Optimal. Leaf size=248 \[ \frac {1}{24} a^2 d^4 \sin (c) \text {Ci}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^3 \cos (c+d x)}{24 x}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {1}{2} b^2 d^2 \sin (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)-\frac {b^2 \sin (c+d x)}{2 x^2}-\frac {b^2 d \cos (c+d x)}{2 x} \]
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Rubi [A] time = 0.48, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6742, 3297, 3303, 3299, 3302} \[ \frac {1}{24} a^2 d^4 \sin (c) \text {CosIntegral}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {1}{3} a b d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {1}{2} b^2 d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)-\frac {b^2 \sin (c+d x)}{2 x^2}-\frac {b^2 d \cos (c+d x)}{2 x} \]
Antiderivative was successfully verified.
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Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 6742
Rubi steps
\begin {align*} \int \frac {(a+b x)^2 \sin (c+d x)}{x^5} \, dx &=\int \left (\frac {a^2 \sin (c+d x)}{x^5}+\frac {2 a b \sin (c+d x)}{x^4}+\frac {b^2 \sin (c+d x)}{x^3}\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x^5} \, dx+(2 a b) \int \frac {\sin (c+d x)}{x^4} \, dx+b^2 \int \frac {\sin (c+d x)}{x^3} \, dx\\ &=-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {1}{4} \left (a^2 d\right ) \int \frac {\cos (c+d x)}{x^4} \, dx+\frac {1}{3} (2 a b d) \int \frac {\cos (c+d x)}{x^3} \, dx+\frac {1}{2} \left (b^2 d\right ) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}-\frac {1}{12} \left (a^2 d^2\right ) \int \frac {\sin (c+d x)}{x^3} \, dx-\frac {1}{3} \left (a b d^2\right ) \int \frac {\sin (c+d x)}{x^2} \, dx-\frac {1}{2} \left (b^2 d^2\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{24} \left (a^2 d^3\right ) \int \frac {\cos (c+d x)}{x^2} \, dx-\frac {1}{3} \left (a b d^3\right ) \int \frac {\cos (c+d x)}{x} \, dx-\frac {1}{2} \left (b^2 d^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx-\frac {1}{2} \left (b^2 d^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4\right ) \int \frac {\sin (c+d x)}{x} \, dx-\frac {1}{3} \left (a b d^3 \cos (c)\right ) \int \frac {\cos (d x)}{x} \, dx+\frac {1}{3} \left (a b d^3 \sin (c)\right ) \int \frac {\sin (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx+\frac {1}{24} \left (a^2 d^4 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)+\frac {1}{24} a^2 d^4 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)\\ \end {align*}
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Mathematica [A] time = 0.49, size = 204, normalized size = 0.82 \[ \frac {d^2 x^4 \text {Ci}(d x) \left (\sin (c) \left (a^2 d^2-12 b^2\right )-8 a b d \cos (c)\right )+d^2 x^4 \text {Si}(d x) \left (a^2 d^2 \cos (c)+8 a b d \sin (c)-12 b^2 \cos (c)\right )+a^2 d^3 x^3 \cos (c+d x)+a^2 d^2 x^2 \sin (c+d x)-6 a^2 \sin (c+d x)-2 a^2 d x \cos (c+d x)+8 a b d^2 x^3 \sin (c+d x)-8 a b d x^2 \cos (c+d x)-16 a b x \sin (c+d x)-12 b^2 d x^3 \cos (c+d x)-12 b^2 x^2 \sin (c+d x)}{24 x^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 222, normalized size = 0.90 \[ -\frac {2 \, {\left (8 \, a b d x^{2} + 2 \, a^{2} d x - {\left (a^{2} d^{3} - 12 \, b^{2} d\right )} x^{3}\right )} \cos \left (d x + c\right ) + 2 \, {\left (4 \, a b d^{3} x^{4} \operatorname {Ci}\left (d x\right ) + 4 \, a b d^{3} x^{4} \operatorname {Ci}\left (-d x\right ) - {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) - 2 \, {\left (8 \, a b d^{2} x^{3} - 16 \, a b x + {\left (a^{2} d^{2} - 12 \, b^{2}\right )} x^{2} - 6 \, a^{2}\right )} \sin \left (d x + c\right ) - {\left (16 \, a b d^{3} x^{4} \operatorname {Si}\left (d x\right ) + {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{48 \, x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 0.43, size = 1712, normalized size = 6.90 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 201, normalized size = 0.81 \[ d^{4} \left (\frac {b^{2} \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )}{d^{2}}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )}{d}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{4 x^{4} d^{4}}-\frac {\cos \left (d x +c \right )}{12 x^{3} d^{3}}+\frac {\sin \left (d x +c \right )}{24 x^{2} d^{2}}+\frac {\cos \left (d x +c \right )}{24 x d}+\frac {\Si \left (d x \right ) \cos \relax (c )}{24}+\frac {\Ci \left (d x \right ) \sin \relax (c )}{24}\right )\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 6.00, size = 188, normalized size = 0.76 \[ -\frac {{\left ({\left (a^{2} {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} - {\left (8 \, a b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - a b {\left (8 i \, \Gamma \left (-4, i \, d x\right ) - 8 i \, \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{5} + {\left (b^{2} {\left (-12 i \, \Gamma \left (-4, i \, d x\right ) + 12 i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - 12 \, b^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4}\right )} x^{4} + 6 \, b^{2} \sin \left (d x + c\right ) + 2 \, {\left (b^{2} d x + 2 \, a b d\right )} \cos \left (d x + c\right )}{2 \, d^{2} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (a+b\,x\right )}^2}{x^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{2} \sin {\left (c + d x \right )}}{x^{5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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