Integrand size = 30, antiderivative size = 114 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\frac {2^{\frac {3}{2}-m} c^2 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)} \]
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Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2824, 2768, 72, 71} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\frac {c^2 2^{\frac {3}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m-1),\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2824
Rubi steps \begin{align*} \text {integral}& = \left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{1-2 m} \, dx \\ & = \frac {\left (c^2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int (c-c x)^{1-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (2^{\frac {1}{2}-m} c^3 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 m)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{1-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {2^{\frac {3}{2}-m} c^2 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\frac {2^{\frac {1}{2}+m} 3^m c \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {3}{2}-m,\frac {5}{2}-m,\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (-1+\sin (e+f x))^2 \sqrt {1+\sin (e+f x)} (c-c \sin (e+f x))^{-m}}{f (-3+2 m)} \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{1-m}d x\]
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\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{1 - m}\, dx \]
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\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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\[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} \,d x } \]
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Timed out. \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{1-m} \,d x \]
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