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

Theorem

$\displaystyle \int \frac {\mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3} = \frac {-1} {2 a^2 x^2 \sqrt {a^2 - x^2} } + \frac 3 {2 a^4 \sqrt {a^2 - x^2} } - \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C$

Proof

 $\displaystyle \int \frac {\mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3}$ $=$ $\displaystyle \int \frac {a^2 \ \mathrm d x} {a^2 x^3 \left({\sqrt {a^2 - x^2} }\right)^3}$ $\displaystyle$ $=$ $\displaystyle \int \frac {\left({a^2 - x^2 + x^2}\right) \ \mathrm d x} {a^2 x^3 \left({\sqrt {a^2 - x^2} }\right)^3}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a^2} \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3} + \frac 1 {a^2} \int \frac {x^2 \ \mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac 1 {a^2} \int \frac {\mathrm d x} {x^3 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3}$ simplifying $\displaystyle$ $=$ $\displaystyle \frac 1 {a^2} \left({\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right) + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 - a^2} }\right)^3} + C$ Primitive of $\dfrac 1 {x^3 \sqrt {a^2 - x^2} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a^2} \left({\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right)$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac 1 {a^2} \left({\frac 1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right) + C$ Primitive of $\dfrac 1 {x \left({\sqrt {a^2 - x^2} }\right)^3}$ $\displaystyle$ $=$ $\displaystyle \frac {-1} {2 a^2 x^2 \sqrt {a^2 - x^2} } + \frac 3 {2 a^4 \sqrt {a^2 - x^2} } - \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C$ simplification

$\blacksquare$