Integrand size = 33, antiderivative size = 170 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {2^{\frac {1}{2} (1+2 m+p)} a (B m+A (1+m+p)) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m-p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2} (1-2 m-p)} (a+a \sin (e+f x))^{-1+m}}{f g (1+p) (1+m+p)}-\frac {B (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g (1+m+p)} \]
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Time = 0.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2939, 2768, 72, 71} \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a 2^{\frac {1}{2} (2 m+p+1)} (A (m+p+1)+B m) (a \sin (e+f x)+a)^{m-1} (g \cos (e+f x))^{p+1} (\sin (e+f x)+1)^{\frac {1}{2} (-2 m-p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2 m-p+1),\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f g (p+1) (m+p+1)}-\frac {B (a \sin (e+f x)+a)^m (g \cos (e+f x))^{p+1}}{f g (m+p+1)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps \begin{align*} \text {integral}& = -\frac {B (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g (1+m+p)}+\left (A+\frac {B m}{1+m+p}\right ) \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m \, dx \\ & = -\frac {B (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g (1+m+p)}+\frac {\left (a^2 \left (A+\frac {B m}{1+m+p}\right ) (g \cos (e+f x))^{1+p} (a-a \sin (e+f x))^{\frac {1}{2} (-1-p)} (a+a \sin (e+f x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (-1+p)} (a+a x)^{m+\frac {1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{f g} \\ & = -\frac {B (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g (1+m+p)}+\frac {\left (2^{-\frac {1}{2}+m+\frac {p}{2}} a^2 \left (A+\frac {B m}{1+m+p}\right ) (g \cos (e+f x))^{1+p} (a-a \sin (e+f x))^{\frac {1}{2} (-1-p)} (a+a \sin (e+f x))^{-\frac {1}{2}+m+\frac {1}{2} (-1-p)+\frac {p}{2}} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m-\frac {p}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{m+\frac {1}{2} (-1+p)} (a-a x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{f g} \\ & = -\frac {2^{\frac {1}{2} (1+2 m+p)} a \left (A+\frac {B m}{1+m+p}\right ) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m-p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2} (1-2 m-p)} (a+a \sin (e+f x))^{-1+m}}{f g (1+p)}-\frac {B (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g (1+m+p)} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (g \cos (e+f x))^p (1+\sin (e+f x))^{\frac {1}{2} (-1-2 m-p)} (a (1+\sin (e+f x)))^m \left (2^{\frac {1}{2} (1+2 m+p)} (B m+A (1+m+p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m-p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x))\right )+B (1+p) (1+\sin (e+f x))^{\frac {1}{2} (1+2 m+p)}\right )}{f (1+p) (1+m+p)} \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (g \cos {\left (e + f x \right )}\right )^{p} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
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