Integrand size = 38, antiderivative size = 182 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f \left (35+24 m+4 m^2\right )}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^3 f (7+2 m) \left (15+16 m+4 m^2\right )} \]
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Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2822, 2821} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^3 f (2 m+7) \left (4 m^2+16 m+15\right )}+\frac {2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c^2 f \left (4 m^2+24 m+35\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{a c f (2 m+7)} \]
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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))^{-4-m} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac {2 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m} \, dx}{a c^2 (7+2 m)} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f (5+2 m) (7+2 m)}+\frac {2 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \, dx}{a c^3 (5+2 m) (7+2 m)} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f (5+2 m) (7+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^3 f (3+2 m) (5+2 m) (7+2 m)} \\ \end{align*}
Time = 6.61 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.58 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos ^3(e+f x) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m} \left (-4 \left (6+5 m+m^2\right )+\cos (2 (e+f x))+2 (5+2 m) \sin (e+f x)\right )}{c^5 f (3+2 m) (5+2 m) (7+2 m) (-1+\sin (e+f x))^5} \]
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\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-5-m}d x\]
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.58 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=-\frac {{\left (2 \, \cos \left (f x + e\right )^{5} + 2 \, {\left (2 \, m + 5\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - {\left (4 \, m^{2} + 20 \, m + 25\right )} \cos \left (f x + e\right )^{3}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5}}{8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 5} \cos ^{2}{\left (e + f x \right )}\, dx \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5} \cos \left (f x + e\right )^{2} \,d x } \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5} \cos \left (f x + e\right )^{2} \,d x } \]
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Time = 17.22 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.84 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos \left (e+f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (24\,m^2+120\,m+140\right )}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {\cos \left (5\,e+5\,f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\cos \left (3\,e+3\,f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (8\,m^2+40\,m+45\right )}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\sin \left (4\,e+4\,f\,x\right )\,\left (m\,4{}\mathrm {i}+10{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,1{}\mathrm {i}}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\sin \left (2\,e+2\,f\,x\right )\,\left (m\,8{}\mathrm {i}+20{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,1{}\mathrm {i}}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )} \]
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