Primitive of Power of a x + b over Power of p x + q/Formulation 1

Theorem

 * $\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac {\paren {a x + b}^{m + 1} } {\paren {p x + q}^{n - 1} } + \paren {n - m - 2} a \int \frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } \rd x}$

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


 * $\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - \paren {m + n + 2} a \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$

Setting $n := -n$: