3.7 \(\int x S(b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0240392, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6426, 3385, 3352} \[ -\frac{\text{FresnelC}(b x)}{2 \pi b^2}+\frac{x \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac{1}{2} x^2 S(b x) \]

Antiderivative was successfully verified.

[In]

Int[x*FresnelS[b*x],x]

[Out]

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

Rule 6426

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*FresnelS[b*x])/(d*(m + 1)), x] -
Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Sin[(Pi*b^2*x^2)/2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 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 S(b x) \, dx &=\frac{1}{2} x^2 S(b x)-\frac{1}{2} b \int x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{x \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{2 b \pi }+\frac{1}{2} x^2 S(b x)-\frac{\int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=\frac{x \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{2 b \pi }-\frac{C(b x)}{2 b^2 \pi }+\frac{1}{2} x^2 S(b x)\\ \end{align*}

Mathematica [A]  time = 0.0110994, size = 49, normalized size = 1. \[ -\frac{\text{FresnelC}(b x)}{2 \pi b^2}+\frac{x \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac{1}{2} x^2 S(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*FresnelS[b*x],x]

[Out]

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

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Maple [A]  time = 0.049, size = 44, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{{b}^{2}{x}^{2}{\it FresnelS} \left ( bx \right ) }{2}}+{\frac{bx}{2\,\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{{\it FresnelC} \left ( bx \right ) }{2\,\pi }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*FresnelS(b*x),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x*fresnels(b*x), 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 ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x*fresnels(b*x), x)

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Sympy [A]  time = 0.524181, size = 53, normalized size = 1.08 \begin{align*} \frac{\pi b^{3} x^{5} \Gamma \left (\frac{3}{4}\right ) \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac{7}{4}\right ) \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnels(b*x),x)

[Out]

pi*b**3*x**5*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4), -pi**2*b**4*x**4/16)/(32*gamma(7/4)*gamm
a(9/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x*fresnels(b*x), x)