Optimal. Leaf size=92 \[ \frac{2 b \left (2 a^2-b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} a x \left (2 a^2-b^2\right )+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{6 d}-\frac{b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]
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Rubi [A] time = 0.112424, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3016, 2753, 2734} \[ \frac{2 b \left (2 a^2-b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} a x \left (2 a^2-b^2\right )+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{6 d}-\frac{b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 3016
Rule 2753
Rule 2734
Rubi steps
\begin{align*} \int (a+b \cos (c+d x)) \left (a^2-b^2 \cos ^2(c+d x)\right ) \, dx &=-\int (-a+b \cos (c+d x)) (a+b \cos (c+d x))^2 \, dx\\ &=-\frac{b (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}-\frac{1}{3} \int (a+b \cos (c+d x)) \left (-3 a^2+2 b^2-a b \cos (c+d x)\right ) \, dx\\ &=\frac{1}{2} a \left (2 a^2-b^2\right ) x+\frac{2 b \left (2 a^2-b^2\right ) \sin (c+d x)}{3 d}+\frac{a b^2 \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.168031, size = 75, normalized size = 0.82 \[ -\frac{\left (9 b^3-12 a^2 b\right ) \sin (c+d x)-12 a^3 d x+3 a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x+b^3 \sin (3 (c+d x))}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.018, size = 75, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}-a{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{a}^{2}b\sin \left ( dx+c \right ) +{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.994795, size = 99, normalized size = 1.08 \begin{align*} \frac{12 \,{\left (d x + c\right )} a^{3} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} + 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{3} + 12 \, a^{2} b \sin \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.38727, size = 149, normalized size = 1.62 \begin{align*} \frac{3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x -{\left (2 \, b^{3} \cos \left (d x + c\right )^{2} + 3 \, a b^{2} \cos \left (d x + c\right ) - 6 \, a^{2} b + 4 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0.962799, size = 131, normalized size = 1.42 \begin{align*} \begin{cases} a^{3} x + \frac{a^{2} b \sin{\left (c + d x \right )}}{d} - \frac{a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} - \frac{a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{a b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{2 b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right ) \left (a^{2} - b^{2} \cos ^{2}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.38413, size = 100, normalized size = 1.09 \begin{align*} -\frac{b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{a b^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{1}{2} \,{\left (2 \, a^{3} - a b^{2}\right )} x + \frac{{\left (4 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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