Optimal. Leaf size=248 \[ \frac{15 i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{15 i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{15 \text{FresnelC}(b x) S(b x)}{2 \pi ^3 b^7}+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^5 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{15 x S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac{5 x^4}{8 \pi ^2 b^3}+\frac{7 x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}-\frac{x^4 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{11 \cos \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7} \]
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Rubi [A] time = 0.254643, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6454, 6462, 3379, 3309, 30, 3296, 2638, 6446} \[ \frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi ^3 b^5}-\frac{15 \text{FresnelC}(b x) S(b x)}{2 \pi ^3 b^7}+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^5 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{15 x S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac{5 x^4}{8 \pi ^2 b^3}+\frac{7 x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}-\frac{x^4 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{11 \cos \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7} \]
Antiderivative was successfully verified.
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Rule 6454
Rule 6462
Rule 3379
Rule 3309
Rule 30
Rule 3296
Rule 2638
Rule 6446
Rubi steps
\begin{align*} \int x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx &=-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{5 \int x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^2 \pi }+\frac{\int x^5 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac{15 \int x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}-\frac{5 \int x^3 \sin ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}+\frac{\operatorname{Subst}\left (\int x^2 \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=-\frac{x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{15 x \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac{15 \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^6 \pi ^3}-\frac{15 \int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^5 \pi ^3}+\frac{\operatorname{Subst}\left (\int x \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2}-\frac{5 \operatorname{Subst}\left (\int x \sin ^2\left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2}\\ &=-\frac{x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{15 x \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac{15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{x^2 \sin \left (b^2 \pi x^2\right )}{2 b^5 \pi ^3}-\frac{\operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3}-\frac{15 \operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}-\frac{5 \operatorname{Subst}\left (\int x \, dx,x,x^2\right )}{4 b^3 \pi ^2}+\frac{5 \operatorname{Subst}\left (\int x \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}\\ &=-\frac{5 x^4}{8 b^3 \pi ^2}+\frac{17 \cos \left (b^2 \pi x^2\right )}{4 b^7 \pi ^4}-\frac{x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{15 x \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac{15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{7 x^2 \sin \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}-\frac{5 \operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}\\ &=-\frac{5 x^4}{8 b^3 \pi ^2}+\frac{11 \cos \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}-\frac{x^4 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{15 x \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac{15 C(b x) S(b x)}{2 b^7 \pi ^3}+\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{15 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac{5 x^3 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{7 x^2 \sin \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}\\ \end{align*}
Mathematica [F] time = 0.468768, size = 0, normalized size = 0. \[ \int x^6 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{x}^{6}{\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^{6}{\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^{6}{\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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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^{6}{\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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