Optimal. Leaf size=132 \[ -\frac{a^3 \cos ^7(c+d x)}{7 d}+\frac{a^3 \cos ^5(c+d x)}{d}-\frac{4 a^3 \cos ^3(c+d x)}{3 d}-\frac{a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16} \]
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Rubi [A] time = 0.30451, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2873, 2568, 2635, 8, 2565, 14, 270} \[ -\frac{a^3 \cos ^7(c+d x)}{7 d}+\frac{a^3 \cos ^5(c+d x)}{d}-\frac{4 a^3 \cos ^3(c+d x)}{3 d}-\frac{a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16} \]
Antiderivative was successfully verified.
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Rule 2873
Rule 2568
Rule 2635
Rule 8
Rule 2565
Rule 14
Rule 270
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^2(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^3(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^4(c+d x)+a^3 \cos ^2(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+a^3 \int \cos ^2(c+d x) \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac{a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac{1}{4} a^3 \int \cos ^2(c+d x) \, dx+\frac{1}{2} \left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac{1}{8} a^3 \int 1 \, dx+\frac{1}{8} \left (3 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a^3 x}{8}-\frac{4 a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos ^5(c+d x)}{d}-\frac{a^3 \cos ^7(c+d x)}{7 d}+\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac{1}{16} \left (3 a^3\right ) \int 1 \, dx\\ &=\frac{5 a^3 x}{16}-\frac{4 a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos ^5(c+d x)}{d}-\frac{a^3 \cos ^7(c+d x)}{7 d}+\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.647043, size = 86, normalized size = 0.65 \[ \frac{a^3 (-63 \sin (2 (c+d x))-105 \sin (4 (c+d x))+21 \sin (6 (c+d x))-609 \cos (c+d x)-91 \cos (3 (c+d x))+63 \cos (5 (c+d x))-3 \cos (7 (c+d x))+420 c+420 d x)}{1344 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 194, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{7}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{35}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{105}} \right ) +3\,{a}^{3} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1/8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +3\,{a}^{3} \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 ) +{a}^{3} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.11474, size = 174, normalized size = 1.32 \begin{align*} -\frac{64 \,{\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 1344 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 105 \,{\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 )} a^{3} - 210 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{6720 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.7867, size = 248, normalized size = 1.88 \begin{align*} -\frac{48 \, a^{3} \cos \left (d x + c\right )^{7} - 336 \, a^{3} \cos \left (d x + c\right )^{5} + 448 \, a^{3} \cos \left (d x + c\right )^{3} - 105 \, a^{3} d x - 21 \,{\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 18 \, a^{3} \cos \left (d x + c\right )^{3} + 5 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 9.70729, size = 379, normalized size = 2.87 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac{a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{4 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac{3 a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{8 a^{3} \cos ^{7}{\left (c + d x \right )}}{105 d} - \frac{2 a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \sin ^{2}{\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 = 1.34728, size = 166, normalized size = 1.26 \begin{align*} \frac{5}{16} \, a^{3} x - \frac{a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{3 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac{13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{29 \, a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac{a^{3} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac{5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac{3 \, a^{3} \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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