Integrand size = 40, antiderivative size = 170 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\frac {2^{\frac {1}{2}-m} c^2 (2 A-B (1-2 m)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f} \]
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Time = 0.22 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3052, 2824, 2768, 72, 71} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\frac {c^2 2^{\frac {1}{2}-m} (2 A-B (1-2 m)) \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m-1),\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{1-m}}{2 f} \]
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Rule 71
Rule 72
Rule 2768
Rule 2824
Rule 3052
Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f}+\frac {(2 A c+B c (-1+2 m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx}{2 c} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f}+\frac {\left ((2 A c+B c (-1+2 m)) \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{1-2 m} \, dx}{2 c} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f}+\frac {\left (c (2 A c+B c (-1+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int (c-c x)^{1-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{2 f} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f}+\frac {\left (2^{-\frac {1}{2}-m} c^2 (2 A c+B c (-1+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 m)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{1-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {2^{\frac {1}{2}-m} c^2 (2 A-B (1-2 m)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{2 f} \\ \end{align*}
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{1-m}d x\]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{1 - m} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-m} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{1-m} \,d x \]
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