Primitive of Reciprocal of x cubed plus a cubed squared

Theorem

 * $\ds \int \frac {\d x} {\paren {x^3 + a^3}^2} = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof
Then from Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:
 * $\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$

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

So: