How to integrate $\int\frac{1}{\sqrt{x^{3}+2}}dx$?

Question

How to integrate $$\int \frac{1}{\sqrt{x^{3}+2}}dx$$? I am trying this question by substituting $x=(2\tan^{2}\theta)^{\frac{1}{3}}$. Therefore, $dx=\frac{(16)^{\frac{1}{3}}}{3}(\cot\theta)^{\frac{1}{3}}(\sec^{2}\theta)d\theta$. Therefore, we can write the above integral as $$\frac{(2)^{\frac{5}{6}}}{3}\int (\sec\theta)(\cot\theta)^{\frac{1}{3}}d\theta$$. Therefore, after this step I can write the above integral as $$\frac{(2)^{\frac{5}{6}}}{3}\int \frac{1}{[(\sin\theta)(\cos^{2}\theta)]^{\frac{1}{3}}}d\theta$$. But after this step, I can't proceed further. So, please help me out with this integral.

Answer 1

A cubic polynomial inside a radical usually means that an elliptic integral function is required---and in particular that an antiderivative cannot be expressed in closed form in terms of elementary functions---and that's the case here.

In general, $$\int \frac{dx}{\sqrt{x^3 + a^6}} = -\frac{2}{3^{1/4} \zeta a} F \left(\frac{1}{3^{1/4}} \zeta \cdot \frac{\sqrt{x + a^2}}{a}; \zeta^{-2}\right) + C ,$$ where $\zeta := \exp \frac{\pi i}{12}$ and $F(u; k)$ is the incomplete elliptic integral of the first kind. Taking $a = 2^{1/6}$ gives $\int \frac{dx}{\sqrt{x^3 + 2}}$.