Optimal. Leaf size=105 \[ \frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^3(c+d x)}{3 d}-\frac{b \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac{b \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{b \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b x}{16} \]
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Rubi [A] time = 0.163955, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2838, 2565, 14, 2568, 2635, 8} \[ \frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^3(c+d x)}{3 d}-\frac{b \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac{b \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{b \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b x}{16} \]
Antiderivative was successfully verified.
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Rule 2838
Rule 2565
Rule 14
Rule 2568
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+b \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac{b \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac{1}{2} b \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{b \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac{1}{8} b \int \cos ^2(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos ^5(c+d x)}{5 d}+\frac{b \cos (c+d x) \sin (c+d x)}{16 d}-\frac{b \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{b \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}+\frac{1}{16} b \int 1 \, dx\\ &=\frac{b x}{16}-\frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos ^5(c+d x)}{5 d}+\frac{b \cos (c+d x) \sin (c+d x)}{16 d}-\frac{b \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{b \cos ^3(c+d x) \sin ^3(c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 0.186917, size = 77, normalized size = 0.73 \[ \frac{-120 a \cos (c+d x)-20 a \cos (3 (c+d x))+12 a \cos (5 (c+d x))-15 b \sin (2 (c+d x))-15 b \sin (4 (c+d x))+5 b \sin (6 (c+d x))+60 b d x}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.03, size = 95, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +b \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 = 1.10122, size = 88, normalized size = 0.84 \begin{align*} \frac{64 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a - 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}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.52557, size = 193, normalized size = 1.84 \begin{align*} \frac{48 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, b d x + 5 \,{\left (8 \, b \cos \left (d x + c\right )^{5} - 14 \, b \cos \left (d x + c\right )^{3} + 3 \, b \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 = 4.25816, size = 192, normalized size = 1.83 \begin{align*} \begin{cases} - \frac{a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac{b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{b \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{b \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 ) \sin ^{3}{\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 = 1.30691, size = 124, normalized size = 1.18 \begin{align*} \frac{1}{16} \, b x + \frac{a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{a \cos \left (d x + c\right )}{8 \, d} + \frac{b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac{b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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