Integrand size = 36, antiderivative size = 138 \[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+n-\frac {p}{2}} c \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m-p),\frac {1}{2} (1-2 n+p),\frac {1}{2} (3+2 m-p),\frac {1}{2} (1+\sin (e+f x))\right ) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1}{2} (1-2 n+p)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m-p)} \]
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Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3005, 2932, 2768, 72, 71} \[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {c 2^{n-\frac {p}{2}+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1}{2} (-2 n+p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m-p+1),\frac {1}{2} (-2 n+p+1),\frac {1}{2} (2 m-p+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m-p+1)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2932
Rule 3005
Rubi steps \begin{align*} \text {integral}& = \left ((g \cos (e+f x))^p (g \sec (e+f x))^p\right ) \int (g \cos (e+f x))^{-p} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \\ & = \left ((g \cos (e+f x))^{-2 m+p} (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int (g \cos (e+f x))^{2 m-p} (c-c \sin (e+f x))^{-m+n} \, dx \\ & = \frac {\left (c^2 \cos (e+f x) (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 m+p)} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m+p)}\right ) \text {Subst}\left (\int (c-c x)^{-m+n+\frac {1}{2} (-1+2 m-p)} (c+c x)^{\frac {1}{2} (-1+2 m-p)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (2^{-\frac {1}{2}+n-\frac {p}{2}} c^2 \cos (e+f x) (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+m+n-\frac {p}{2}+\frac {1}{2} (-1-2 m+p)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}-n+\frac {p}{2}} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m+p)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-m+n+\frac {1}{2} (-1+2 m-p)} (c+c x)^{\frac {1}{2} (-1+2 m-p)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {2^{\frac {1}{2}+n-\frac {p}{2}} c \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m-p),\frac {1}{2} (1-2 n+p),\frac {1}{2} (3+2 m-p),\frac {1}{2} (1+\sin (e+f x))\right ) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1}{2} (1-2 n+p)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m-p)} \\ \end{align*}
\[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \]
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\[\int \left (g \sec \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
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\[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \]
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\[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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\[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int {\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \]
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