Optimal. Leaf size=132 \[ -\frac{1}{6} a d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a d^3 \sin (c) \text{Si}(d x)+\frac{a d^2 \sin (c+d x)}{6 x}-\frac{a \sin (c+d x)}{3 x^3}-\frac{a d \cos (c+d x)}{6 x^2}-\frac{1}{2} b d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} b d^2 \cos (c) \text{Si}(d x)-\frac{b \sin (c+d x)}{2 x^2}-\frac{b d \cos (c+d x)}{2 x} \]
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Rubi [A] time = 0.324551, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3299, 3302} \[ -\frac{1}{6} a d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a d^3 \sin (c) \text{Si}(d x)+\frac{a d^2 \sin (c+d x)}{6 x}-\frac{a \sin (c+d x)}{3 x^3}-\frac{a d \cos (c+d x)}{6 x^2}-\frac{1}{2} b d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} b d^2 \cos (c) \text{Si}(d x)-\frac{b \sin (c+d x)}{2 x^2}-\frac{b d \cos (c+d x)}{2 x} \]
Antiderivative was successfully verified.
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Rule 6742
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{(a+b x) \sin (c+d x)}{x^4} \, dx &=\int \left (\frac{a \sin (c+d x)}{x^4}+\frac{b \sin (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac{\sin (c+d x)}{x^4} \, dx+b \int \frac{\sin (c+d x)}{x^3} \, dx\\ &=-\frac{a \sin (c+d x)}{3 x^3}-\frac{b \sin (c+d x)}{2 x^2}+\frac{1}{3} (a d) \int \frac{\cos (c+d x)}{x^3} \, dx+\frac{1}{2} (b d) \int \frac{\cos (c+d x)}{x^2} \, dx\\ &=-\frac{a d \cos (c+d x)}{6 x^2}-\frac{b d \cos (c+d x)}{2 x}-\frac{a \sin (c+d x)}{3 x^3}-\frac{b \sin (c+d x)}{2 x^2}-\frac{1}{6} \left (a d^2\right ) \int \frac{\sin (c+d x)}{x^2} \, dx-\frac{1}{2} \left (b d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx\\ &=-\frac{a d \cos (c+d x)}{6 x^2}-\frac{b d \cos (c+d x)}{2 x}-\frac{a \sin (c+d x)}{3 x^3}-\frac{b \sin (c+d x)}{2 x^2}+\frac{a d^2 \sin (c+d x)}{6 x}-\frac{1}{6} \left (a d^3\right ) \int \frac{\cos (c+d x)}{x} \, dx-\frac{1}{2} \left (b d^2 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx-\frac{1}{2} \left (b d^2 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{a d \cos (c+d x)}{6 x^2}-\frac{b d \cos (c+d x)}{2 x}-\frac{1}{2} b d^2 \text{Ci}(d x) \sin (c)-\frac{a \sin (c+d x)}{3 x^3}-\frac{b \sin (c+d x)}{2 x^2}+\frac{a d^2 \sin (c+d x)}{6 x}-\frac{1}{2} b d^2 \cos (c) \text{Si}(d x)-\frac{1}{6} \left (a d^3 \cos (c)\right ) \int \frac{\cos (d x)}{x} \, dx+\frac{1}{6} \left (a d^3 \sin (c)\right ) \int \frac{\sin (d x)}{x} \, dx\\ &=-\frac{a d \cos (c+d x)}{6 x^2}-\frac{b d \cos (c+d x)}{2 x}-\frac{1}{6} a d^3 \cos (c) \text{Ci}(d x)-\frac{1}{2} b d^2 \text{Ci}(d x) \sin (c)-\frac{a \sin (c+d x)}{3 x^3}-\frac{b \sin (c+d x)}{2 x^2}+\frac{a d^2 \sin (c+d x)}{6 x}-\frac{1}{2} b d^2 \cos (c) \text{Si}(d x)+\frac{1}{6} a d^3 \sin (c) \text{Si}(d x)\\ \end{align*}
Mathematica [A] time = 0.338596, size = 110, normalized size = 0.83 \[ -\frac{d^2 x^3 \text{CosIntegral}(d x) (a d \cos (c)+3 b \sin (c))+d^2 x^3 \text{Si}(d x) (3 b \cos (c)-a d \sin (c))-a d^2 x^2 \sin (c+d x)+2 a \sin (c+d x)+a d x \cos (c+d x)+3 b d x^2 \cos (c+d x)+3 b x \sin (c+d x)}{6 x^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.012, size = 117, normalized size = 0.9 \begin{align*}{d}^{3} \left ({\frac{b}{d} \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 ) }+a \left ( -{\frac{\sin \left ( dx+c \right ) }{3\,{d}^{3}{x}^{3}}}-{\frac{\cos \left ( dx+c \right ) }{6\,{d}^{2}{x}^{2}}}+{\frac{\sin \left ( dx+c \right ) }{6\,dx}}+{\frac{{\it Si} \left ( dx \right ) \sin \left ( c \right ) }{6}}-{\frac{{\it Ci} \left ( dx \right ) \cos \left ( c \right ) }{6}} \right ) \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 2.09614, size = 149, normalized size = 1.13 \begin{align*} -\frac{{\left ({\left (a{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + a{\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4} +{\left (b{\left (3 i \, \Gamma \left (-3, i \, d x\right ) - 3 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + 3 \, b{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3}\right )} x^{3} + 2 \, b \cos \left (d x + c\right )}{2 \, d x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.7507, size = 404, normalized size = 3.06 \begin{align*} -\frac{2 \,{\left (3 \, b d x^{2} + a d x\right )} \cos \left (d x + c\right ) +{\left (a d^{3} x^{3} \operatorname{Ci}\left (d x\right ) + a d^{3} x^{3} \operatorname{Ci}\left (-d x\right ) + 6 \, b d^{2} x^{3} \operatorname{Si}\left (d x\right )\right )} \cos \left (c\right ) - 2 \,{\left (a d^{2} x^{2} - 3 \, b x - 2 \, a\right )} \sin \left (d x + c\right ) -{\left (2 \, a d^{3} x^{3} \operatorname{Si}\left (d x\right ) - 3 \, b d^{2} x^{3} \operatorname{Ci}\left (d x\right ) - 3 \, b d^{2} x^{3} \operatorname{Ci}\left (-d x\right )\right )} \sin \left (c\right )}{12 \, x^{3}} \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\right ) \sin{\left (c + d x \right )}}{x^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [C] time = 1.14349, size = 1297, normalized size = 9.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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