Integrand size = 40, antiderivative size = 191 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (5+2 m) \left (3+8 m+4 m^2\right )} \]
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Time = 0.23 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3051, 2822, 2821} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {(2 A-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c^2 f (2 m+5) \left (4 m^2+8 m+3\right )}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)}+\frac {(2 A-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{c f (2 m+3) (2 m+5)} \]
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Rule 2821
Rule 2822
Rule 3051
Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx}{c (5+2 m)} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c^2 (3+2 m) (5+2 m)} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (1+2 m) (3+2 m) (5+2 m)} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.74 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\sec (e+f x) (a (1+\sin (e+f x)))^{1+m} (c-c \sin (e+f x))^{-m} \left (-B (3+2 m)+A \left (7+12 m+4 m^2\right )+(3+2 m) (-2 A+B (3+2 m)) \sin (e+f x)+(2 A-B (3+2 m)) \sin ^2(e+f x)\right )}{a c^3 f (1+2 m) (3+2 m) (5+2 m) (-1+\sin (e+f x))^2} \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-3-m}d x\]
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Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.72 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {{\left ({\left (2 \, B m - 2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} + {\left (4 \, B m^{2} - 4 \, {\left (A - 3 \, B\right )} m - 6 \, A + 9 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (4 \, A m^{2} + 4 \, {\left (3 \, A - B\right )} m + 9 \, A - 6 \, B\right )} \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f} \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 3} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \,d x } \]
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Time = 15.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.25 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (30\,A\,\cos \left (e+f\,x\right )-15\,B\,\cos \left (e+f\,x\right )-2\,A\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (3\,e+3\,f\,x\right )-12\,A\,\sin \left (2\,e+2\,f\,x\right )+18\,B\,\sin \left (2\,e+2\,f\,x\right )+8\,B\,m^2\,\sin \left (2\,e+2\,f\,x\right )+48\,A\,m\,\cos \left (e+f\,x\right )-10\,B\,m\,\cos \left (e+f\,x\right )+16\,A\,m^2\,\cos \left (e+f\,x\right )+2\,B\,m\,\cos \left (3\,e+3\,f\,x\right )-8\,A\,m\,\sin \left (2\,e+2\,f\,x\right )+24\,B\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^3\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (8\,m^3+36\,m^2+46\,m+15\right )\,\left (15\,\sin \left (e+f\,x\right )+6\,\cos \left (2\,e+2\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )-10\right )} \]
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