Integrand size = 27, antiderivative size = 132 \[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\frac {e (e \cos (c+d x))^{-2-m} \operatorname {Hypergeometric2F1}\left (1+m,\frac {2+m}{2},2+m,\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{(a+b) d (1+m)} \]
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Time = 0.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2777} \[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\frac {e (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,\frac {m+2}{2},m+2,\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d (m+1) (a+b)} \]
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Rule 2777
Rubi steps \begin{align*} \text {integral}& = \frac {e (e \cos (c+d x))^{-2-m} \operatorname {Hypergeometric2F1}\left (1+m,\frac {2+m}{2},2+m,\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{(a+b) d (1+m)} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00 \[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=-\frac {e (e \cos (c+d x))^{-2-m} \operatorname {Hypergeometric2F1}\left (1+m,\frac {2+m}{2},2+m,-\frac {2 (a+b \sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right ) (1+\sin (c+d x)) \left (\frac {(a+b) (1+\sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{(a-b) d (1+m)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-1-m} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 1} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{- m - 1} \left (a + b \sin {\left (c + d x \right )}\right )^{m}\, dx \]
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\[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 1} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 1} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+1}} \,d x \]
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