Integrand size = 38, antiderivative size = 140 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt {a+a \sin (e+f x)}}-\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f} \]
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Time = 0.38 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 c f} \]
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Rule 2817
Rule 2819
Rule 2920
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx}{a c} \\ & = -\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}+\frac {2 \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx}{3 c} \\ & = -\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}+\frac {(4 a) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{15 c} \\ & = -\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt {a+a \sin (e+f x)}}-\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f} \\ \end{align*}
Time = 7.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2 (-1+\sin (e+f x))^2 (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (75 \cos (2 (e+f x))+30 \cos (4 (e+f x))+5 \cos (6 (e+f x))+600 \sin (e+f x)+100 \sin (3 (e+f x))+12 \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
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Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.66
method | result | size |
default | \(\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{2} a \left (5 \left (\cos ^{5}\left (f x +e \right )\right )+6 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 \cos \left (f x +e \right ) \sin \left (f x +e \right )+16 \tan \left (f x +e \right )-5 \sec \left (f x +e \right )\right )}{30 f}\) | \(92\) |
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Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (5 \, a c^{2} \cos \left (f x + e\right )^{6} - 5 \, a c^{2} + 2 \, {\left (3 \, a c^{2} \cos \left (f x + e\right )^{4} + 4 \, a c^{2} \cos \left (f x + e\right )^{2} + 8 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]
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Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2} \,d x } \]
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Time = 0.31 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {16 \, {\left (10 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 24 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 15 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \]
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Time = 12.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,\cos \left (e+f\,x\right )+105\,\cos \left (3\,e+3\,f\,x\right )+35\,\cos \left (5\,e+5\,f\,x\right )+5\,\cos \left (7\,e+7\,f\,x\right )+700\,\sin \left (2\,e+2\,f\,x\right )+112\,\sin \left (4\,e+4\,f\,x\right )+12\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]
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