Integrand size = 34, antiderivative size = 77 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=-\frac {2^{\frac {3}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{c f} \]
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Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2919, 2731, 2730} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=-\frac {2^{m+\frac {3}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m-\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{c f} \]
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Rule 2730
Rule 2731
Rule 2919
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{1+m} \, dx}{a c} \\ & = \frac {\left ((1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^{1+m} \, dx}{c} \\ & = -\frac {2^{\frac {3}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {3}{2},\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 f} \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 2.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\frac {2^{1+m} B_{\frac {1}{2} (1+\sin (e+f x))}\left (\frac {3}{2}+m,\frac {1}{2}\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m}{c f} \]
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\[\int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{c -c \sin \left (f x +e \right )}d x\]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=- \frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m} \cos ^{2}{\left (e + f x \right )}}{\sin {\left (e + f x \right )} - 1}\, dx}{c} \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{c-c\,\sin \left (e+f\,x\right )} \,d x \]
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