Optimal. Leaf size=112 \[ \frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 \text {Si}(2 b x)}{3 b^3}+\frac {5 \sin (b x) \cos (b x)}{6 b^3}-\frac {4 x \text {Ci}(b x) \cos (b x)}{3 b^2}+\frac {x}{2 b^2}+\frac {x \sin ^2(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b} \]
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Rubi [A] time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6508, 6514, 12, 3443, 2635, 8, 6520, 6512, 4406, 3299} \[ \frac {4 \text {CosIntegral}(b x) \sin (b x)}{3 b^3}-\frac {4 x \text {CosIntegral}(b x) \cos (b x)}{3 b^2}-\frac {2 \text {Si}(2 b x)}{3 b^3}+\frac {x}{2 b^2}+\frac {x \sin ^2(b x)}{3 b^2}+\frac {5 \sin (b x) \cos (b x)}{6 b^3}+\frac {1}{3} x^3 \text {CosIntegral}(b x)^2-\frac {2 x^2 \text {CosIntegral}(b x) \sin (b x)}{3 b} \]
Antiderivative was successfully verified.
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Rule 8
Rule 12
Rule 2635
Rule 3299
Rule 3443
Rule 4406
Rule 6508
Rule 6512
Rule 6514
Rule 6520
Rubi steps
\begin {align*} \int x^2 \text {Ci}(b x)^2 \, dx &=\frac {1}{3} x^3 \text {Ci}(b x)^2-\frac {2}{3} \int x^2 \cos (b x) \text {Ci}(b x) \, dx\\ &=\frac {1}{3} x^3 \text {Ci}(b x)^2-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {2}{3} \int \frac {x \cos (b x) \sin (b x)}{b} \, dx+\frac {4 \int x \text {Ci}(b x) \sin (b x) \, dx}{3 b}\\ &=-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {4 \int \cos (b x) \text {Ci}(b x) \, dx}{3 b^2}+\frac {2 \int x \cos (b x) \sin (b x) \, dx}{3 b}+\frac {4 \int \frac {\cos ^2(b x)}{b} \, dx}{3 b}\\ &=-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2+\frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {x \sin ^2(b x)}{3 b^2}-\frac {\int \sin ^2(b x) \, dx}{3 b^2}+\frac {4 \int \cos ^2(b x) \, dx}{3 b^2}-\frac {4 \int \frac {\cos (b x) \sin (b x)}{b x} \, dx}{3 b^2}\\ &=-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2+\frac {5 \cos (b x) \sin (b x)}{6 b^3}+\frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \int \frac {\cos (b x) \sin (b x)}{x} \, dx}{3 b^3}-\frac {\int 1 \, dx}{6 b^2}+\frac {2 \int 1 \, dx}{3 b^2}\\ &=\frac {x}{2 b^2}-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2+\frac {5 \cos (b x) \sin (b x)}{6 b^3}+\frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \int \frac {\sin (2 b x)}{2 x} \, dx}{3 b^3}\\ &=\frac {x}{2 b^2}-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2+\frac {5 \cos (b x) \sin (b x)}{6 b^3}+\frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {x \sin ^2(b x)}{3 b^2}-\frac {2 \int \frac {\sin (2 b x)}{x} \, dx}{3 b^3}\\ &=\frac {x}{2 b^2}-\frac {4 x \cos (b x) \text {Ci}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)^2+\frac {5 \cos (b x) \sin (b x)}{6 b^3}+\frac {4 \text {Ci}(b x) \sin (b x)}{3 b^3}-\frac {2 x^2 \text {Ci}(b x) \sin (b x)}{3 b}+\frac {x \sin ^2(b x)}{3 b^2}-\frac {2 \text {Si}(2 b x)}{3 b^3}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 78, normalized size = 0.70 \[ \frac {4 b^3 x^3 \text {Ci}(b x)^2-8 \text {Ci}(b x) \left (\left (b^2 x^2-2\right ) \sin (b x)+2 b x \cos (b x)\right )-8 \text {Si}(2 b x)+8 b x+5 \sin (2 b x)-2 b x \cos (2 b x)}{12 b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 4.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Ci}\left (b x\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 0.25, size = 149, normalized size = 1.33 \[ \frac {1}{3} \, x^{3} \operatorname {Ci}\left (b x\right )^{2} - \frac {2}{3} \, {\left (\frac {2 \, x \cos \left (b x\right )}{b^{2}} + \frac {{\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{3}}\right )} \operatorname {Ci}\left (b x\right ) + \frac {5 \, b x \tan \left (b x\right )^{2} - 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \left (b x\right )^{2} + 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \left (b x\right )^{2} - 4 \, \operatorname {Si}\left (2 \, b x\right ) \tan \left (b x\right )^{2} + 3 \, b x - 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) + 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) - 4 \, \operatorname {Si}\left (2 \, b x\right ) + 5 \, \tan \left (b x\right )}{6 \, {\left (b^{3} \tan \left (b x\right )^{2} + b^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 84, normalized size = 0.75 \[ \frac {\frac {b^{3} x^{3} \Ci \left (b x \right )^{2}}{3}-2 \Ci \left (b x \right ) \left (\frac {b^{2} x^{2} \sin \left (b x \right )}{3}-\frac {2 \sin \left (b x \right )}{3}+\frac {2 b x \cos \left (b x \right )}{3}\right )-\frac {b x \left (\cos ^{2}\left (b x \right )\right )}{3}+\frac {5 \sin \left (b x \right ) \cos \left (b x \right )}{6}+\frac {5 b x}{6}-\frac {2 \Si \left (2 b x \right )}{3}}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Ci}\left (b x\right )^{2}\,{d x} \]
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 \[ \int x^2\,{\mathrm {cosint}\left (b\,x\right )}^2 \,d x \]
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 \[ \int x^{2} \operatorname {Ci}^{2}{\left (b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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