Integrand size = 38, antiderivative size = 160 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\frac {2^{\frac {1}{2}-\frac {p}{2}} c^2 (2 A-B (1-p)) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1+p}{2}} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g} \]
[Out]
Time = 0.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2939, 2768, 72, 71} \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\frac {c^2 2^{\frac {1}{2}-\frac {p}{2}} (2 A-B (1-p)) (1-\sin (e+f x))^{\frac {p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {p-1}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)}-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g} \]
[In]
[Out]
Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps \begin{align*} \text {integral}& = -\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g}-\frac {(-2 A c+B c (1-p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{1-p} \, dx}{2 c} \\ & = -\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g}-\frac {\left (c (-2 A c+B c (1-p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{\frac {1}{2} (-1-p)} (c+c \sin (e+f x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int (c-c x)^{1+\frac {1}{2} (-1+p)-p} (c+c x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{2 f g} \\ & = -\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g}-\frac {\left (2^{-\frac {1}{2}-\frac {p}{2}} c^2 (-2 A c+B c (1-p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-p)-\frac {p}{2}} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+\frac {p}{2}} (c+c \sin (e+f x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{1+\frac {1}{2} (-1+p)-p} (c+c x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{f g} \\ & = \frac {2^{\frac {1}{2}-\frac {p}{2}} c^2 (2 A-B (1-p)) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+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+p}{2}} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\frac {2^{\frac {1}{2} (-3-p)} c \cos (e+f x) (g \cos (e+f x))^p \left (-4 (2 A+B (-1+p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1+p}{2}}+2^{\frac {1+p}{2}} B (1+p) (-1+\sin (e+f x))^2\right ) (c-c \sin (e+f x))^{-p}}{f (1+p) (-1+\sin (e+f x))} \]
[In]
[Out]
\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{1-p}d x\]
[In]
[Out]
\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \]
[In]
[Out]
\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{1 - p} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
[In]
[Out]
\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \]
[In]
[Out]
\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \]
[In]
[Out]
Timed out. \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{1-p} \,d x \]
[In]
[Out]