Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=\frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))} \]
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Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2743, 12, 2738, 213} \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=\frac {5 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}+\frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \]
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Rule 12
Rule 213
Rule 2738
Rule 2743
Rubi steps \begin{align*} \text {integral}& = \frac {5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}+\frac {1}{16} \int \frac {3}{-3-5 \cos (c+d x)} \, dx \\ & = \frac {5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}+\frac {3}{16} \int \frac {1}{-3-5 \cos (c+d x)} \, dx \\ & = \frac {5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}+\frac {3 \text {Subst}\left (\int \frac {1}{-8+2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \\ & = \frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=\frac {9 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (c+d x) \left (\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-9 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+20 \sin (c+d x)}{64 d (3+5 \cos (c+d x))} \]
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Time = 0.78 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64}}{d}\) | \(64\) |
default | \(\frac {-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64}}{d}\) | \(64\) |
norman | \(-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-4\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64 d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64 d}\) | \(65\) |
risch | \(\frac {i \left (3 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}+\frac {4 i}{5}\right )}{64 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}-\frac {4 i}{5}\right )}{64 d}\) | \(85\) |
parallelrisch | \(\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right ) \cos \left (d x +c \right )-15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \cos \left (d x +c \right )+20 \sin \left (d x +c \right )+9 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )-9 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64 d \left (3+5 \cos \left (d x +c \right )\right )}\) | \(95\) |
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Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=-\frac {3 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 3 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 40 \, \sin \left (d x + c\right )}{128 \, {\left (5 \, d \cos \left (d x + c\right ) + 3 \, d\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (78) = 156\).
Time = 0.68 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.57 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=\begin {cases} \frac {x}{\left (-3 - 5 \cos {\left (2 \operatorname {atan}{\left (2 \right )} \right )}\right )^{2}} & \text {for}\: c = - d x - 2 \operatorname {atan}{\left (2 \right )} \vee c = - d x + 2 \operatorname {atan}{\left (2 \right )} \\\frac {x}{\left (- 5 \cos {\left (c \right )} - 3\right )^{2}} & \text {for}\: d = 0 \\\frac {3 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} - \frac {12 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{64 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} - \frac {3 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {12 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{64 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} - \frac {20 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} & \text {otherwise} \end {cases} \]
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Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=-\frac {\frac {20 \, \sin \left (d x + c\right )}{{\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 4\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + 3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{64 \, d} \]
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Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=-\frac {\frac {20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4} + 3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) - 3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{64 \, d} \]
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Time = 0.00 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(-3-5 \cos (c+d x))^2} \, dx=-\frac {3\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{32\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\right )} \]
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