What is the integral of $\int\frac{dx}{\sqrt{x^3+a^3}}$? 
What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$

I came across this integration in a physics problem. I suspect role of complex numbers here.
'$a$' is a constant

As the other users said, this integral is unlikely to have an elementary form, however, it can be expressed in terms of the well known Gauss hypergeometric function, which can be easily evaluated by most CAS or even Wolfram Alpha.

First, let's consider the case $|x|<|a|$, then we can substitute:
$$x=at, \qquad |t|<1$$
$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{dt}{\sqrt{1+t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{3k} dt=$$
$$=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{3k+1}}{3k+1}=a^{-1/2} \Gamma \left( \frac{1}{2} \right) \sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k+1}}{3k+1}=$$
$$=\frac{1}{3}\sqrt{\frac{\pi}{a}}~t~\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}$$
To find the hypergeometric form of the series above, we consider the ratio of the successive terms:
$$\frac{c_{k+1}}{c_k}=\frac{\left(-1/2-k \right)(k+1/3)}{ (k+4/3)} \frac{t^3}{k+1}=\frac{\left(k+1/2 \right)(k+1/3)}{ (k+4/3)} \frac{-t^3}{k+1}$$
$$c_0=\frac{3}{\sqrt{\pi}}$$
Which, by definition makes the series:
$$\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}=\frac{3}{\sqrt{\pi}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)$$
Which makes the integral:

$$\int\frac{dx}{\sqrt{x^3+a^3}}=\frac{t}{\sqrt{a}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)=\frac{x}{\sqrt{a^3}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -\frac{x^3}{a^3} \right)$$

This is a correct answer for $|x|<|a|$, as can be checked by numerical experiments.
Mathematica, or other advanced software, can directly evaluate and plot hypergeometric function, which makes this form more useful than the original integral.

For $|x|>|a|$ we can use the same method of binomial expansion to get the hypergeometric form.
$$x=at, \qquad |t|>1$$
$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{t^{-3/2} dt}{\sqrt{1+1/t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{-3k-3/2} dt=$$
$$=-a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{-3k-1/2}}{3k+1/2}$$
It's straightforward to continue in the same way and obtain another hypergeometric function.
