Integrand size = 45, antiderivative size = 134 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\frac {2^{3-\frac {m}{2}+\frac {n}{2}} c^3 (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-4+m-n),\frac {m-n}{2},\frac {1}{2} (2+m-n),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {m-n}{2}} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (m-n)} \]
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Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {2932, 2768, 72, 71} \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\frac {c^3 2^{-\frac {m}{2}+\frac {n}{2}+3} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac {m-n}{2}} (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (m-n-4),\frac {m-n}{2},\frac {1}{2} (m-n+2),\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)} \]
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Rule 71
Rule 72
Rule 2768
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 (g \cos (e+f x))^{-1+m-n} (c-c \sin (e+f x))^{3-m+n} \, dx \\ & = \frac {\left (c^2 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-m+n)} (c+c \sin (e+f x))^{\frac {1}{2} (-m+n)}\right ) \text {Subst}\left (\int (c-c x)^{3-m+\frac {1}{2} (-2+m-n)+n} (c+c x)^{\frac {1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g} \\ & = \frac {\left (2^{2-\frac {m}{2}+\frac {n}{2}} c^4 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {m}{2}+\frac {n}{2}+\frac {1}{2} (-m+n)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {m}{2}-\frac {n}{2}} (c+c \sin (e+f x))^{\frac {1}{2} (-m+n)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{3-m+\frac {1}{2} (-2+m-n)+n} (c+c x)^{\frac {1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g} \\ & = \frac {2^{3-\frac {m}{2}+\frac {n}{2}} c^3 (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-4+m-n),\frac {m-n}{2},\frac {1}{2} (2+m-n),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {m-n}{2}} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (m-n)} \\ \end{align*}
\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{-1-m -n} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{3+n}d x\]
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\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\text {Timed out} \]
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\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
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\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{n+3}}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{m+n+1}} \,d x \]
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