Optimal. Leaf size=96 \[ -\frac{a^2 S(a+b x)}{2 b^2}-\frac{\text{FresnelC}(a+b x)}{2 \pi b^2}-\frac{a \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac{(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac{1}{2} x^2 S(a+b x) \]
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Rubi [A] time = 0.0672627, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {6428, 3433, 3351, 3379, 2638, 3385, 3352} \[ -\frac{a^2 S(a+b x)}{2 b^2}-\frac{\text{FresnelC}(a+b x)}{2 \pi b^2}-\frac{a \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac{(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac{1}{2} x^2 S(a+b x) \]
Antiderivative was successfully verified.
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Rule 6428
Rule 3433
Rule 3351
Rule 3379
Rule 2638
Rule 3385
Rule 3352
Rubi steps
\begin{align*} \int x S(a+b x) \, dx &=\frac{1}{2} x^2 S(a+b x)-\frac{1}{2} b \int x^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac{1}{2} x^2 S(a+b x)-\frac{\operatorname{Subst}\left (\int \left (a^2 \sin \left (\frac{\pi x^2}{2}\right )-2 a x \sin \left (\frac{\pi x^2}{2}\right )+x^2 \sin \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{1}{2} x^2 S(a+b x)-\frac{\operatorname{Subst}\left (\int x^2 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2}+\frac{a \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^2}-\frac{a^2 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }-\frac{a^2 S(a+b x)}{2 b^2}+\frac{1}{2} x^2 S(a+b x)+\frac{a \operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2 \pi }\\ &=-\frac{a \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^2 \pi }+\frac{(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }-\frac{C(a+b x)}{2 b^2 \pi }-\frac{a^2 S(a+b x)}{2 b^2}+\frac{1}{2} x^2 S(a+b x)\\ \end{align*}
Mathematica [A] time = 0.166339, size = 51, normalized size = 0.53 \[ -\frac{\text{FresnelC}(a+b x)+(a-b x) \left (\pi (a+b x) S(a+b x)+\cos \left (\frac{1}{2} \pi (a+b x)^2\right )\right )}{2 \pi b^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.051, size = 80, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{2}} \left ({\it FresnelS} \left ( bx+a \right ) \left ({\frac{ \left ( bx+a \right ) ^{2}}{2}}-a \left ( bx+a \right ) \right ) +{\frac{bx+a}{2\,\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{{\it FresnelC} \left ( bx+a \right ) }{2\,\pi }}-{\frac{a}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \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{\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{\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 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{\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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