Primitive of Power of a x + b over Power of p x + q/Formulation 2

Theorem

 * $\displaystyle \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \ \mathrm d x = \frac {-1} {\left({n - m - 1}\right) p} \left({\frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^{n-1} } + m \left({b p - a q}\right) \int \frac {\left({a x + b}\right)^{m-1}} { \left({p x + q}\right)^n} \ \mathrm d x}\right)$

Proof
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:


 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^m \left({p x + q}\right)^{n+1}} {\left({m + n + 1}\right) a} + \frac {m \left({b p - a q}\right)} {\left({m + n + 1}\right) p} \int \left({a x + b}\right)^{m-1} \left({p x + q}\right)^n \ \mathrm d x$

Setting $n := -n$: