Integrand size = 33, antiderivative size = 145 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {2^{\frac {3}{2}+\frac {p}{2}} a \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {1}{2} (-1-p),-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{1+p} (1+\sin (e+f x))^{\frac {1}{2} (-1-p)} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f g (1+p)} \]
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Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2947, 144, 143} \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {a g 2^{\frac {p+3}{2}} (1-\sin (e+f x)) (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {p+1}{2},\frac {1}{2} (-p-1),-n,\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f (p+1)} \]
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Rule 143
Rule 144
Rule 2947
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac {1-p}{2}} (1+\sin (e+f x))^{\frac {1-p}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{\frac {1+p}{2}} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (a g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac {1-p}{2}} (1+\sin (e+f x))^{\frac {1-p}{2}} (c+d \sin (e+f x))^n \left (-\frac {c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{\frac {1+p}{2}} \left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^n \, dx,x,\sin (e+f x)\right )}{f} \\ & = -\frac {2^{\frac {3+p}{2}} a g \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {1}{2} (-1-p),-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{-1+p} (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac {1-p}{2}} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f (1+p)} \\ \end{align*}
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=a \left (\int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (c + d \sin {\left (e + f x \right )}\right )^{n}\, dx + \int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (c + d \sin {\left (e + f x \right )}\right )^{n} \sin {\left (e + f x \right )}\, dx\right ) \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]
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