Integrand size = 27, antiderivative size = 201 \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m)} \]
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Time = 0.20 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2778, 2999, 134} \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\frac {a 2^{\frac {1}{2}-\frac {m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+1}{2}} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-m-1),\frac {m+1}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (m+1) \left (a^2-b^2\right )}-\frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (a-b)} \]
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Rule 134
Rule 2778
Rule 2999
Rubi steps \begin{align*} \text {integral}& = -\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {a \int \frac {(e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2} \\ & = -\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {\left (a (e \cos (c+d x))^{-1-m} (1-\sin (c+d x))^{\frac {1+m}{2}} (1+\sin (c+d x))^{\frac {1+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-1-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e} \\ & = -\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m)} \\ \end{align*}
Time = 0.74 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.84 \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=-\frac {2^{\frac {1}{2} (-1-m)} (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m} \left (2^{\frac {1+m}{2}} (a+b)-2 a \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},-\frac {(a-b) (-1+\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}}\right )}{(a-b) (a+b) d e (1+m)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-2-m} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{- m - 2} \left (a + b \sin {\left (c + d x \right )}\right )^{m}\, dx \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+2}} \,d x \]
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