Integrand size = 30, antiderivative size = 96 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{4} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d} \]
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Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3045, 2718, 2715, 8, 2713} \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac {a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac {1}{4} a^3 A x \]
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Rule 8
Rule 2713
Rule 2715
Rule 2718
Rule 3045
Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 A \sin (c+d x)+2 a^3 A \sin ^2(c+d x)-2 a^3 A \sin ^4(c+d x)-a^3 A \sin ^5(c+d x)\right ) \, dx \\ & = \left (a^3 A\right ) \int \sin (c+d x) \, dx-\left (a^3 A\right ) \int \sin ^5(c+d x) \, dx+\left (2 a^3 A\right ) \int \sin ^2(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^4(c+d x) \, dx \\ & = -\frac {a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{d}+\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}+\left (a^3 A\right ) \int 1 \, dx-\frac {1}{2} \left (3 a^3 A\right ) \int \sin ^2(c+d x) \, dx+\frac {\left (a^3 A\right ) \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac {1}{4} \left (3 a^3 A\right ) \int 1 \, dx \\ & = \frac {1}{4} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \cos (c+d x) \left (-30 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(c+d x)} \left (-28-15 \sin (c+d x)+16 \sin ^2(c+d x)+30 \sin ^3(c+d x)+12 \sin ^4(c+d x)\right )\right )}{60 d \sqrt {\cos ^2(c+d x)}} \]
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Time = 1.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59
method | result | size |
parallelrisch | \(-\frac {A \,a^{3} \left (-60 d x +90 \cos \left (d x +c \right )-3 \cos \left (5 d x +5 c \right )+15 \sin \left (4 d x +4 c \right )+25 \cos \left (3 d x +3 c \right )+112\right )}{240 d}\) | \(57\) |
risch | \(\frac {a^{3} A x}{4}-\frac {3 a^{3} A \cos \left (d x +c \right )}{8 d}+\frac {A \,a^{3} \cos \left (5 d x +5 c \right )}{80 d}-\frac {A \,a^{3} \sin \left (4 d x +4 c \right )}{16 d}-\frac {5 A \,a^{3} \cos \left (3 d x +3 c \right )}{48 d}\) | \(78\) |
derivativedivides | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A \,a^{3} \cos \left (d x +c \right )}{d}\) | \(117\) |
default | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A \,a^{3} \cos \left (d x +c \right )}{d}\) | \(117\) |
parts | \(-\frac {a^{3} A \cos \left (d x +c \right )}{d}+\frac {2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5 d}\) | \(125\) |
norman | \(\frac {-\frac {14 A \,a^{3}}{15 d}+\frac {a^{3} A x}{4}-\frac {4 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {3 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{3} A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(282\) |
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Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {12 \, A a^{3} \cos \left (d x + c\right )^{5} - 40 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} d x - 15 \, {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (88) = 176\).
Time = 0.25 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.78 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + A a^{3} x \sin ^{2}{\left (c + d x \right )} - \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + A a^{3} x \cos ^{2}{\left (c + d x \right )} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {4 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {8 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {A a^{3} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} A a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 120 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 240 \, A a^{3} \cos \left (d x + c\right )}{240 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{4} \, A a^{3} x + \frac {A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, A a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac {A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \]
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Time = 14.73 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.04 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,x}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{12}-\frac {A\,a^3\,\left (75\,c+75\,d\,x-120\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{12}-\frac {A\,a^3\,\left (75\,c+75\,d\,x-160\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (150\,c+150\,d\,x-80\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (150\,c+150\,d\,x-480\right )}{60}\right )+\frac {A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{60}-\frac {A\,a^3\,\left (15\,c+15\,d\,x-56\right )}{60}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
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