Integrand size = 38, antiderivative size = 158 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\frac {2^{\frac {1}{2}-m} c (A+2 B m) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f} \]
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Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3052, 2824, 2768, 72, 71} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\frac {c 2^{\frac {1}{2}-m} (A+2 B m) \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m}}{f} \]
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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))^{-m}}{f}+(A+2 B m) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\left ((A+2 B 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))^{-2 m} \, dx \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\frac {\left (c^2 (A+2 B 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)^{-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 {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\frac {\left (2^{-\frac {1}{2}-m} c^2 (A+2 B 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 )^{-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 (A+2 B 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))^{-m}}{f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 6.02 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.54 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=-\frac {i 2^{-1-m} \left (a \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )^m (c-c \sin (e+f x))^{-m} (\cosh (m \log (2))+\sinh (m \log (2))) \left (\frac {A \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac {\cos (e+f x)+i (-1+\sin (e+f x))}{\cos (e+f x)+i (1+\sin (e+f x))}\right )}{m}-\frac {2 B \operatorname {Hypergeometric2F1}\left (2,1-2 m,2-2 m,-\frac {\cos (e+f x)+i (-1+\sin (e+f x))}{\cos (e+f x)+i (1+\sin (e+f x))}\right ) (\cos (e+f x)+i (-1+\sin (e+f x)))}{(-1+2 m) (\cos (e+f x)+i (1+\sin (e+f x)))}+(1-i \cos (e+f x)+\sin (e+f x))^{-2 m} \left (\frac {B \operatorname {Hypergeometric2F1}\left (2 m,1+2 m,2+2 m,\frac {1}{2} (1-i \cos (e+f x)+\sin (e+f x))\right ) (\cos (e+f x) (\cos (e+f x)+i \sin (e+f x)))^{2 m} (1-i \cos (e+f x)+\sin (e+f x))}{1+2 m}-\frac {4^{-m} A \operatorname {Hypergeometric2F1}\left (-2 m,-2 m,1-2 m,\frac {1}{2} (1+i \cos (e+f x)-\sin (e+f x))\right ) (\cosh (m \log (16))+\sinh (m \log (16)))}{m}\right )\right )}{f} \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{-m}d x\]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{m}} \,d x } \]
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Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\text {Timed out} \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{m}} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{m}} \,d x } \]
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Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m} \,d x \]
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