course, my teacher told us that the following integral is convergent and used the comparison test to prove it; my question is how to find the antiderivative in closed form? It seems to exist; if, however, it doesn't exist, can someone prove it?  

$$\int\sqrt{\dfrac1{1+x^3}}\mathrm dx$$

REPLY [4 votes]: For positive values of $x$, the integral is solved by
$$
\int\sqrt{\dfrac1{1+x^3}}\mathrm \;dx = x \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-x^3\right)
$$
which contains a hypergeometric function. In general for positive $a$
$$
\int\sqrt{\dfrac1{a+x^3}}\mathrm \;dx = \frac{x}{\sqrt{a}} \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-\frac{x^3}{a}\right)
$$
To get this use a Mellin transform under the integral by introducing a virtual parameter (my pet adaptation of the Feynman trick):
$$
\mathcal{M}_{a \to s}\left[\sqrt{\dfrac1{a+x^3}}\right] =  \frac{\left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right)
    \Gamma (s)}{\sqrt{\pi } \sqrt{x^3}}
$$
$$
\int\mathcal{M}_{a \to s}\left[\sqrt{\dfrac1{a+x^3}}\right] \mathrm\; dx =  \frac{x \left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right)
    \Gamma (s)}{\sqrt{\pi } \left(3 s-\frac{1}{2}\right) \sqrt{x^3}}
$$
Then take the inverse Mellin transform, and simplify
$$
\mathcal{M}^{-1}_{s\to a}\left[ \frac{x \left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right)
    \Gamma (s)}{\sqrt{\pi } \left(3 s-\frac{1}{2}\right) \sqrt{x^3}}\right]  = \frac{x}{\sqrt{a}} \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-\frac{x^3}{a}\right)
$$