Optimal. Leaf size=118 \[ \frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {a^2 \cos (a+b x)}{3 b^3}+\frac {a \sin (a+b x)}{3 b^3}-\frac {2 \cos (a+b x)}{3 b^3}-\frac {2 x \sin (a+b x)}{3 b^2}-\frac {a x \cos (a+b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(a+b x)+\frac {x^2 \cos (a+b x)}{3 b} \]
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Rubi [A] time = 0.28, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6503, 6742, 2638, 3296, 2637, 3299} \[ \frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {a^2 \cos (a+b x)}{3 b^3}+\frac {a \sin (a+b x)}{3 b^3}-\frac {2 x \sin (a+b x)}{3 b^2}-\frac {a x \cos (a+b x)}{3 b^2}-\frac {2 \cos (a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {Si}(a+b x)+\frac {x^2 \cos (a+b x)}{3 b} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 2638
Rule 3296
Rule 3299
Rule 6503
Rule 6742
Rubi steps
\begin {align*} \int x^2 \text {Si}(a+b x) \, dx &=\frac {1}{3} x^3 \text {Si}(a+b x)-\frac {1}{3} b \int \frac {x^3 \sin (a+b x)}{a+b x} \, dx\\ &=\frac {1}{3} x^3 \text {Si}(a+b x)-\frac {1}{3} b \int \left (\frac {a^2 \sin (a+b x)}{b^3}-\frac {a x \sin (a+b x)}{b^2}+\frac {x^2 \sin (a+b x)}{b}-\frac {a^3 \sin (a+b x)}{b^3 (a+b x)}\right ) \, dx\\ &=\frac {1}{3} x^3 \text {Si}(a+b x)-\frac {1}{3} \int x^2 \sin (a+b x) \, dx-\frac {a^2 \int \sin (a+b x) \, dx}{3 b^2}+\frac {a^3 \int \frac {\sin (a+b x)}{a+b x} \, dx}{3 b^2}+\frac {a \int x \sin (a+b x) \, dx}{3 b}\\ &=\frac {a^2 \cos (a+b x)}{3 b^3}-\frac {a x \cos (a+b x)}{3 b^2}+\frac {x^2 \cos (a+b x)}{3 b}+\frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {Si}(a+b x)+\frac {a \int \cos (a+b x) \, dx}{3 b^2}-\frac {2 \int x \cos (a+b x) \, dx}{3 b}\\ &=\frac {a^2 \cos (a+b x)}{3 b^3}-\frac {a x \cos (a+b x)}{3 b^2}+\frac {x^2 \cos (a+b x)}{3 b}+\frac {a \sin (a+b x)}{3 b^3}-\frac {2 x \sin (a+b x)}{3 b^2}+\frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {Si}(a+b x)+\frac {2 \int \sin (a+b x) \, dx}{3 b^2}\\ &=-\frac {2 \cos (a+b x)}{3 b^3}+\frac {a^2 \cos (a+b x)}{3 b^3}-\frac {a x \cos (a+b x)}{3 b^2}+\frac {x^2 \cos (a+b x)}{3 b}+\frac {a \sin (a+b x)}{3 b^3}-\frac {2 x \sin (a+b x)}{3 b^2}+\frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {Si}(a+b x)\\ \end {align*}
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Mathematica [A] time = 0.18, size = 63, normalized size = 0.53 \[ \frac {\left (a^3+b^3 x^3\right ) \text {Si}(a+b x)+\left (a^2-a b x+b^2 x^2-2\right ) \cos (a+b x)+(a-2 b x) \sin (a+b x)}{3 b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Si}\left (b x + a\right ), x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 0.83, size = 252, normalized size = 2.14 \[ \frac {1}{3} \, x^{3} \operatorname {Si}\left (b x + a\right ) - \frac {{\left (2 \, b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - a^{3} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + a^{3} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a^{3} \operatorname {Si}\left (b x + a\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, b^{2} x^{2} - a^{3} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) + a^{3} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) - 2 \, a^{3} \operatorname {Si}\left (b x + a\right ) + 2 \, a^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a b x + 8 \, b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 2 \, a^{2} - 4 \, a \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 4 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 4\right )} b}{6 \, {\left (b^{4} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 99, normalized size = 0.84 \[ \frac {\frac {b^{3} x^{3} \Si \left (b x +a \right )}{3}+\frac {\left (b x +a \right )^{2} \cos \left (b x +a \right )}{3}-\frac {2 \cos \left (b x +a \right )}{3}-\frac {2 \left (b x +a \right ) \sin \left (b x +a \right )}{3}+a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )+a^{2} \cos \left (b x +a \right )+\frac {a^{3} \Si \left (b x +a \right )}{3}}{b^{3}} \]
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 \[ \int x^{2} {\rm Si}\left (b x + a\right )\,{d x} \]
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 \[ \int x^2\,\mathrm {sinint}\left (a+b\,x\right ) \,d x \]
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 \[ \int x^{2} \operatorname {Si}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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