Optimal. Leaf size=163 \[ -\frac{a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}+\frac{b \left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} b x \left (6 a^2+b^2\right )-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d} \]
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Rubi [A] time = 0.294542, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac{a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}+\frac{b \left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} b x \left (6 a^2+b^2\right )-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d} \]
Antiderivative was successfully verified.
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Rule 2862
Rule 2669
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac{1}{6} \int \cos ^2(c+d x) (3 b+3 a \sin (c+d x)) (a+b \sin (c+d x))^2 \, dx\\ &=-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac{1}{30} \int \cos ^2(c+d x) (a+b \sin (c+d x)) \left (21 a b+3 \left (2 a^2+5 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac{1}{120} \int \cos ^2(c+d x) \left (15 b \left (6 a^2+b^2\right )+3 a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac{1}{8} \left (b \left (6 a^2+b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}+\frac{b \left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac{1}{16} \left (b \left (6 a^2+b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{16} b \left (6 a^2+b^2\right ) x-\frac{a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}+\frac{b \left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac{a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac{\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}\\ \end{align*}
Mathematica [A] time = 0.76512, size = 138, normalized size = 0.85 \[ \frac{-120 a \left (2 a^2+3 b^2\right ) \cos (c+d x)-20 \left (4 a^3+3 a b^2\right ) \cos (3 (c+d x))+b \left (5 \left (-3 \left (6 a^2+b^2\right ) \sin (4 (c+d x))+72 a^2 c+72 a^2 d x-3 b^2 \sin (2 (c+d x))+b^2 \sin (6 (c+d x))+18 b^2 c+12 b^2 d x\right )+36 a b \cos (5 (c+d x))\right )}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.046, size = 158, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+3\,{a}^{2}b \left ( -1/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/8\,dx+c/8 \right ) +3\,a{b}^{2} \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +{b}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{8}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.13537, size = 146, normalized size = 0.9 \begin{align*} -\frac{320 \, a^{3} \cos \left (d x + c\right )^{3} - 90 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b - 192 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b^{2} + 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.47875, size = 281, normalized size = 1.72 \begin{align*} \frac{144 \, a b^{2} \cos \left (d x + c\right )^{5} - 80 \,{\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (6 \, a^{2} b + b^{3}\right )} d x + 5 \,{\left (8 \, b^{3} \cos \left (d x + c\right )^{5} - 2 \,{\left (18 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 4.75514, size = 340, normalized size = 2.09 \begin{align*} \begin{cases} - \frac{a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 a^{2} b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{2} b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a^{2} b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{a b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac{2 a b^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{b^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{b^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{b^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{b^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \sin{\left (c \right )} \cos ^{2}{\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 = 1.19758, size = 188, normalized size = 1.15 \begin{align*} \frac{3 \, a b^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{b^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{b^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{1}{16} \,{\left (6 \, a^{2} b + b^{3}\right )} x - \frac{{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac{{\left (6 \, a^{2} b + b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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