Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \]
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Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3050, 2817} \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \]
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Rule 2817
Rule 3050
Rubi steps \begin{align*} \text {integral}& = \frac {B \int (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}
Time = 2.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.26 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {a^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (4 (60 A+23 B) \sin (e+f x)+\cos (4 (e+f x)) (5 A+15 B+4 B \sin (e+f x))-4 \cos (2 (e+f x)) (5 (7 A+5 B)+4 (5 A+6 B) \sin (e+f x)))}{160 f} \]
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Time = 2.99 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.36
method | result | size |
default | \(-\frac {a^{3} \tan \left (f x +e \right ) \left (4 B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-15 B \left (\sin ^{3}\left (f x +e \right )\right )+20 A \left (\cos ^{2}\left (f x +e \right )\right )-24 B \left (\sin ^{2}\left (f x +e \right )\right )-35 A \sin \left (f x +e \right )-10 B \sin \left (f x +e \right )-40 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{20 f}\) | \(131\) |
parts | \(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{3} \left (\cos ^{3}\left (f x +e \right )-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \cos \left (f x +e \right )+8 \tan \left (f x +e \right )+7 \sec \left (f x +e \right )\right )}{4 f}+\frac {B \sec \left (f x +e \right ) \left (4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+15 \left (\cos ^{2}\left (f x +e \right )\right )-24 \sin \left (f x +e \right )-25\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{3} \left (\cos ^{2}\left (f x +e \right )-1\right )}{20 f}\) | \(169\) |
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Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.45 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {{\left (5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (f x + e\right )^{2} + 5 \, {\left (7 \, A + 5 \, B\right )} a^{3} + 4 \, {\left (B a^{3} \cos \left (f x + e\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, f \cos \left (f x + e\right )} \]
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Timed out. \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\text {Timed out} \]
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\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \]
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Time = 0.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.56 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {4 \, {\left (8 \, B a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, A a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{5 \, f} \]
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Time = 15.87 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.80 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (140\,A\,\cos \left (e+f\,x\right )+100\,B\,\cos \left (e+f\,x\right )+135\,A\,\cos \left (3\,e+3\,f\,x\right )-5\,A\,\cos \left (5\,e+5\,f\,x\right )+85\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )-240\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )-90\,B\,\sin \left (2\,e+2\,f\,x\right )+48\,B\,\sin \left (4\,e+4\,f\,x\right )-2\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{160\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]
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