Primitive of Reciprocal of Power of p x + q by Root of a x + b

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Theorem

$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$


Proof

From Primitive of Reciprocal of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:

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


Setting $m := \dfrac 1 2$:

\(\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} }\) \(=\) \(\ds \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{1/2 - 1} \paren {p x + q}^{n - 1} } + a \paren {\frac 1 2 + n - 2} \int \frac {\d x} {\paren {a x + b}^{1/2} \paren {p x + q}^{n - 1} } }\)
\(\ds \) \(=\) \(\ds \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }\)

$\blacksquare$


Sources