Integrand size = 27, antiderivative size = 70 \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-2 (1+m)} \operatorname {Hypergeometric2F1}\left (2,-1-m,-m,\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{1+m}}{4 a d e (1+m)} \]
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Time = 0.05 (sec) , antiderivative size = 70, 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 = {2768, 7, 70} \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-2 (m+1)} \operatorname {Hypergeometric2F1}\left (2,-m-1,-m,\frac {1}{2} (1-\sin (c+d x))\right )}{4 a d e (m+1)} \]
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Rule 7
Rule 70
Rule 2768
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 (e \cos (c+d x))^{-2-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (2+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (2+2 m)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (-4-2 m)} (a+a x)^{\frac {1}{2} (-4-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (a^2 (e \cos (c+d x))^{-2-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (2+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (2+2 m)}\right ) \text {Subst}\left (\int \frac {(a-a x)^{\frac {1}{2} (-4-2 m)}}{(a+a x)^2} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {(e \cos (c+d x))^{-2 (1+m)} \operatorname {Hypergeometric2F1}\left (2,-1-m,-m,\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{1+m}}{4 a d e (1+m)} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09 \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-2 m} \operatorname {Hypergeometric2F1}\left (2,-1-m,-m,\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^2(c+d x) (a (1+\sin (c+d x)))^{1+m}}{4 a d e^3 (1+m)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-3-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- 2 m - 3}\, dx \]
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\[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{2\,m+3}} \,d x \]
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