Optimal. Leaf size=137 \[ -\frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {\text {FresnelC}(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi } \]
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Rubi [A]
time = 0.04, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6589, 6581,
3460, 2718} \begin {gather*} -\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac {\text {FresnelC}(b x) S(b x)}{2 \pi b^3}-\frac {x S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2718
Rule 3460
Rule 6581
Rule 6589
Rubi steps
\begin {align*} \int x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx &=-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {\int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^2 \pi }+\frac {\int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {C(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {\text {Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=-\frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac {C(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }\\ \end {align*}
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Mathematica [F]
time = 0.16, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{2} \mathrm {S}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^2\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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