Integrand size = 38, antiderivative size = 113 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {2^{-\frac {1}{2}-m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)} \]
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Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2920, 2824, 2768, 72, 71} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {2^{-m-\frac {1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2824
Rule 2920
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-1-m} \, dx}{a c} \\ & = \left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 (1+m)}(e+f x) (c-c \sin (e+f x))^{-2-2 m} \, dx \\ & = \frac {\left (c^2 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 (1+m))} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \text {Subst}\left (\int (c-c x)^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (2^{-\frac {3}{2}-m} c \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 (1+m))} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {2^{-\frac {1}{2}-m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)} \\ \end{align*}
Result contains complex when optimal does not.
Time = 14.20 (sec) , antiderivative size = 453, normalized size of antiderivative = 4.01 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {2^{-1-m} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-2-m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {1}{1+\cos (e+f x)}}}\right )^{2 m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}\right )^{-2 m} \left (\frac {i \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )}{m}-\frac {i \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )}{m}+2^{1+2 m} \left (\frac {2 \operatorname {Hypergeometric2F1}\left (-2 m,-2 (1+m),1-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{m}-\frac {4 \operatorname {Hypergeometric2F1}\left (-1-2 m,-2 (1+m),-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{(1+2 m) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}-\frac {2 \operatorname {Hypergeometric2F1}\left (-1-2 m,1-2 m,2-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-1+2 m}-\frac {\operatorname {Hypergeometric2F1}\left (1-2 m,-2 m,2-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-1+2 m}\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^{-2 m}\right )}{f} \]
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\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m}d x\]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \cos \left (f x + e\right )^{2} \,d x } \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 2} \cos ^{2}{\left (e + f x \right )}\, dx \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \cos \left (f x + e\right )^{2} \,d x } \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \cos \left (f x + e\right )^{2} \,d x } \]
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Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \]
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