Integrand size = 36, antiderivative size = 107 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\frac {8 c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f \left (15+16 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f (5+2 m)} \]
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Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\frac {8 c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+16 m+15\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+5)} \]
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Rule 2817
Rule 2819
Rule 2920
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2} \, dx}{a c} \\ & = \frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f (5+2 m)}+\frac {4 \int (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)} \, dx}{a (5+2 m)} \\ & = \frac {8 c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f \left (15+16 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f (5+2 m)} \\ \end{align*}
Time = 1.55 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} (-7-2 m+(3+2 m) \sin (e+f x))}{f (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
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\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}d x\]
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none
Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.45 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 \, {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, m - 1\right )} \cos \left (f x + e\right )^{2} + {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 8\right )} \sin \left (f x + e\right ) + 4 \, \cos \left (f x + e\right ) + 8\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 16 \, f m + {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \cos ^{2}{\left (e + f x \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (103) = 206\).
Time = 0.37 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.92 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (a^{m} \sqrt {c} {\left (2 \, m + 7\right )} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 7\right )} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 16 \, m + \frac {2 \, {\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} f \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\int { \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2} \,d x } \]
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Time = 2.02 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (25\,\cos \left (e+f\,x\right )+3\,\cos \left (3\,e+3\,f\,x\right )+8\,\sin \left (2\,e+2\,f\,x\right )+6\,m\,\cos \left (e+f\,x\right )+2\,m\,\cos \left (3\,e+3\,f\,x\right )\right )}{2\,f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (4\,m^2+16\,m+15\right )} \]
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