Optimal. Leaf size=96 \[ \frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a^3 A x \]
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Rubi [A] time = 0.116473, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2966, 2638, 2635, 8, 2633} \[ \frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a^3 A x \]
Antiderivative was successfully verified.
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Rule 2966
Rule 2638
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\int \left (a^3 A \sin (c+d x)+2 a^3 A \sin ^2(c+d x)-2 a^3 A \sin ^4(c+d x)-a^3 A \sin ^5(c+d x)\right ) \, dx\\ &=\left (a^3 A\right ) \int \sin (c+d x) \, dx-\left (a^3 A\right ) \int \sin ^5(c+d x) \, dx+\left (2 a^3 A\right ) \int \sin ^2(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^4(c+d x) \, dx\\ &=-\frac{a^3 A \cos (c+d x)}{d}-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{d}+\frac{a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}+\left (a^3 A\right ) \int 1 \, dx-\frac{1}{2} \left (3 a^3 A\right ) \int \sin ^2(c+d x) \, dx+\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{1}{4} \left (3 a^3 A\right ) \int 1 \, dx\\ &=\frac{1}{4} a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.482205, size = 55, normalized size = 0.57 \[ \frac{a^3 A (-90 \cos (c+d x)-25 \cos (3 (c+d x))+3 (-5 \sin (4 (c+d x))+\cos (5 (c+d x))+20 d x))}{240 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 117, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3}A\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-2\,{a}^{3}A \left ( -1/4\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +2\,{a}^{3}A \left ( -1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) -{a}^{3}A\cos \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02804, size = 151, normalized size = 1.57 \begin{align*} \frac{16 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} A a^{3} - 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 120 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 240 \, A a^{3} \cos \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.17571, size = 188, normalized size = 1.96 \begin{align*} \frac{12 \, A a^{3} \cos \left (d x + c\right )^{5} - 40 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} d x - 15 \,{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 6.07396, size = 267, normalized size = 2.78 \begin{align*} \begin{cases} - \frac{3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac{3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + A a^{3} x \sin ^{2}{\left (c + d x \right )} - \frac{3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + A a^{3} x \cos ^{2}{\left (c + d x \right )} + \frac{A a^{3} \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{4 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 A a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac{A a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{8 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{A a^{3} \cos{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (- A \sin{\left (c \right )} + A\right ) \left (a \sin{\left (c \right )} + a\right )^{3} \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.11934, size = 104, normalized size = 1.08 \begin{align*} \frac{1}{4} \, A a^{3} x + \frac{A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{3 \, A a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac{A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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