Integrand size = 12, antiderivative size = 81 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {59 x}{2048}-\frac {59 \arctan \left (\frac {\cos (c+d x)}{3+\sin (c+d x)}\right )}{1024 d}-\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))} \]
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Time = 0.04 (sec) , antiderivative size = 81, 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, 2833, 12, 2737} \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {59 \arctan \left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{1024 d}-\frac {45 \cos (c+d x)}{512 d (3 \sin (c+d x)+5)}-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}-\frac {59 x}{2048} \]
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Rule 12
Rule 2737
Rule 2743
Rule 2833
Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {1}{32} \int \frac {10-3 \sin (c+d x)}{(-5-3 \sin (c+d x))^2} \, dx \\ & = -\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))}+\frac {1}{512} \int \frac {59}{-5-3 \sin (c+d x)} \, dx \\ & = -\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))}+\frac {59}{512} \int \frac {1}{-5-3 \sin (c+d x)} \, dx \\ & = -\frac {59 x}{2048}-\frac {59 \arctan \left (\frac {\cos (c+d x)}{3+\sin (c+d x)}\right )}{1024 d}-\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\frac {-59 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {3 (-182 \cos (c+d x)+3 (59-9 \cos (2 (c+d x))+60 \sin (c+d x)-15 \sin (2 (c+d x))))}{(5+3 \sin (c+d x))^2}}{1024 d} \]
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Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {-\frac {50 \left (\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64000}+\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}+\frac {273}{12800}\right )}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\) | \(91\) |
default | \(\frac {-\frac {50 \left (\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64000}+\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}+\frac {273}{12800}\right )}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\) | \(91\) |
risch | \(-\frac {3 \left (295 i {\mathrm e}^{2 i \left (d x +c \right )}+59 \,{\mathrm e}^{3 i \left (d x +c \right )}-241 \,{\mathrm e}^{i \left (d x +c \right )}-45 i\right )}{256 \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3+10 i {\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}+\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i}{3}\right )}{2048 d}-\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{2048 d}\) | \(109\) |
parallelrisch | \(\frac {1062+59 i \left (-59+9 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3-4 i\right )+59 i \left (59-9 \cos \left (2 d x +2 c \right )+60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3+4 i\right )+2184 \cos \left (d x +c \right )-162 \cos \left (2 d x +2 c \right )+1080 \sin \left (d x +c \right )+540 \sin \left (2 d x +2 c \right )}{2048 d \left (-59+9 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )\right )}\) | \(149\) |
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Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) + 3}{4 \, \cos \left (d x + c\right )}\right ) - 540 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 1092 \, \cos \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} - 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]
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Result contains complex when optimal does not.
Time = 1.93 (sec) , antiderivative size = 921, normalized size of antiderivative = 11.37 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (73) = 146\).
Time = 0.26 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.14 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {\frac {12 \, {\left (\frac {3855 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3913 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 2275\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {86 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 25} + 1475 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {3}{4}\right )}{25600 \, d} \]
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Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {1475 \, d x + 1475 \, c + \frac {24 \, {\left (1605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3913 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2275\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{2}} + 2950 \, \arctan \left (-\frac {3 \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) - 9}\right )}{51200 \, d} \]
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Time = 6.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {59\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {3}{4}\right )}{1024\,d}-\frac {\frac {963\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{1280}+\frac {11739\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400}+\frac {2313\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1280}+\frac {273}{256}}{d\,{\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\right )}^2} \]
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