Integrand size = 27, antiderivative size = 89 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^{1+m}}{a d e \left (1-m^2\right )} \]
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Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-1}}{a d e \left (1-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)} \]
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Rule 2750
Rule 2751
Rubi steps \begin{align*} \text {integral}& = -\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac {\int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (1-m)} \\ & = -\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^{1+m}}{a d e \left (1-m^2\right )} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-1-m} (m-\sin (c+d x)) (a (1+\sin (c+d x)))^m}{d e (-1+m) (1+m)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-2-m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
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none
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {{\left (m \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} - d} \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- m - 2}\, dx \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Time = 5.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=-\frac {\left (\sin \left (2\,c+2\,d\,x\right )-2\,m\,\cos \left (c+d\,x\right )\right )\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{d\,e^2\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-1\right )} \]
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