Optimal. Leaf size=142 \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{2 a b \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]
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Rubi [A] time = 0.218588, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {3339, 2638, 3297, 3303, 3299, 3302, 3296, 2637} \[ -\frac{1}{2} a^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{2 x^2}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{2 a b \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 3339
Rule 2638
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rule 3296
Rule 2637
Rubi steps
\begin{align*} \int \frac{\left (a+b x^3\right )^2 \sin (c+d x)}{x^3} \, dx &=\int \left (2 a b \sin (c+d x)+\frac{a^2 \sin (c+d x)}{x^3}+b^2 x^3 \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^3} \, dx+(2 a b) \int \sin (c+d x) \, dx+b^2 \int x^3 \sin (c+d x) \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{\left (3 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d}+\frac{1}{2} \left (a^2 d\right ) \int \frac{\cos (c+d x)}{x^2} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{\left (6 b^2\right ) \int x \sin (c+d x) \, dx}{d^2}-\frac{1}{2} \left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{a^2 \sin (c+d x)}{2 x^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{1}{2} \left (a^2 d^2 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx-\frac{1}{2} \left (a^2 d^2 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{2 a b \cos (c+d x)}{d}-\frac{a^2 d \cos (c+d x)}{2 x}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d}-\frac{1}{2} a^2 d^2 \text{Ci}(d x) \sin (c)-\frac{6 b^2 \sin (c+d x)}{d^4}-\frac{a^2 \sin (c+d x)}{2 x^2}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{1}{2} a^2 d^2 \cos (c) \text{Si}(d x)\\ \end{align*}
Mathematica [A] time = 0.382464, size = 138, normalized size = 0.97 \[ \frac{1}{2} \left (-a^2 d^2 \sin (c) \text{CosIntegral}(d x)-a^2 d^2 \cos (c) \text{Si}(d x)-\frac{a^2 \sin (c+d x)}{x^2}-\frac{a^2 d \cos (c+d x)}{x}-\frac{4 a b \cos (c+d x)}{d}+\frac{6 b^2 x^2 \sin (c+d x)}{d^2}-\frac{12 b^2 \sin (c+d x)}{d^4}+\frac{12 b^2 x \cos (c+d x)}{d^3}-\frac{2 b^2 x^3 \cos (c+d x)}{d}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.03, size = 251, normalized size = 1.8 \begin{align*}{d}^{2} \left ({\frac{ \left ( 10\,{c}^{3}+6\,{c}^{2}+3\,c+1 \right ){b}^{2} \left ( - \left ( dx+c \right ) ^{3}\cos \left ( dx+c \right ) +3\, \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) -6\,\sin \left ( dx+c \right ) +6\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{6}}}-6\,{\frac{c{b}^{2} \left ( 6\,{c}^{2}+3\,c+1 \right ) \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{6}}}+15\,{\frac{ \left ( 1+3\,c \right ){c}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{6}}}-2\,{\frac{ab\cos \left ( dx+c \right ) }{{d}^{3}}}+20\,{\frac{{b}^{2}{c}^{3}\cos \left ( dx+c \right ) }{{d}^{6}}}+{a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) }{2\,{d}^{2}{x}^{2}}}-{\frac{\cos \left ( dx+c \right ) }{2\,dx}}-{\frac{{\it Si} \left ( dx \right ) \cos \left ( c \right ) }{2}}-{\frac{{\it Ci} \left ( dx \right ) \sin \left ( c \right ) }{2}} \right ) \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 12.6194, size = 149, normalized size = 1.05 \begin{align*} \frac{{\left (a^{2}{\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 2 \,{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} - 6 \, b^{2} d x\right )} \cos \left (d x + c\right ) + 6 \,{\left (b^{2} d^{2} x^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.9102, size = 355, normalized size = 2.5 \begin{align*} -\frac{2 \, a^{2} d^{6} x^{2} \cos \left (c\right ) \operatorname{Si}\left (d x\right ) + 2 \,{\left (2 \, b^{2} d^{3} x^{5} + a^{2} d^{5} x + 4 \, a b d^{3} x^{2} - 12 \, b^{2} d x^{3}\right )} \cos \left (d x + c\right ) - 2 \,{\left (6 \, b^{2} d^{2} x^{4} - a^{2} d^{4} - 12 \, b^{2} x^{2}\right )} \sin \left (d x + c\right ) +{\left (a^{2} d^{6} x^{2} \operatorname{Ci}\left (d x\right ) + a^{2} d^{6} x^{2} \operatorname{Ci}\left (-d x\right )\right )} \sin \left (c\right )}{4 \, d^{4} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{2} \sin{\left (c + d x \right )}}{x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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