Optimal. Leaf size=229 \[ -\frac{3 a^2 \text{FresnelC}(a+b x)}{2 \pi b^4}-\frac{a^4 S(a+b x)}{4 b^4}-\frac{a^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac{3 a^2 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac{3 S(a+b x)}{4 \pi ^2 b^4}+\frac{2 a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac{3 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}-\frac{a (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac{(a+b x)^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}+\frac{1}{4} x^4 S(a+b x) \]
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Rubi [A] time = 0.184451, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6428, 3433, 3351, 3379, 2638, 3385, 3352, 3296, 2637, 3386} \[ -\frac{3 a^2 \text{FresnelC}(a+b x)}{2 \pi b^4}-\frac{a^4 S(a+b x)}{4 b^4}-\frac{a^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac{3 a^2 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac{3 S(a+b x)}{4 \pi ^2 b^4}+\frac{2 a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac{3 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}-\frac{a (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac{(a+b x)^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}+\frac{1}{4} x^4 S(a+b x) \]
Antiderivative was successfully verified.
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Rule 6428
Rule 3433
Rule 3351
Rule 3379
Rule 2638
Rule 3385
Rule 3352
Rule 3296
Rule 2637
Rule 3386
Rubi steps
\begin{align*} \int x^3 S(a+b x) \, dx &=\frac{1}{4} x^4 S(a+b x)-\frac{1}{4} b \int x^4 \sin \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac{1}{4} x^4 S(a+b x)-\frac{\operatorname{Subst}\left (\int \left (a^4 \sin \left (\frac{\pi x^2}{2}\right )-4 a^3 x \sin \left (\frac{\pi x^2}{2}\right )+6 a^2 x^2 \sin \left (\frac{\pi x^2}{2}\right )-4 a x^3 \sin \left (\frac{\pi x^2}{2}\right )+x^4 \sin \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac{1}{4} x^4 S(a+b x)-\frac{\operatorname{Subst}\left (\int x^4 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}+\frac{a \operatorname{Subst}\left (\int x^3 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int x^2 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4}+\frac{a^3 \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac{a^4 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac{3 a^2 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }+\frac{(a+b x)^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac{a^4 S(a+b x)}{4 b^4}+\frac{1}{4} x^4 S(a+b x)+\frac{a \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}+\frac{a^3 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}-\frac{3 \operatorname{Subst}\left (\int x^2 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi }-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4 \pi }\\ &=-\frac{a^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac{3 a^2 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac{a (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac{(a+b x)^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac{3 a^2 C(a+b x)}{2 b^4 \pi }-\frac{a^4 S(a+b x)}{4 b^4}+\frac{1}{4} x^4 S(a+b x)-\frac{3 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}+\frac{3 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi ^2}+\frac{a \operatorname{Subst}\left (\int \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{b^4 \pi }\\ &=-\frac{a^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac{3 a^2 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac{a (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac{(a+b x)^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac{3 a^2 C(a+b x)}{2 b^4 \pi }-\frac{a^4 S(a+b x)}{4 b^4}+\frac{3 S(a+b x)}{4 b^4 \pi ^2}+\frac{1}{4} x^4 S(a+b x)+\frac{2 a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac{3 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}\\ \end{align*}
Mathematica [A] time = 0.323819, size = 166, normalized size = 0.72 \[ \frac{\left (-\pi ^2 a^4+\pi ^2 b^4 x^4+3\right ) S(a+b x)-6 \pi a^2 \text{FresnelC}(a+b x)-\pi a^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi a^2 b x \cos \left (\frac{1}{2} \pi (a+b x)^2\right )-\pi a b^2 x^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^3 x^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+5 a \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-3 b x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.052, size = 189, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\it FresnelS} \left ( bx+a \right ){b}^{4}{x}^{4}}{4}}+{\frac{ \left ( bx+a \right ) ^{3}}{4\,\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{3}{4\,\pi } \left ({\frac{bx+a}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{{\it FresnelS} \left ( bx+a \right ) }{\pi }} \right ) }-{\frac{a \left ( bx+a \right ) ^{2}}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+2\,{\frac{a\sin \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{{\pi }^{2}}}+{\frac{3\,{a}^{2} \left ( bx+a \right ) }{2\,\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{3\,{a}^{2}{\it FresnelC} \left ( bx+a \right ) }{2\,\pi }}-{\frac{{a}^{3}}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{{a}^{4}{\it FresnelS} \left ( bx+a \right ) }{4}} \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^{3}{\rm fresnels}\left (b x + a\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^{3}{\rm fresnels}\left (b x + a\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^{3} S\left (a + 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^{3}{\rm fresnels}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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