Integrand size = 34, antiderivative size = 139 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {2^{\frac {1}{2}+m} a^2 c (B (1-m)-A (2+m)) \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{3 f (2+m)}-\frac {a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)} \]
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Time = 0.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3046, 2939, 2768, 72, 71} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {a^2 c 2^{m+\frac {1}{2}} (B (1-m)-A (m+2)) \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 f (m+2)}-\frac {a B c \cos ^3(e+f x) (a \sin (e+f x)+a)^{m-1}}{f (m+2)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rule 3046
Rubi steps \begin{align*} \text {integral}& = (a c) \int \cos ^2(e+f x) (a+a \sin (e+f x))^{-1+m} (A+B \sin (e+f x)) \, dx \\ & = -\frac {a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\left (a c \left (A-\frac {B (1-m)}{2+m}\right )\right ) \int \cos ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \, dx \\ & = -\frac {a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\frac {\left (a^3 c \left (A-\frac {B (1-m)}{2+m}\right ) \cos ^3(e+f x)\right ) \text {Subst}\left (\int \sqrt {a-a x} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}} \\ & = -\frac {a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\frac {\left (2^{-\frac {1}{2}+m} a^3 c \left (A-\frac {B (1-m)}{2+m}\right ) \cos ^3(e+f x) (a+a \sin (e+f x))^{-2+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m} \sqrt {a-a x} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2}} \\ & = -\frac {2^{\frac {1}{2}+m} a^2 c \left (A-\frac {B (1-m)}{2+m}\right ) \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{3 f}-\frac {a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)} \\ \end{align*}
Result contains complex when optimal does not.
Time = 3.18 (sec) , antiderivative size = 468, normalized size of antiderivative = 3.37 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {2^{-2-m} c e^{i f m x} \left (1-i e^{i (e+f x)}\right )^{-2 m} \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^m \left (\frac {i B e^{-i (2 e+f (2+m) x)} \operatorname {Hypergeometric2F1}\left (-2-m,-2 m,-1-m,i e^{i (e+f x)}\right )}{2+m}+\frac {2 (A-B) e^{-i (e+f (1+m) x)} \operatorname {Hypergeometric2F1}\left (-1-m,-2 m,-m,i e^{i (e+f x)}\right )}{1+m}-\frac {2 A e^{i (e-f (-1+m) x)} \operatorname {Hypergeometric2F1}\left (1-m,-2 m,2-m,i e^{i (e+f x)}\right )}{-1+m}+\frac {2 B e^{i (e-f (-1+m) x)} \operatorname {Hypergeometric2F1}\left (1-m,-2 m,2-m,i e^{i (e+f x)}\right )}{-1+m}+\frac {i B e^{2 i e-i f (-2+m) x} \operatorname {Hypergeometric2F1}\left (2-m,-2 m,3-m,i e^{i (e+f x)}\right )}{-2+m}+\frac {4 i A e^{-i f m x} \operatorname {Hypergeometric2F1}\left (-2 m,-m,1-m,i e^{i (e+f x)}\right )}{m}-\frac {2 i B e^{-i f m x} \operatorname {Hypergeometric2F1}\left (-2 m,-m,1-m,i e^{i (e+f x)}\right )}{m}\right ) (-1+\sin (e+f x))}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )d x\]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\int { -{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=- c \left (\int \left (- A \left (a \sin {\left (e + f x \right )} + a\right )^{m}\right )\, dx + \int A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\right )\, dx + \int B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\int { -{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\int { -{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c-c\,\sin \left (e+f\,x\right )\right ) \,d x \]
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