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

Theorem

 * $\displaystyle \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C$

for $\size x \ge a$.

Proof
Let:

Also see

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