Primitive of Reciprocal of x cubed plus a cubed squared

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\left({x^3 + a^3}\right)^2} = \frac x {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {9 a^5} \ln \left({\frac {\left({x + a}\right)^2} {x^2 - a x + a^2} }\right) + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof
Then from Primitive of $\dfrac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}$:
 * $\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$

Here we have $a = 1, b = 0, m = \dfrac 2 3, p = 1, q = a^3, n = 2$.

So: