Optimal. Leaf size=135 \[ \frac {a^2 \sin (c+d x)}{d^2}-\frac {a^2 x \cos (c+d x)}{d}+\frac {4 a b \cos (c+d x)}{d^3}+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d} \]
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Rubi [A] time = 0.19, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6742, 3296, 2637, 2638} \[ \frac {a^2 \sin (c+d x)}{d^2}-\frac {a^2 x \cos (c+d x)}{d}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 a b \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {b^2 x^3 \cos (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 2638
Rule 3296
Rule 6742
Rubi steps
\begin {align*} \int x (a+b x)^2 \sin (c+d x) \, dx &=\int \left (a^2 x \sin (c+d x)+2 a b x^2 \sin (c+d x)+b^2 x^3 \sin (c+d x)\right ) \, dx\\ &=a^2 \int x \sin (c+d x) \, dx+(2 a b) \int x^2 \sin (c+d x) \, dx+b^2 \int x^3 \sin (c+d x) \, dx\\ &=-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+\frac {a^2 \int \cos (c+d x) \, dx}{d}+\frac {(4 a b) \int x \cos (c+d x) \, dx}{d}+\frac {\left (3 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d}\\ &=-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {(4 a b) \int \sin (c+d x) \, dx}{d^2}-\frac {\left (6 b^2\right ) \int x \sin (c+d x) \, dx}{d^2}\\ &=\frac {4 a b \cos (c+d x)}{d^3}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {\left (6 b^2\right ) \int \cos (c+d x) \, dx}{d^3}\\ &=\frac {4 a b \cos (c+d x)}{d^3}+\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}\\ \end {align*}
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Mathematica [A] time = 0.22, size = 87, normalized size = 0.64 \[ \frac {\left (a^2 d^2+4 a b d^2 x+3 b^2 \left (d^2 x^2-2\right )\right ) \sin (c+d x)-d \left (a^2 d^2 x+2 a b \left (d^2 x^2-2\right )+b^2 x \left (d^2 x^2-6\right )\right ) \cos (c+d x)}{d^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 95, normalized size = 0.70 \[ -\frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} - 4 \, a b d + {\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) - {\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.75, size = 95, normalized size = 0.70 \[ -\frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x - 6 \, b^{2} d x - 4 \, a b d\right )} \cos \left (d x + c\right )}{d^{4}} + \frac {{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.03, size = 281, normalized size = 2.08 \[ \frac {\frac {b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}+\frac {2 a b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}-\frac {3 b^{2} c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+a^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )-\frac {4 a b c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}+a^{2} c \cos \left (d x +c \right )-\frac {2 a b \,c^{2} \cos \left (d x +c \right )}{d}+\frac {b^{2} c^{3} \cos \left (d x +c \right )}{d^{2}}}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.75, size = 259, normalized size = 1.92 \[ \frac {a^{2} c \cos \left (d x + c\right ) + \frac {b^{2} c^{3} \cos \left (d x + c\right )}{d^{2}} - \frac {2 \, a b c^{2} \cos \left (d x + c\right )}{d} - {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a^{2} - \frac {3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b^{2} c^{2}}{d^{2}} + \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a b c}{d} + \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b^{2} c}{d^{2}} - \frac {2 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a b}{d} - \frac {{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b^{2}}{d^{2}}}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.76, size = 128, normalized size = 0.95 \[ \frac {3\,b^2\,x^2\,\sin \left (c+d\,x\right )}{d^2}-\frac {b^2\,x^3\,\cos \left (c+d\,x\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (6\,b^2-a^2\,d^2\right )}{d^4}+\frac {4\,a\,b\,\cos \left (c+d\,x\right )}{d^3}+\frac {x\,\cos \left (c+d\,x\right )\,\left (6\,b^2-a^2\,d^2\right )}{d^3}-\frac {2\,a\,b\,x^2\,\cos \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,x\,\sin \left (c+d\,x\right )}{d^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.37, size = 172, normalized size = 1.27 \[ \begin {cases} - \frac {a^{2} x \cos {\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d^{2}} - \frac {2 a b x^{2} \cos {\left (c + d x \right )}}{d} + \frac {4 a b x \sin {\left (c + d x \right )}}{d^{2}} + \frac {4 a b \cos {\left (c + d x \right )}}{d^{3}} - \frac {b^{2} x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 b^{2} x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 b^{2} x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 b^{2} \sin {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a^{2} x^{2}}{2} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{4}}{4}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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