Optimal. Leaf size=120 \[ \frac {2 b \left (-2 a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (-2 a^3 C+2 a^2 b B+a b^2 C+b^3 B\right )+\frac {b^2 (3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]
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Rubi [A] time = 0.21, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3015, 2753, 2734} \[ \frac {2 b \left (-2 a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (2 a^2 b B-2 a^3 C+a b^2 C+b^3 B\right )+\frac {b^2 (3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 2734
Rule 2753
Rule 3015
Rubi steps
\begin {align*} \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \cos (c+d x))^2 \left (b^2 (b B-a C)+b^3 C \cos (c+d x)\right ) \, dx}{b^2}\\ &=\frac {b C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {\int (a+b \cos (c+d x)) \left (b^2 \left (2 b^2 C+3 a (b B-a C)\right )+b^3 (3 b B-a C) \cos (c+d x)\right ) \, dx}{3 b^2}\\ &=\frac {1}{2} \left (2 a^2 b B+b^3 B-2 a^3 C+a b^2 C\right ) x+\frac {2 b \left (3 a b B-2 a^2 C+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 b B-a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 102, normalized size = 0.85 \[ \frac {3 b \left (-4 a^2 C+8 a b B+3 b^2 C\right ) \sin (c+d x)-6 (c+d x) \left (2 a^3 C-2 a^2 b B-a b^2 C-b^3 B\right )+3 b^2 (a C+b B) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.42, size = 100, normalized size = 0.83 \[ -\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} d x - {\left (2 \, C b^{3} \cos \left (d x + c\right )^{2} - 6 \, C a^{2} b + 12 \, B a b^{2} + 4 \, C b^{3} + 3 \, {\left (C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 107, normalized size = 0.89 \[ \frac {C b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {1}{2} \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} x + \frac {{\left (C a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (4 \, C a^{2} b - 8 \, B a b^{2} - 3 \, C b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 131, normalized size = 1.09 \[ \frac {\frac {b^{3} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a \,b^{2} \sin \left (d x +c \right )-C \,a^{2} b \sin \left (d x +c \right )+B \left (d x +c \right ) a^{2} b -C \,a^{3} \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 125, normalized size = 1.04 \[ -\frac {12 \, {\left (d x + c\right )} C a^{3} - 12 \, {\left (d x + c\right )} B a^{2} b - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 12 \, C a^{2} b \sin \left (d x + c\right ) - 24 \, B a b^{2} \sin \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.93, size = 132, normalized size = 1.10 \[ \frac {B\,b^3\,x}{2}-C\,a^3\,x+B\,a^2\,b\,x+\frac {C\,a\,b^2\,x}{2}+\frac {3\,C\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{d}-\frac {C\,a^2\,b\,\sin \left (c+d\,x\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.67, size = 241, normalized size = 2.01 \[ \begin {cases} B a^{2} b x + \frac {2 B a b^{2} \sin {\left (c + d x \right )}}{d} + \frac {B b^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - C a^{3} x - \frac {C a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {C a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right ) \left (B a b + B b^{2} \cos {\relax (c )} - C a^{2} + C b^{2} \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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