Integrand size = 27, antiderivative size = 81 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^6(c+d x)}{6 d}+\frac {b \cos ^8(c+d x)}{8 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^7(c+d x)}{7 d} \]
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Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2913, 2644, 276, 2645, 14} \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^7(c+d x)}{7 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {b \cos ^8(c+d x)}{8 d}-\frac {b \cos ^6(c+d x)}{6 d} \]
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Rule 14
Rule 276
Rule 2644
Rule 2645
Rule 2913
Rubi steps \begin{align*} \text {integral}& = a \int \cos ^5(c+d x) \sin ^2(c+d x) \, dx+b \int \cos ^5(c+d x) \sin ^3(c+d x) \, dx \\ & = \frac {a \text {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{d}-\frac {b \text {Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {b \text {Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {b \cos ^6(c+d x)}{6 d}+\frac {b \cos ^8(c+d x)}{8 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^7(c+d x)}{7 d} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {2520 b \cos (2 (c+d x))+420 b \cos (4 (c+d x))-280 b \cos (6 (c+d x))-105 b \cos (8 (c+d x))-8400 a \sin (c+d x)+560 a \sin (3 (c+d x))+1008 a \sin (5 (c+d x))+240 a \sin (7 (c+d x))}{107520 d} \]
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Time = 0.49 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {b \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right ) b}{3}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right ) a}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(72\) |
default | \(\frac {\frac {b \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right ) b}{3}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right ) a}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(72\) |
parallelrisch | \(-\frac {\left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (240 \cos \left (4 d x +4 c \right ) a +1728 a \cos \left (2 d x +2 c \right )+1050 b \sin \left (d x +c \right )+105 b \sin \left (5 d x +5 c \right )+595 b \sin \left (3 d x +3 c \right )+2512 a \right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{26880 d}\) | \(109\) |
risch | \(\frac {5 a \sin \left (d x +c \right )}{64 d}+\frac {b \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a \sin \left (7 d x +7 c \right )}{448 d}+\frac {b \cos \left (6 d x +6 c \right )}{384 d}-\frac {3 a \sin \left (5 d x +5 c \right )}{320 d}-\frac {b \cos \left (4 d x +4 c \right )}{256 d}-\frac {a \sin \left (3 d x +3 c \right )}{192 d}-\frac {3 b \cos \left (2 d x +2 c \right )}{128 d}\) | \(119\) |
norman | \(\frac {\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {688 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {688 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {8 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {40 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\) | \(205\) |
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Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {105 \, b \cos \left (d x + c\right )^{8} - 140 \, b \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{840 \, d} \]
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Time = 0.68 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {b \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {b \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {105 \, b \sin \left (d x + c\right )^{8} + 120 \, a \sin \left (d x + c\right )^{7} - 280 \, b \sin \left (d x + c\right )^{6} - 336 \, a \sin \left (d x + c\right )^{5} + 210 \, b \sin \left (d x + c\right )^{4} + 280 \, a \sin \left (d x + c\right )^{3}}{840 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.46 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {b \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {3 \, b \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {a \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {3 \, a \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{64 \, d} \]
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Time = 11.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {b\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {b\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {b\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]
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