Integrand size = 38, antiderivative size = 102 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\frac {(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-2-p}}{f g (3+p)}+\frac {(A-B (2+p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{c f g (1+p) (3+p)} \]
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Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2938, 2750} \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\frac {(A+B) (c-c \sin (e+f x))^{-p-2} (g \cos (e+f x))^{p+1}}{f g (p+3)}+\frac {(A-B (p+2)) (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1}}{c f g (p+1) (p+3)} \]
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Rule 2750
Rule 2938
Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-2-p}}{f g (3+p)}+\frac {(A-B (2+p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{-1-p} \, dx}{c (3+p)} \\ & = \frac {(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-2-p}}{f g (3+p)}+\frac {(A-B (2+p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{c f g (1+p) (3+p)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\frac {\cos (e+f x) (g \cos (e+f x))^p (c-c \sin (e+f x))^{-p} (-B+A (2+p)+(-A+B (2+p)) \sin (e+f x))}{c^2 f (1+p) (3+p) (-1+\sin (e+f x))^2} \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-2-p}d x\]
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none
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\frac {{\left ({\left (B p - A + 2 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (A p + 2 \, A - B\right )} \cos \left (f x + e\right )\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p - 2}}{f p^{2} + 4 \, f p + 3 \, f} \]
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\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- p - 2} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p - 2} \,d x } \]
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\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p - 2} \,d x } \]
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Time = 1.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.26 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-p} \, dx=-\frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (4\,A\,\cos \left (e+f\,x\right )-2\,B\,\cos \left (e+f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+2\,B\,\sin \left (2\,e+2\,f\,x\right )+2\,A\,p\,\cos \left (e+f\,x\right )+B\,p\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^2\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^p\,\left (4\,\sin \left (e+f\,x\right )+\cos \left (2\,e+2\,f\,x\right )-3\right )\,\left (p^2+4\,p+3\right )} \]
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