Integrand size = 12, antiderivative size = 115 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=-\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac {45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))} \]
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Time = 0.12 (sec) , antiderivative size = 115, 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, 212} \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=-\frac {45 \sin (c+d x)}{512 d (5 \cos (c+d x)+3)}+\frac {5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}-\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]
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Rule 12
Rule 212
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}{8-2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{256 d} \\ & = -\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\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 = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=-\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {5}{512 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{2048 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {5}{512 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{2048 d \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
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Time = 0.66 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72
method | result | size |
norman | \(\frac {-\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d}+\frac {85 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{512 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-4\right )^{2}}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048 d}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048 d}\) | \(83\) |
derivativedivides | \(\frac {\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048}-\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048}}{d}\) | \(94\) |
default | \(\frac {\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{2048}-\frac {25}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}+\frac {85}{1024 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{2048}}{d}\) | \(94\) |
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 | \(\frac {\left (-2580 \cos \left (d x +c \right )-1075 \cos \left (2 d x +2 c \right )-1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )+\left (2580 \cos \left (d x +c \right )+1075 \cos \left (2 d x +2 c \right )+1849\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )-440 \sin \left (d x +c \right )-900 \sin \left (2 d x +2 c \right )}{2048 d \left (43+25 \cos \left (2 d x +2 c \right )+60 \cos \left (d x +c \right )\right )}\) | \(117\) |
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Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 40 \, {\left (45 \, \cos \left (d x + c\right ) + 11\right )} \sin \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (102) = 204\).
Time = 1.18 (sec) , antiderivative size = 474, normalized size of antiderivative = 4.12 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\begin {cases} \frac {x}{\left (5 \cos {\left (2 \operatorname {atan}{\left (2 \right )} \right )} + 3\right )^{3}} & \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 )^{3}} & \text {for}\: d = 0 \\- \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 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 )} + 32768 d} + \frac {344 \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 )}}{2048 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 )} + 32768 d} - \frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{2048 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 )} + 32768 d} + \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 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 )} + 32768 d} - \frac {344 \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 )}}{2048 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 )} + 32768 d} + \frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{2048 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 )} + 32768 d} + \frac {340 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 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 )} + 32768 d} - \frac {560 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2048 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 )} + 32768 d} & \text {otherwise} \end {cases} \]
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Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (\frac {28 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {17 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 16} + 43 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 43 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{2048 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {\frac {20 \, {\left (17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4\right )}^{2}} + 43 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) - 43 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{2048 \, d} \]
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Time = 14.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(3+5 \cos (c+d x))^3} \, dx=\frac {43\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{1024\,d}-\frac {\frac {35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{512}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]
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