Optimal. Leaf size=167 \[ \frac {1}{24} a^2 d^4 \sin (c) \text {Ci}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^3 \cos (c+d x)}{24 x}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}+2 a b d \cos (c) \text {Ci}(d x)-2 a b d \sin (c) \text {Si}(d x)-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {b^2 x \cos (c+d x)}{d} \]
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Rubi [A] time = 0.28, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3339, 3297, 3303, 3299, 3302, 3296, 2637} \[ \frac {1}{24} a^2 d^4 \sin (c) \text {CosIntegral}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}+2 a b d \cos (c) \text {CosIntegral}(d x)-2 a b d \sin (c) \text {Si}(d x)-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {b^2 x \cos (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 3296
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 3339
Rubi steps
\begin {align*} \int \frac {\left (a+b x^3\right )^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^2}+b^2 x \sin (c+d x)\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x^5} \, dx+(2 a b) \int \frac {\sin (c+d x)}{x^2} \, dx+b^2 \int x \sin (c+d x) \, dx\\ &=-\frac {b^2 x \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{x}+\frac {b^2 \int \cos (c+d x) \, dx}{d}+\frac {1}{4} \left (a^2 d\right ) \int \frac {\cos (c+d x)}{x^4} \, dx+(2 a b d) \int \frac {\cos (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {b^2 x \cos (c+d x)}{d}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{x}-\frac {1}{12} \left (a^2 d^2\right ) \int \frac {\sin (c+d x)}{x^3} \, dx+(2 a b d \cos (c)) \int \frac {\cos (d x)}{x} \, dx-(2 a b d \sin (c)) \int \frac {\sin (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \sin (c) \text {Si}(d x)-\frac {1}{24} \left (a^2 d^3\right ) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \sin (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}-2 a b d \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^2 d^3 \cos (c+d x)}{24 x}-\frac {b^2 x \cos (c+d x)}{d}+2 a b d \cos (c) \text {Ci}(d x)+\frac {1}{24} a^2 d^4 \text {Ci}(d x) \sin (c)+\frac {b^2 \sin (c+d x)}{d^2}-\frac {a^2 \sin (c+d x)}{4 x^4}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {2 a b \sin (c+d x)}{x}+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)-2 a b d \sin (c) \text {Si}(d x)\\ \end {align*}
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Mathematica [A] time = 0.61, size = 148, normalized size = 0.89 \[ \frac {1}{24} \left (\frac {a^2 d^3 \cos (c+d x)}{x}+\frac {a^2 d^2 \sin (c+d x)}{x^2}-\frac {6 a^2 \sin (c+d x)}{x^4}-\frac {2 a^2 d \cos (c+d x)}{x^3}+a d \text {Ci}(d x) \left (a d^3 \sin (c)+48 b \cos (c)\right )+a d \text {Si}(d x) \left (a d^3 \cos (c)-48 b \sin (c)\right )-\frac {48 a b \sin (c+d x)}{x}+\frac {24 b^2 \sin (c+d x)}{d^2}-\frac {24 b^2 x \cos (c+d x)}{d}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 186, normalized size = 1.11 \[ \frac {2 \, {\left (a^{2} d^{5} x^{3} - 24 \, b^{2} d x^{5} - 2 \, a^{2} d^{3} x\right )} \cos \left (d x + c\right ) + 2 \, {\left (a^{2} d^{6} x^{4} \operatorname {Si}\left (d x\right ) + 24 \, a b d^{3} x^{4} \operatorname {Ci}\left (d x\right ) + 24 \, a b d^{3} x^{4} \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) + 2 \, {\left (a^{2} d^{4} x^{2} - 48 \, a b d^{2} x^{3} + 24 \, b^{2} x^{4} - 6 \, a^{2} d^{2}\right )} \sin \left (d x + c\right ) + {\left (a^{2} d^{6} x^{4} \operatorname {Ci}\left (d x\right ) + a^{2} d^{6} x^{4} \operatorname {Ci}\left (-d x\right ) - 96 \, a b d^{3} x^{4} \operatorname {Si}\left (d x\right )\right )} \sin \relax (c)}{48 \, d^{2} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 1.33, size = 1255, normalized size = 7.51 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 167, normalized size = 1.00 \[ d^{4} \left (\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d^{3}}+\frac {6 c \,b^{2} \cos \left (d x +c \right )}{d^{6}}+\frac {\left (1+5 c \right ) b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{6}}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{4 x^{4} d^{4}}-\frac {\cos \left (d x +c \right )}{12 x^{3} d^{3}}+\frac {\sin \left (d x +c \right )}{24 x^{2} d^{2}}+\frac {\cos \left (d x +c \right )}{24 x d}+\frac {\Si \left (d x \right ) \cos \relax (c )}{24}+\frac {\Ci \left (d x \right ) \sin \relax (c )}{24}\right )\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 16.43, size = 166, normalized size = 0.99 \[ \frac {{\left ({\left (a^{2} {\left (-i \, \Gamma \left (-4, i \, d x\right ) + i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - a^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{7} - {\left (48 \, a b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - a b {\left (48 i \, \Gamma \left (-4, i \, d x\right ) - 48 i \, \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4}\right )} x^{4} - 2 \, {\left (b^{2} d^{2} x^{5} + 2 \, a b d^{2} x^{2} - 12 \, a b\right )} \cos \left (d x + c\right ) + 2 \, {\left (b^{2} d x^{4} - 4 \, a b d x\right )} \sin \left (d x + c\right )}{2 \, d^{3} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^3+a\right )}^2}{x^5} \,d x \]
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 \[ \int \frac {\left (a + b x^{3}\right )^{2} \sin {\left (c + d x \right )}}{x^{5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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