Primitive of Reciprocal of p plus q by Cosine of a x

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {p + q \cos a x} = \begin{cases}

\displaystyle \frac 2 {a \sqrt {p^2 - q^2} } \arctan \left({\sqrt {\frac {p - q} {p + q} } \tan \dfrac {a x} 2}\right) + C & : p^2 > q^2 \\ \displaystyle \frac 1 {a \sqrt {q^2 - p^2} } \ln \left\vert{\frac {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } }\right\vert + C & : p^2 < q^2 \\ \end{cases}$

Also see

 * For $p = q$: Primitive of $\dfrac 1 {1 + \cos a x}$


 * For $p = -q$: Primitive of $\dfrac 1 {1 - \cos a x}$


 * Primitive of $\dfrac 1 {p + q \sin a x}$


 * Primitive of $\dfrac 1 {p + q \tan a x}$


 * Primitive of $\dfrac 1 {p + q \cot a x}$


 * Primitive of $\dfrac 1 {q + p \sec a x}$


 * Primitive of $\dfrac 1 {q + p \csc a x}$