Integrand size = 27, antiderivative size = 87 \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=-\frac {2^{-\frac {1}{2}-m} (e \cos (c+d x))^{-1-2 m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (3+2 m),\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{\frac {1}{2}+m} (a+a \sin (c+d x))^m}{d e} \]
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Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2768, 7, 72, 71} \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=-\frac {2^{-m-\frac {1}{2}} (1-\sin (c+d x))^{m+\frac {1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-2 m-1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (2 m+3),\frac {1}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{d e} \]
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Rule 7
Rule 71
Rule 72
Rule 2768
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (1+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (-3-2 m)} (a+a x)^{\frac {1}{2} (-3-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (a^2 (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (1+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \text {Subst}\left (\int \frac {(a-a x)^{\frac {1}{2} (-3-2 m)}}{(a+a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (2^{-\frac {3}{2}-m} a (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{-\frac {1}{2}-m+\frac {1}{2} (1+2 m)} \left (\frac {a-a \sin (c+d x)}{a}\right )^{\frac {1}{2}+m} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {1}{2} (-3-2 m)}}{(a+a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = -\frac {2^{-\frac {1}{2}-m} (e \cos (c+d x))^{-1-2 m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (3+2 m),\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{\frac {1}{2}+m} (a+a \sin (c+d x))^m}{d e} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-1-2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {1}{2}-m,\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)} (a (1+\sin (c+d x)))^m}{\sqrt {2} e (d+2 d m)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-2-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- 2 m - 2}\, dx \]
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\[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{2\,m+2}} \,d x \]
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