3.78 \(\int x S(b x) \sin (\frac{1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=49 \[ \frac{S\left (\sqrt{2} b x\right )}{2 \sqrt{2} \pi b^2}-\frac{S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2} \]

[Out]

-((Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi)) + FresnelS[Sqrt[2]*b*x]/(2*Sqrt[2]*b^2*Pi)

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Rubi [A]  time = 0.0216188, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6452, 3351} \[ \frac{S\left (\sqrt{2} b x\right )}{2 \sqrt{2} \pi b^2}-\frac{S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

-((Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi)) + FresnelS[Sqrt[2]*b*x]/(2*Sqrt[2]*b^2*Pi)

Rule 6452

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(Cos[d*x^2]*FresnelS[b*x])/(2*d), x] + Dis
t[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int x S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx &=-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{\int \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^2 \pi }+\frac{S\left (\sqrt{2} b x\right )}{2 \sqrt{2} b^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.0244937, size = 44, normalized size = 0.9 \[ \frac{\sqrt{2} S\left (\sqrt{2} b x\right )-4 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(-4*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x] + Sqrt[2]*FresnelS[Sqrt[2]*b*x])/(4*b^2*Pi)

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Maple [A]  time = 0.063, size = 46, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{b\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,b\pi }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

(-cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b/Pi+1/4*FresnelS(b*x*2^(1/2))/b/Pi*2^(1/2))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\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.

[In]

integrate(x*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

integrate(x*fresnels(b*x)*sin(1/2*pi*b^2*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x{\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.

[In]

integrate(x*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")

[Out]

integral(x*fresnels(b*x)*sin(1/2*pi*b^2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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.

[In]

integrate(x*fresnels(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x*sin(pi*b**2*x**2/2)*fresnels(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\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.

[In]

integrate(x*fresnels(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

integrate(x*fresnels(b*x)*sin(1/2*pi*b^2*x^2), x)