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

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Theorem

$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \frac {b p + a q} {2 a p} \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$


Proof

Let:

\(\displaystyle u\) \(=\) \(\displaystyle \sqrt{a x + b}\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \frac {u^2 - b} a\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sqrt{p x + q}\) \(=\) \(\displaystyle \sqrt{p \left({\frac {u^2 - b} a}\right) + q}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac {p \left({u^2 - b}\right) + a q} a}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac {p u^2 - b p + a q} a}\) $\quad$ $\quad$
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac p a} \sqrt{u^2 - \left({\frac {b p - a q} p}\right)}\) $\quad$ $\quad$


Let $c^2 = \dfrac {b p - a q} p$.


Then:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {2 u} a \left({\frac {u^2 - b} a}\right) \frac {\mathrm d u} {u \sqrt {\frac p a} \sqrt{u^2 - c^2} }\) $\quad$ Primitive of Function of Root of $a x + b$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a \sqrt {a p} } \int \frac {u^2 \ \mathrm d u} {\sqrt{u^2 - c^2} } - \frac {2 b} {a \sqrt {a p} } \int \frac {\mathrm d u} {\sqrt{u^2 - c^2} }\) $\quad$ Linear Combination of Integrals $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a \sqrt {a p} } \left({\frac {u \sqrt{u^2 - c^2} } 2 + \frac {c^2} 2 \ln \left({u + \sqrt {u^2 - c^2} }\right) }\right)\) $\quad$ Primitive of $\dfrac {x^2}{\sqrt{x^2 - a^2} }$ $\quad$
\(\displaystyle \) \(=\) \(\, \displaystyle - \, \) \(\displaystyle \frac {2 b} {a \sqrt {a p} } \ln \left({u + \sqrt {u^2 - c^2} }\right)\) $\quad$ ... and Primitive of $\dfrac 1 {\sqrt{x^2 - a^2} }$: Logarithm Form $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a \sqrt {a p} } \left({\frac {u \sqrt{u^2 - c^2} } 2 + \left({\frac {c^2} 2 - b}\right) \ln \left({u + \sqrt {u^2 - c^2} }\right) }\right)\) $\quad$ common factor $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a \sqrt {a p} } \left({\frac {\sqrt{a x + b} \sqrt {\frac a p} \sqrt{p x + q} } 2 + \left({\frac {c^2} 2 - b}\right) \ln \left({\sqrt{a x + b} + \sqrt {\frac a p} \sqrt{p x + q} }\right)}\right)\) $\quad$ substituting for $u$, noting that $\sqrt{p x + q} = \sqrt {\frac p a} \sqrt{u^2 - c^2}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} + \left({\frac {b p - a q} {a p \sqrt {a p} } - \frac {2 b} {a \sqrt {a p} } }\right) \ln \left({\frac {\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} } {\sqrt p} }\right)\) $\quad$ substituting for $c^2$ and simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} + \left({\frac {b p - a q - 2 b p} {a p \sqrt {a p} } }\right) \ln \left({\frac {\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} } {\sqrt p} }\right)\) $\quad$ common denominator $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \left({\frac {b p + a q} {a p \sqrt {a p} } }\right) \frac {\sqrt {a p} } 2 \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }\) $\quad$ Primitive of Reciprocal of $\sqrt{\left({a x + b}\right) \left({p x + q}\right)}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \left({\frac {b p + a q} {2 a p} }\right) \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }\) $\quad$ simplifying $\quad$

$\blacksquare$


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