Optimal. Leaf size=96 \[ \frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x \]
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Rubi [A] time = 0.142691, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 2635, 8, 2633} \[ \frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x \]
Antiderivative was successfully verified.
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Rule 2966
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx &=\int \left (a^2 c \sin ^2(e+f x)+a^2 c \sin ^3(e+f x)-a^2 c \sin ^4(e+f x)-a^2 c \sin ^5(e+f x)\right ) \, dx\\ &=\left (a^2 c\right ) \int \sin ^2(e+f x) \, dx+\left (a^2 c\right ) \int \sin ^3(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^4(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^5(e+f x) \, dx\\ &=-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{2 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac{1}{2} \left (a^2 c\right ) \int 1 \, dx-\frac{1}{4} \left (3 a^2 c\right ) \int \sin ^2(e+f x) \, dx-\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{1}{2} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}-\frac{1}{8} \left (3 a^2 c\right ) \int 1 \, dx\\ &=\frac{1}{8} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}\\ \end{align*}
Mathematica [A] time = 0.0860954, size = 57, normalized size = 0.59 \[ \frac{a^2 c (-15 \sin (4 (e+f x))-60 \cos (e+f x)-10 \cos (3 (e+f x))+6 \cos (5 (e+f x))+60 e+60 f x)}{480 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 126, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ({\frac{{a}^{2}c\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }-{a}^{2}c \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) -{\frac{{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.977755, size = 166, normalized size = 1.73 \begin{align*} \frac{32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} c + 160 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c - 15 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c}{480 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.90476, size = 189, normalized size = 1.97 \begin{align*} \frac{24 \, a^{2} c \cos \left (f x + e\right )^{5} - 40 \, a^{2} c \cos \left (f x + e\right )^{3} + 15 \, a^{2} c f x - 15 \,{\left (2 \, a^{2} c \cos \left (f x + e\right )^{3} - a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.49798, size = 301, normalized size = 3.14 \begin{align*} \begin{cases} - \frac{3 a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{3 a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{3 a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{a^{2} c \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{5 a^{2} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{4 a^{2} c \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{3 a^{2} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{8 a^{2} c \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{2 a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right ) \sin ^{2}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.27511, size = 109, normalized size = 1.14 \begin{align*} \frac{1}{8} \, a^{2} c x + \frac{a^{2} c \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{a^{2} c \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{a^{2} c \cos \left (f x + e\right )}{8 \, f} - \frac{a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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