Integrand size = 29, antiderivative size = 127 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^6(c+d x)}{6 d}-\frac {4 a^2 \sin ^7(c+d x)}{7 d}-\frac {a^2 \sin ^8(c+d x)}{8 d}+\frac {2 a^2 \sin ^9(c+d x)}{9 d}+\frac {a^2 \sin ^{10}(c+d x)}{10 d} \]
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Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^{10}(c+d x)}{10 d}+\frac {2 a^2 \sin ^9(c+d x)}{9 d}-\frac {a^2 \sin ^8(c+d x)}{8 d}-\frac {4 a^2 \sin ^7(c+d x)}{7 d}-\frac {a^2 \sin ^6(c+d x)}{6 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^4(c+d x)}{4 d} \]
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Rule 12
Rule 90
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^3 (a+x)^4}{a^3} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^2 x^3 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = \frac {\text {Subst}\left (\int \left (a^6 x^3+2 a^5 x^4-a^4 x^5-4 a^3 x^6-a^2 x^7+2 a x^8+x^9\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = \frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^6(c+d x)}{6 d}-\frac {4 a^2 \sin ^7(c+d x)}{7 d}-\frac {a^2 \sin ^8(c+d x)}{8 d}+\frac {2 a^2 \sin ^9(c+d x)}{9 d}+\frac {a^2 \sin ^{10}(c+d x)}{10 d} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (-2625+10710 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-1365 \cos (6 (c+d x))-315 \cos (8 (c+d x))+63 \cos (10 (c+d x))-15120 \sin (c+d x)+3360 \sin (3 (c+d x))+2016 \sin (5 (c+d x))-360 \sin (7 (c+d x))-280 \sin (9 (c+d x)))}{322560 d} \]
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Time = 0.57 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}\right )}{d}\) | \(79\) |
default | \(\frac {a^{2} \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}\right )}{d}\) | \(79\) |
parallelrisch | \(-\frac {a^{2} \left (10710 \cos \left (2 d x +2 c \right )-280 \sin \left (9 d x +9 c \right )-315 \cos \left (8 d x +8 c \right )-360 \sin \left (7 d x +7 c \right )+2016 \sin \left (5 d x +5 c \right )-1365 \cos \left (6 d x +6 c \right )-15120 \sin \left (d x +c \right )+3360 \sin \left (3 d x +3 c \right )+1260 \cos \left (4 d x +4 c \right )-10353+63 \cos \left (10 d x +10 c \right )\right )}{322560 d}\) | \(118\) |
risch | \(\frac {3 a^{2} \sin \left (d x +c \right )}{64 d}-\frac {a^{2} \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a^{2} \sin \left (9 d x +9 c \right )}{1152 d}+\frac {a^{2} \cos \left (8 d x +8 c \right )}{1024 d}+\frac {a^{2} \sin \left (7 d x +7 c \right )}{896 d}+\frac {13 a^{2} \cos \left (6 d x +6 c \right )}{3072 d}-\frac {a^{2} \sin \left (5 d x +5 c \right )}{160 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{256 d}-\frac {a^{2} \sin \left (3 d x +3 c \right )}{96 d}-\frac {17 a^{2} \cos \left (2 d x +2 c \right )}{512 d}\) | \(169\) |
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Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {252 \, a^{2} \cos \left (d x + c\right )^{10} - 945 \, a^{2} \cos \left (d x + c\right )^{8} + 840 \, a^{2} \cos \left (d x + c\right )^{6} - 16 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 50 \, a^{2} \cos \left (d x + c\right )^{6} + 3 \, a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \]
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Time = 1.25 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.49 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {16 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a^{2} \cos ^{10}{\left (c + d x \right )}}{60 d} - \frac {a^{2} \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {252 \, a^{2} \sin \left (d x + c\right )^{10} + 560 \, a^{2} \sin \left (d x + c\right )^{9} - 315 \, a^{2} \sin \left (d x + c\right )^{8} - 1440 \, a^{2} \sin \left (d x + c\right )^{7} - 420 \, a^{2} \sin \left (d x + c\right )^{6} + 1008 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4}}{2520 \, d} \]
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Time = 0.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.32 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {13 \, a^{2} \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {a^{2} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {17 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac {a^{2} \sin \left (9 \, d x + 9 \, c\right )}{1152 \, d} + \frac {a^{2} \sin \left (7 \, d x + 7 \, c\right )}{896 \, d} - \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac {3 \, a^{2} \sin \left (d x + c\right )}{64 \, d} \]
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Time = 9.73 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{8}-\frac {4\,a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^2\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{4}}{d} \]
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