Integrand size = 27, antiderivative size = 105 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{16}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin ^3(c+d x)}{6 d} \]
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Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2917, 2645, 14, 2648, 2715, 8} \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {a x}{16} \]
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Rule 8
Rule 14
Rule 2645
Rule 2648
Rule 2715
Rule 2917
Rubi steps \begin{align*} \text {integral}& = a \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+a \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx \\ & = -\frac {a \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac {1}{2} a \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx-\frac {a \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac {1}{8} a \int \cos ^2(c+d x) \, dx-\frac {a \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac {1}{16} a \int 1 \, dx \\ & = \frac {a x}{16}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin ^3(c+d x)}{6 d} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a (60 d x-120 \cos (c+d x)-20 \cos (3 (c+d x))+12 \cos (5 (c+d x))-15 \sin (2 (c+d x))-15 \sin (4 (c+d x))+5 \sin (6 (c+d x)))}{960 d} \]
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Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(-\frac {a \left (-60 d x +120 \cos \left (d x +c \right )-12 \cos \left (5 d x +5 c \right )+20 \cos \left (3 d x +3 c \right )+15 \sin \left (4 d x +4 c \right )+15 \sin \left (2 d x +2 c \right )-5 \sin \left (6 d x +6 c \right )+128\right )}{960 d}\) | \(76\) |
risch | \(\frac {a x}{16}-\frac {a \cos \left (d x +c \right )}{8 d}+\frac {a \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (4 d x +4 c \right )}{64 d}-\frac {a \cos \left (3 d x +3 c \right )}{48 d}-\frac {a \sin \left (2 d x +2 c \right )}{64 d}\) | \(93\) |
derivativedivides | \(\frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(95\) |
default | \(\frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(95\) |
norman | \(\frac {\frac {a x}{16}-\frac {4 a}{15 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {17 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {19 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {19 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {17 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {3 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {5 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {3 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(269\) |
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Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {48 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, a d x + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{5} - 14 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (92) = 184\).
Time = 0.34 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.83 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {2 a \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{3}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.62 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a}{960 \, d} \]
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Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{16} \, a x + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {a \cos \left (d x + c\right )}{8 \, d} + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]
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Time = 13.56 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.15 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,x}{16}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\left (\frac {a\,\left (225\,c+225\,d\,x-960\right )}{240}-\frac {15\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\left (\frac {a\,\left (300\,c+300\,d\,x-640\right )}{240}-\frac {5\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\left (\frac {a\,\left (90\,c+90\,d\,x-384\right )}{240}-\frac {3\,a\,\left (c+d\,x\right )}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {a\,\left (15\,c+15\,d\,x-64\right )}{240}-\frac {a\,\left (c+d\,x\right )}{16}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]
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