Optimal. Leaf size=158 \[ \frac{4 x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^4 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{8 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac{43 S\left (\sqrt{2} b x\right )}{8 \sqrt{2} \pi ^3 b^6}-\frac{2 x^3}{3 \pi ^2 b^3}+\frac{11 x \sin \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{x^3 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]
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Rubi [A] time = 0.156292, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6454, 6462, 3391, 30, 3386, 3351, 6452, 3385} \[ \frac{4 x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^4 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{8 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac{43 S\left (\sqrt{2} b x\right )}{8 \sqrt{2} \pi ^3 b^6}-\frac{2 x^3}{3 \pi ^2 b^3}+\frac{11 x \sin \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{x^3 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]
Antiderivative was successfully verified.
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Rule 6454
Rule 6462
Rule 3391
Rule 30
Rule 3386
Rule 3351
Rule 6452
Rule 3385
Rubi steps
\begin{align*} \int x^5 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx &=-\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{4 \int x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{b^2 \pi }+\frac{\int x^4 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{4 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac{8 \int x S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}+\frac{3 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}-\frac{4 \int x^2 \sin ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}\\ &=-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{8 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{4 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{3 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{3 \int \sin \left (b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}-\frac{4 \int \sin \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac{2 \int x^2 \, dx}{b^3 \pi ^2}+\frac{2 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}\\ &=-\frac{2 x^3}{3 b^3 \pi ^2}-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{8 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac{3 S\left (\sqrt{2} b x\right )}{8 \sqrt{2} b^6 \pi ^3}-\frac{2 \sqrt{2} S\left (\sqrt{2} b x\right )}{b^6 \pi ^3}+\frac{4 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{11 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{\int \sin \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}\\ &=-\frac{2 x^3}{3 b^3 \pi ^2}-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{8 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^6 \pi ^3}-\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }-\frac{11 S\left (\sqrt{2} b x\right )}{8 \sqrt{2} b^6 \pi ^3}-\frac{2 \sqrt{2} S\left (\sqrt{2} b x\right )}{b^6 \pi ^3}+\frac{4 x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{11 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}\\ \end{align*}
Mathematica [A] time = 0.172595, size = 120, normalized size = 0.76 \[ -\frac{48 S(b x) \left (\left (\pi ^2 b^4 x^4-8\right ) \cos \left (\frac{1}{2} \pi b^2 x^2\right )-4 \pi b^2 x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )\right )+32 \pi b^3 x^3-66 b x \sin \left (\pi b^2 x^2\right )+12 \pi b^3 x^3 \cos \left (\pi b^2 x^2\right )+129 \sqrt{2} S\left (\sqrt{2} b x\right )}{48 \pi ^3 b^6} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.078, size = 202, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelS} \left ( bx \right ) }{{b}^{5}} \left ( -{\frac{{x}^{4}{b}^{4}}{\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+4\,{\frac{1}{\pi } \left ({\frac{{b}^{2}{x}^{2}\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+2\,{\frac{\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) } \right ) }-{\frac{1}{{b}^{5}} \left ({\frac{2\,{x}^{3}{b}^{3}}{3\,{\pi }^{2}}}-2\,{\frac{1}{{\pi }^{2}} \left ( 1/2\,{\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{\pi }}-1/4\,{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{\pi }} \right ) }-{\frac{1}{2\,{\pi }^{3}} \left ( -{\frac{\pi \,{b}^{3}{x}^{3}\cos \left ({b}^{2}\pi \,{x}^{2} \right ) }{2}}+{\frac{3\,\pi }{2} \left ({\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}-{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) }-4\,\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) \right ) } \right ) } \right ) } \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^{5}{\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^{5}{\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^{5} \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^{5}{\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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