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

Theorem

Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.

Then:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}

\dfrac 1 {\sqrt {p \paren {b p - a q} } } \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} } } + C & : p \paren {b p - a q} > 0 \\ \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C & : p \paren {b p - a q} < 0 \\ \end {cases}$

Proof

Case $1$: $p \paren {b p - a q} > 0$

Let $p \paren {b p - a q} > 0$.

Then:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$

$\Box$

Case $2$: $p \paren {b p - a q} < 0$

Let $p \paren {b p - a q} > 0$.

Then:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C$

$\blacksquare$

Sources

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions
• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.114$