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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right) \sqrt{a x + b} } = \begin{cases}

\dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \left\vert{\dfrac {\sqrt {p \left({a x + b}\right)} - \sqrt {b p - a q} } {\sqrt {p \left({a x + b}\right)} + \sqrt {b p - a q} } }\right\vert & : b p - a q > 0 \\ \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \left({a x + b}\right)} {a q - b p} } & : b p - a q < 0 \\ \end{cases}$

Proof
Let:

Then:

Now let $z = \sqrt p u$

Thus:

Let $b p - a q > 0$.

Let $d = \sqrt {b p - a q}$.

Then:

Let $b p - a q < 0$.

Let $d = \sqrt {-\left({b p - a q}\right)} = \sqrt {a q - b p}$.

Then: