Integrand size = 24, antiderivative size = 33 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\log \left (a+b \tan \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{2 b e} \]
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Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3201, 31} \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\log \left (a+b \tan \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )\right )}{2 b e} \]
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Rule 31
Rule 3201
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{2 a+2 b x} \, dx,x,\tan \left (\frac {\pi }{4}+\frac {1}{2} (d+e x)\right )\right )}{e} \\ & = \frac {\log \left (a+b \tan \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{2 b e} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(96\) vs. \(2(33)=66\).
Time = 0.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.91 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=-\frac {\log \left (\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )\right )}{2 b e}+\frac {\log \left (a \cos \left (\frac {1}{2} (d+e x)\right )+b \cos \left (\frac {1}{2} (d+e x)\right )-a \sin \left (\frac {1}{2} (d+e x)\right )+b \sin \left (\frac {1}{2} (d+e x)\right )\right )}{2 b e} \]
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Time = 0.88 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42
method | result | size |
parallelrisch | \(\frac {-\ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )+\ln \left (-a -b +\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 b e}\) | \(47\) |
derivativedivides | \(\frac {-\frac {\ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b}+\frac {\ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b}}{2 e}\) | \(59\) |
default | \(\frac {-\frac {\ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b}+\frac {\ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b}}{2 e}\) | \(59\) |
norman | \(-\frac {\ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{2 b e}+\frac {\ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{2 b e}\) | \(61\) |
risch | \(\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a -b}{i b -a}\right )}{2 b e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}-i\right )}{2 b e}\) | \(61\) |
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\log \left (2 \, a b \cos \left (e x + d\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \sin \left (e x + d\right )\right ) - \log \left (-\sin \left (e x + d\right ) + 1\right )}{4 \, b e} \]
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Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (22) = 44\).
Time = 0.98 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.30 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\begin {cases} \frac {\tilde {\infty } x}{\cos {\left (d \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge e = 0 \\- \frac {1}{a e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} - a e} & \text {for}\: b = 0 \\\frac {x}{- 2 a \sin {\left (d \right )} + 2 a + 2 b \cos {\left (d \right )}} & \text {for}\: e = 0 \\- \frac {\log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} - 1 \right )}}{2 b e} & \text {for}\: a = b \\- \frac {\log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} - 1 \right )}}{2 b e} + \frac {\log {\left (- \frac {a}{a - b} - \frac {b}{a - b} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{2 b e} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (25) = 50\).
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.88 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\frac {\log \left (a + b - \frac {{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b} - \frac {\log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right )}{b}}{2 \, e} \]
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Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.39 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a - 2 \, {\left | b \right |} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a + 2 \, {\left | b \right |} \right |}}\right )}{2 \, e {\left | b \right |}} \]
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Time = 27.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx=\frac {\mathrm {atanh}\left (\frac {a-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2}}{b}\right )}{b\,e} \]
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