Integrand size = 40, antiderivative size = 88 \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {c^2 (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (3,-2-m+n,-1-m+n,\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n}}{8 f g^5 (2+m-n)} \]
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Time = 0.16 (sec) , antiderivative size = 88, 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} \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {c^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (3,-m+n-2,-m+n-1,\frac {1}{2} (1-\sin (e+f x))\right )}{8 f g^5 (m-n+2)} \]
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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 ^5(e+f x) (c-c \sin (e+f x))^{-m+n}}{g^5} \, 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 ^5(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx}{g^5} \\ & = -\frac {\left (c^5 (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)^{-3-m+n}}{(c-x)^3} \, dx,x,-c \sin (e+f x)\right )}{f g^5} \\ & = \frac {c^2 (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (3,-2-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))^{-2+n}}{8 f g^5 (2+m-n)} \\ \end{align*}
\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{-5-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
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\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \]
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\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+5}} \,d x \]
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