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

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Theorem

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


Proof

Let:

\(\displaystyle u\) \(=\) \(\displaystyle \sqrt {a x + b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \frac {u^2 - b} a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt {p x + q}\) \(=\) \(\displaystyle \sqrt {p \paren {\frac {u^2 - b} a} + q}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {p \paren {u^2 - b} + a q} a}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {p u^2 - b p + a q} a}\)
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} }\)


Then:

\(\displaystyle \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\displaystyle \int \frac {2 u \rd u} {a \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} } u}\) Primitive of Function of Root of $a x + b$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } }\) Primitive of Constant Multiple of Function
Setting $c^2 := \dfrac {b p - a q} p$:
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \int \frac {\rd u} {\sqrt{u^2 - c^2} }\)


Let $c^2 > 0$.

Then:

\(\displaystyle \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt{u^2 - c^2} }\) from $(2)$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {u + \sqrt {u^2 - c^2} } + C\) Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: Logarithm Form
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C\) substituting for $u$ and $c$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {\frac a p} \sqrt {p x + q} } + C\) substituting from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {\frac {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } {\sqrt p} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } - \ln {\sqrt p} + C\) Difference of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C\) subsuming $-\ln {\sqrt p}$ into the arbitrary constant

$\Box$


Let $c^2 < 0$.

Then:

\(\displaystyle \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 + c^2} }\) from $(2)$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \sinh^{-1} {\frac u c} + C\) Primitive of Reciprocal of $\sqrt {x^2 + a^2}$: $\sinh^{-1}$ form
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \sinh^{-1} \frac {\sqrt {a x + b} } {\sqrt {\frac {b p - a q} p} } + C\) substituting for $u$ and $c$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\frac {p \paren {a x + b} } {b p - a q} } + C\) simplifying

$\blacksquare$


Sources

(in which a mistake apppears)