Optimal. Leaf size=137 \[ -\frac{i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi b}+\frac{i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi b}+\frac{\text{FresnelC}(b x) S(b x)}{2 \pi b^3}-\frac{x S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]
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Rubi [A] time = 0.0606421, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6454, 6446, 3379, 2638} \[ -\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac{\text{FresnelC}(b x) S(b x)}{2 \pi b^3}-\frac{x S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]
Antiderivative was successfully verified.
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Rule 6454
Rule 6446
Rule 3379
Rule 2638
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{\operatorname{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*}
Mathematica [F] time = 0.215816, size = 0, normalized size = 0. \[ \int x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.095, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}{\it FresnelS} \left ( bx \right ) \sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2}{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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