How to integrate $\int\frac{1}{\sqrt{1+x^3}}\mathrm dx$?

The first thing to do is to note that

$$x^3+1=(x+1)(x^2-x+1)$$

(one real and two complex conjugate roots). Using Jacobian elliptic functions requires having a quartic within the square root (the alternative of using Weierstrass elliptic functions is fine with square roots of cubics, but I'll leave that approach to someone else); the good thing is that by choosing a proper Möbius transformation, one can turn a cubic into a quartic (the algebraic geometers here might want to say a bit more than I have).

For the integral in question, the Möbius substitution needed is $x=\frac{-1+\sqrt{(-1)^2-(-1)+1}+(-1-\sqrt{(-1)^2-(-1)+1})v}{1+v}=\frac{2\sqrt{3}}{1+v}-(1+\sqrt{3})$; we then have

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=-2\int\frac{\mathrm dv}{\sqrt{(1-v^2)(2\sqrt{3}-3+(2\sqrt{3}+3)v^2)}}$$

At this point, making use of the Jacobian elliptic function identity $\mathrm{sn}^2(u|m)+\mathrm{cn}^2(u|m)=1$ (nothing more than the usual Pythagorean identity in elliptic function garb), we could make either of the substitutions $v=\mathrm{sn}(u|m)$ or $v=\mathrm{cn}(u|m)$. The latter is a bit more convenient, since $\mathrm dv=-\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du$, which can conveniently get rid of the negative sign in the integral. Thus, the integral turns into

$$2\int\frac{\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du}{\sqrt{(1-\mathrm{cn}^2(u|m))(2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m))}}$$

or (by using the Pythagorean identity)

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}$$

Here, one now chooses a proper value of $m$ such that the integrand reduces to a constant. Skipping the details, we let $m=\frac{2+\sqrt{3}}{4}$ such that

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}=\int\frac{\mathrm du}{\sqrt[4]{3}}$$

To undo the substitutions, we note that $u=F(\arccos(v)|m)$ and $v=\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1$, giving the final result

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$

This result can be verified by differentiating the right hand side (remember that $\frac{\mathrm d}{\mathrm d\phi}F(\phi|m)=\frac1{\sqrt{1-m\sin^2\phi}}$) and noting that it is the same as the integrand.

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 $$ \mathcal{M}^{-1}_{s\to a}\left[ \frac{x \left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s)}{\sqrt{\pi } \left(3 s-\frac{1}{2}\right) \sqrt{x^3}}\right] = \frac{x}{\sqrt{a}} \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-\frac{x^3}{a}\right) $$