Integrand size = 38, antiderivative size = 114 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^2 f \left (15+16 m+4 m^2\right )} \]
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Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2822, 2821} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^2 f \left (4 m^2+16 m+15\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c f (2 m+5)} \]
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Rule 2821
Rule 2822
Rule 2920
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \, dx}{a c^2 (5+2 m)} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^2 f (3+2 m) (5+2 m)} \\ \end{align*}
Time = 5.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\cos ^3(e+f x) (4+2 m-\sin (e+f x)) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m}}{c^4 f (3+2 m) (5+2 m) (-1+\sin (e+f x))^4} \]
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\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-4-m}d x\]
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none
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {{\left (2 \, {\left (m + 2\right )} \cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4}}{4 \, f m^{2} + 16 \, f m + 15 \, f} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 4} \cos ^{2}{\left (e + f x \right )}\, dx \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2} \,d x } \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2} \,d x } \]
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Time = 11.74 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.55 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+48\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+16\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+12\,m\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+4\,m\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )-32\right )}{c^4\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+16\,m+15\right )\,\left (56\,{\sin \left (e+f\,x\right )}^2-56\,\sin \left (e+f\,x\right )-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+8\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]
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