Integrand size = 40, antiderivative size = 81 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {(g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (1,-m+n,1-m+n,\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{2 f g (m-n)} \]
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Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2932, 12, 2746, 70} \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (1,n-m,-m+n+1,\frac {1}{2} (1-\sin (e+f x))\right )}{2 f g (m-n)} \]
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Rule 12
Rule 70
Rule 2746
Rule 2932
Rubi steps \begin{align*} \text {integral}& = \left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \frac {\sec (e+f x) (c-c \sin (e+f x))^{-m+n}}{g} \, dx \\ & = \frac {\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \sec (e+f x) (c-c \sin (e+f x))^{-m+n} \, dx}{g} \\ & = -\frac {\left (c (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \text {Subst}\left (\int \frac {(c+x)^{-1-m+n}}{c-x} \, dx,x,-c \sin (e+f x)\right )}{f g} \\ & = \frac {(g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (1,-m+n,1-m+n,\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{2 f g (m-n)} \\ \end{align*}
\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{-1-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
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\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \]
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\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+1}} \,d x \]
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