Optimal. Leaf size=180 \[ \frac{2 c (A (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}+\frac{4 c C (2 m+1) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.56964, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {3040, 2971, 2738} \[ \frac{2 c (A (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}+\frac{4 c C (2 m+1) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3040
Rule 2971
Rule 2738
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \left (A+C \sin ^2(e+f x)\right ) \, dx &=\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}-\frac{2 \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \left (-\frac{1}{2} a c (C (3-2 m)+A (5+2 m))-a c C (1+2 m) \sin (e+f x)\right ) \, dx}{a c (5+2 m)}\\ &=\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}+\frac{(2 C (1+2 m)) \int (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)} \, dx}{a (5+2 m)}+\frac{(C-6 C m+A (5+2 m)) \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx}{5+2 m}\\ &=\frac{2 c (C-6 C m+A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{4 c C (1+2 m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}\\ \end{align*}
Mathematica [A] time = 0.800659, size = 160, normalized size = 0.89 \[ -\frac{\sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m \left (-8 A m^2-32 A m-30 A+C \left (4 m^2+8 m+3\right ) \cos (2 (e+f x))+8 C (2 m+1) \sin (e+f x)-4 C m^2-8 C m-19 C\right )}{f (2 m+1) (2 m+3) (2 m+5) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.675, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\sqrt{c-c\sin \left ( fx+e \right ) } \left ( A+C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.62967, size = 595, normalized size = 3.31 \begin{align*} \frac{2 \,{\left (\frac{4 \,{\left (\frac{4 \, a^{m} \sqrt{c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt{c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{4 \, a^{m} \sqrt{c} m \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2 \, a^{m} \sqrt{c} - \frac{2 \, a^{m} \sqrt{c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} C 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 (8 \, m^{3} + 36 \, m^{2} + 46 \, m + \frac{2 \,{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac{{\left (a^{m} \sqrt{c} + \frac{a^{m} \sqrt{c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A 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 (2 \, m + 1\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.79447, size = 659, normalized size = 3.66 \begin{align*} -\frac{2 \,{\left ({\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 4 \,{\left (A + C\right )} m^{2} +{\left (4 \, C m^{2} - C\right )} \cos \left (f x + e\right )^{2} - 16 \, A m -{\left (4 \,{\left (A + C\right )} m^{2} + 8 \,{\left (2 \, A + C\right )} m + 15 \, A + 11 \, C\right )} \cos \left (f x + e\right ) -{\left (4 \,{\left (A + C\right )} m^{2} -{\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{2} + 16 \, A m - 4 \,{\left (2 \, C m + C\right )} \cos \left (f x + e\right ) + 15 \, A + 7 \, C\right )} \sin \left (f x + e\right ) - 15 \, A - 7 \, C\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m +{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) -{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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