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

Theorem

$\displaystyle \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Proof

$\displaystyle \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$

Putting $n = \dfrac 1 2$:

 $\displaystyle \int \sqrt {\frac {p x + q} {a x + b} } \rd x$ $=$ $\displaystyle \frac {2 \paren {p x + q}^{1/2} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 1} a} + \frac {2 \cdot \frac 1 2 \paren {a q - b p} } {\paren {2 \cdot \frac 1 2 + 1} a} \int \frac {\paren {p x + q}^{1/2 - 1} } {\sqrt {a x + b} } \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

$\blacksquare$