Optimal. Leaf size=131 \[ -\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (4 a^2+3 b^2\right )-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d} \]
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Rubi [A] time = 0.193578, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2692, 2862, 2669, 2635, 8} \[ -\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (4 a^2+3 b^2\right )-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d} \]
Antiderivative was successfully verified.
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Rule 2692
Rule 2862
Rule 2669
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{5} \int \cos ^2(c+d x) (a+b \sin (c+d x)) \left (5 a^2+2 b^2+7 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{20} \int \cos ^2(c+d x) \left (5 a \left (4 a^2+3 b^2\right )+b \left (27 a^2+8 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{4} \left (a \left (4 a^2+3 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{8} \left (a \left (4 a^2+3 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{8} a \left (4 a^2+3 b^2\right ) x-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\\ \end{align*}
Mathematica [A] time = 0.539743, size = 107, normalized size = 0.82 \[ \frac{15 a \left (4 \left (4 a^2+3 b^2\right ) (c+d x)+8 a^2 \sin (2 (c+d x))-3 b^2 \sin (4 (c+d x))\right )-60 b \left (6 a^2+b^2\right ) \cos (c+d x)-10 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))+6 b^3 \cos (5 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.05, size = 123, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{3} \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 ) +3\,a{b}^{2} \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 ) -{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{a}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.961946, size = 126, normalized size = 0.96 \begin{align*} -\frac{480 \, a^{2} b \cos \left (d x + c\right )^{3} - 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 45 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 32 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.23542, size = 232, normalized size = 1.77 \begin{align*} \frac{24 \, b^{3} \cos \left (d x + c\right )^{5} - 40 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} d x - 15 \,{\left (6 \, a b^{2} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.59479, size = 236, normalized size = 1.8 \begin{align*} \begin{cases} \frac{a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{a^{2} b \cos ^{3}{\left (c + d x \right )}}{d} + \frac{3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \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 = 1.08993, size = 153, normalized size = 1.17 \begin{align*} \frac{b^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{1}{8} \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} x - \frac{{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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