Integrand size = 8, antiderivative size = 37 \[ \int \frac {1}{1+a \cos (x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt {1-a} \tan \left (\frac {x}{2}\right )}{\sqrt {1+a}}\right )}{\sqrt {1-a^2}} \]
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Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2738, 211} \[ \int \frac {1}{1+a \cos (x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt {1-a} \tan \left (\frac {x}{2}\right )}{\sqrt {a+1}}\right )}{\sqrt {1-a^2}} \]
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Rule 211
Rule 2738
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{1+a+(1-a) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {2 \arctan \left (\frac {\sqrt {1-a} \tan \left (\frac {x}{2}\right )}{\sqrt {1+a}}\right )}{\sqrt {1-a^2}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {1}{1+a \cos (x)} \, dx=\frac {2 \text {arctanh}\left (\frac {(-1+a) \tan \left (\frac {x}{2}\right )}{\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}} \]
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Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {2 \,\operatorname {arctanh}\left (\frac {\left (a -1\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (1+a \right ) \left (a -1\right )}}\right )}{\sqrt {\left (1+a \right ) \left (a -1\right )}}\) | \(30\) |
risch | \(\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a^{2}+\sqrt {a^{2}-1}-i}{a \sqrt {a^{2}-1}}\right )}{\sqrt {a^{2}-1}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {-i a^{2}+\sqrt {a^{2}-1}+i}{a \sqrt {a^{2}-1}}\right )}{\sqrt {a^{2}-1}}\) | \(87\) |
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Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.00 \[ \int \frac {1}{1+a \cos (x)} \, dx=\left [\frac {\log \left (-\frac {{\left (a^{2} - 2\right )} \cos \left (x\right )^{2} - 2 \, \sqrt {a^{2} - 1} {\left (a + \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \, a^{2} - 2 \, a \cos \left (x\right ) + 1}{a^{2} \cos \left (x\right )^{2} + 2 \, a \cos \left (x\right ) + 1}\right )}{2 \, \sqrt {a^{2} - 1}}, -\frac {\sqrt {-a^{2} + 1} \arctan \left (\frac {\sqrt {-a^{2} + 1} {\left (a + \cos \left (x\right )\right )}}{{\left (a^{2} - 1\right )} \sin \left (x\right )}\right )}{a^{2} - 1}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (29) = 58\).
Time = 1.45 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.97 \[ \int \frac {1}{1+a \cos (x)} \, dx=\begin {cases} \tan {\left (\frac {x}{2} \right )} & \text {for}\: a = 1 \\- \frac {1}{\tan {\left (\frac {x}{2} \right )}} & \text {for}\: a = -1 \\- \frac {\log {\left (- \sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}} + \tan {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}} - \sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}}} + \frac {\log {\left (\sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}} + \tan {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}} - \sqrt {\frac {a}{a - 1} + \frac {1}{a - 1}}} & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int \frac {1}{1+a \cos (x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {1}{1+a \cos (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) - \tan \left (\frac {1}{2} \, x\right )}{\sqrt {-a^{2} + 1}}\right )\right )}}{\sqrt {-a^{2} + 1}} \]
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Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {1}{1+a \cos (x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a-1}}{\sqrt {a+1}}\right )}{\sqrt {a-1}\,\sqrt {a+1}} \]
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