Primitive of x squared by Arcsecant of x over a

Theorem

 * $\displaystyle \int x^2 \operatorname{arcsec} \frac x a \ \mathrm d x = \begin{cases}

\displaystyle \frac {x^3} 3 \operatorname{arcsec} \frac x a - \frac {a x \sqrt{x^2 - a^2} } 6 - \frac {a^3} 6 \ln \left({x + \sqrt {x^2 - a^2} }\right) + C & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \displaystyle \frac {x^3} 3 \operatorname{arcsec} \frac x a + \frac {a x \sqrt{x^2 - a^2} } 6 + \frac {a^3} 6 \ln \left({x + \sqrt {x^2 - a^2} }\right) + C & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}$

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

First let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$.

Then:

Similarly, let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

Then:

Also see

 * Primitive of $x^2 \arcsin \dfrac x a$


 * Primitive of $x^2 \arccos \dfrac x a$


 * Primitive of $x^2 \arctan \dfrac x a$


 * Primitive of $x^2 \operatorname{arccot} \dfrac x a$


 * Primitive of $x^2 \operatorname{arccsc} \dfrac x a$