Integrand size = 12, antiderivative size = 113 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))} \]
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Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2743, 2833, 12, 2738, 213} \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}-\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]
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Rule 12
Rule 213
Rule 2738
Rule 2743
Rule 2833
Rubi steps \begin{align*} \text {integral}& = -\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {1}{32} \int \frac {-6-5 \cos (c+d x)}{(3-5 \cos (c+d x))^2} \, dx \\ & = -\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}+\frac {1}{512} \int \frac {43}{3-5 \cos (c+d x)} \, dx \\ & = -\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}+\frac {43}{512} \int \frac {1}{3-5 \cos (c+d x)} \, dx \\ & = -\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}+\frac {43 \text {Subst}\left (\int \frac {1}{-2+8 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{256 d} \\ & = \frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.87 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {5}{512 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{1024 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {5}{512 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{1024 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
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Time = 0.70 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79
method | result | size |
norman | \(\frac {-\frac {85 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{512 d}+\frac {35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d}}{{\left (4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{2}}+\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048 d}-\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2048 d}\) | \(89\) |
derivativedivides | \(\frac {-\frac {25}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {35}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048}+\frac {25}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {35}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2048}}{d}\) | \(106\) |
default | \(\frac {-\frac {25}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {35}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048}+\frac {25}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {35}{2048 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2048}}{d}\) | \(106\) |
risch | \(\frac {i \left (215 \,{\mathrm e}^{3 i \left (d x +c \right )}-387 \,{\mathrm e}^{2 i \left (d x +c \right )}+325 \,{\mathrm e}^{i \left (d x +c \right )}-225\right )}{256 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )^{2}}+\frac {43 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}-\frac {4 i}{5}\right )}{2048 d}-\frac {43 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}+\frac {4 i}{5}\right )}{2048 d}\) | \(107\) |
parallelrisch | \(\frac {\left (2580 \cos \left (d x +c \right )-1075 \cos \left (2 d x +2 c \right )-1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )+\left (-2580 \cos \left (d x +c \right )+1075 \cos \left (2 d x +2 c \right )+1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{2}\right )-440 \sin \left (d x +c \right )+900 \sin \left (2 d x +2 c \right )}{2048 d \left (-25 \cos \left (2 d x +2 c \right )-43+60 \cos \left (d x +c \right )\right )}\) | \(117\) |
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Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=-\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 40 \, {\left (45 \, \cos \left (d x + c\right ) - 11\right )} \sin \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} - 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (102) = 204\).
Time = 1.19 (sec) , antiderivative size = 490, normalized size of antiderivative = 4.34 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\begin {cases} \frac {x}{\left (3 - 5 \cos {\left (2 \operatorname {atan}{\left (\frac {1}{2} \right )} \right )}\right )^{3}} & \text {for}\: c = - d x - 2 \operatorname {atan}{\left (\frac {1}{2} \right )} \vee c = - d x + 2 \operatorname {atan}{\left (\frac {1}{2} \right )} \\\frac {x}{\left (3 - 5 \cos {\left (c \right )}\right )^{3}} & \text {for}\: d = 0 \\\frac {688 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {344 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {43 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {688 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {344 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {43 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {560 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {340 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} & \text {otherwise} \end {cases} \]
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Time = 0.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (\frac {17 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {28 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {16 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1} - 43 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 43 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{2048 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - 43 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 43 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{2048 \, d} \]
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Time = 14.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(3-5 \cos (c+d x))^3} \, dx=-\frac {43\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{1024\,d}-\frac {\frac {85\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8192}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2048}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {1}{16}\right )} \]
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