Integrand size = 27, antiderivative size = 65 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \]
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Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 76} \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin ^6(c+d x)}{6 d}-\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d} \]
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Rule 12
Rule 76
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x) x^2 (a+x)^2}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int (a-x) x^2 (a+x)^2 \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (a^3 x^2+a^2 x^3-a x^4-x^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-45 \cos (2 (c+d x))+5 \cos (6 (c+d x))+32 (7+3 \cos (2 (c+d x))) \sin ^3(c+d x)\right )}{960 d} \]
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Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(48\) |
default | \(-\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(48\) |
risch | \(\frac {a \sin \left (d x +c \right )}{8 d}+\frac {a \cos \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (3 d x +3 c \right )}{48 d}-\frac {3 a \cos \left (2 d x +2 c \right )}{64 d}\) | \(74\) |
parallelrisch | \(-\frac {a \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (28+12 \cos \left (2 d x +2 c \right )+5 \sin \left (3 d x +3 c \right )+15 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}\) | \(81\) |
norman | \(\frac {\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(137\) |
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Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {10 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \]
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Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \cos ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \]
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Time = 0.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \]
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Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]
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