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

Theorem

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

Proof

Let:

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

Then:

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

Let $c^2 > 0$.

Then:

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

$\Box$

Let $c^2 < 0$.

Then:

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

$\blacksquare$

Sources

(in which a mistake apppears)