3.207 \(\int x \text{FresnelC}(b x) \sin (\frac{1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=60 \[ -\frac{\text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{\text{FresnelC}\left (\sqrt{2} b x\right )}{2 \sqrt{2} \pi b^2}+\frac{x}{2 \pi b} \]

[Out]

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

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Rubi [A]  time = 0.0317032, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6461, 3358, 3352} \[ -\frac{\text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{\text{FresnelC}\left (\sqrt{2} b x\right )}{2 \sqrt{2} \pi b^2}+\frac{x}{2 \pi b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6461

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

Rule 3358

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Cos[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[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 C(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 ) C(b x)}{b^2 \pi }+\frac{\int \cos ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac{\int \left (\frac{1}{2}+\frac{1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{b \pi }\\ &=\frac{x}{2 b \pi }-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac{\int \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=\frac{x}{2 b \pi }-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac{C\left (\sqrt{2} b x\right )}{2 \sqrt{2} b^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.0263026, size = 48, normalized size = 0.8 \[ \frac{-4 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )+\sqrt{2} \text{FresnelC}\left (\sqrt{2} b x\right )+2 b x}{4 \pi b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 52, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{{\it FresnelC} \left ( bx \right ) }{b\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{1}{b\pi } \left ({\frac{bx}{2}}+{\frac{\sqrt{2}{\it FresnelC} \left ( bx\sqrt{2} \right ) }{4}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnelc}\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*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

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

[Out]

integral(x*fresnelc(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 )} C\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnelc}\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*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

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