Primitive of Reciprocal of x cubed by Root of x squared minus a squared cubed

Theorem

 * $\displaystyle \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C$

for $\size x > a$.

Also see

 * Primitive of $\dfrac 1 {x^3 \paren {\sqrt {x^2 + a^2} }^3}$
 * Primitive of $\dfrac 1 {x^3 \paren {\sqrt {a^2 - x^2} }^3}$