Integrand size = 29, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^4(c+d x)}{4 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^6(c+d x)}{6 a^2 d} \]
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Time = 0.08 (sec) , antiderivative size = 55, 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, 45} \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^6(c+d x)}{6 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^4(c+d x)}{4 a^2 d} \]
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Rule 12
Rule 45
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^3}{a^3} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^2 x^3 \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 x^3-2 a x^4+x^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d} \\ & = \frac {\sin ^4(c+d x)}{4 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^6(c+d x)}{6 a^2 d} \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^4(c+d x) \left (15-24 \sin (c+d x)+10 \sin ^2(c+d x)\right )}{60 a^2 d} \]
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Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\) | \(39\) |
default | \(\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\) | \(39\) |
parallelrisch | \(\frac {-5 \cos \left (6 d x +6 c \right )-195 \cos \left (2 d x +2 c \right )-24 \sin \left (5 d x +5 c \right )+120 \sin \left (3 d x +3 c \right )-240 \sin \left (d x +c \right )+60 \cos \left (4 d x +4 c \right )+140}{960 d \,a^{2}}\) | \(74\) |
risch | \(-\frac {\sin \left (d x +c \right )}{4 a^{2} d}-\frac {\cos \left (6 d x +6 c \right )}{192 d \,a^{2}}-\frac {\sin \left (5 d x +5 c \right )}{40 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right )}{16 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right )}{8 d \,a^{2}}-\frac {13 \cos \left (2 d x +2 c \right )}{64 d \,a^{2}}\) | \(101\) |
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Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {10 \, \cos \left (d x + c\right )^{6} - 45 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{60 \, a^{2} d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (46) = 92\).
Time = 53.78 (sec) , antiderivative size = 682, normalized size of antiderivative = 12.40 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {60 \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {280 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {60 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]
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Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]
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Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]
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Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^4\,\left (10\,{\sin \left (c+d\,x\right )}^2-24\,\sin \left (c+d\,x\right )+15\right )}{60\,a^2\,d} \]
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