Primitive of Power of p x + q over Root of a x + b

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Theorem

$\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$


Proof

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

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


Setting $m := -\dfrac 1 2$:

\(\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x\) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{-1/2 + 1} \left({p x + q}\right)^n} {\left({-\frac 1 2 + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({-\frac 1 2 + n + 1}\right) a} \int \left({a x + b}\right)^{-1/2} \left({p x + q}\right)^{n-1} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x\)

$\blacksquare$


Sources