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

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Theorem

$\displaystyle \int \frac {\sqrt{a x + b} } {p x + q} \rd x = \begin{cases} \dfrac {2 \sqrt{a x + b} } p + \dfrac {\sqrt {b p - a q} } {p \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac {2 \sqrt{a x + b} } p - \dfrac {\sqrt {a q - b p} } {p \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end{cases}$


Proof

From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 2:

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


Setting $m := \dfrac 1 2$ and $n = 1$:

\(\displaystyle \int \frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^n} \rd x\) \(=\) \(\displaystyle \frac {-1} {\paren {1 - \frac 1 2 - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^0} + \frac 1 2 \paren {b p - a q} \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q} } \rd x}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \sqrt {a x + b} } p + \frac {b p - a q} p \int \frac {\d x} {\sqrt {a x + b} \paren {p x + q} }\)


From Primitive of Reciprocal of $p x + q$ by Root of $a x + b$:

$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} } = \begin{cases} \dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end{cases}$


The result follows by substitution.

$\blacksquare$


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