3.11 \(\int \frac{S(b x)}{x^3} \, dx\)

Optimal. Leaf size=44 \[ \frac{1}{2} \pi b^2 \text{FresnelC}(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 x}-\frac{S(b x)}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0287987, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6426, 3387, 3352} \[ \frac{1}{2} \pi b^2 \text{FresnelC}(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 x}-\frac{S(b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^3,x]

[Out]

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

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 3387

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[((e*x)^(m + 1)*Sin[c + d*x^n])/(e*(m + 1
)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[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 \frac{S(b x)}{x^3} \, dx &=-\frac{S(b x)}{2 x^2}+\frac{1}{2} b \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac{S(b x)}{2 x^2}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{2 x}+\frac{1}{2} \left (b^3 \pi \right ) \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{2} b^2 \pi C(b x)-\frac{S(b x)}{2 x^2}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{2 x}\\ \end{align*}

Mathematica [A]  time = 0.0112385, size = 44, normalized size = 1. \[ \frac{1}{2} \pi b^2 \text{FresnelC}(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 x}-\frac{S(b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^3,x]

[Out]

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

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Maple [A]  time = 0.048, size = 43, normalized size = 1. \begin{align*}{b}^{2} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{2\,{b}^{2}{x}^{2}}}-{\frac{1}{2\,bx}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi \,{\it FresnelC} \left ( bx \right ) }{2}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^3,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm fresnels}\left (b x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^3,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^3, x)

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Sympy [A]  time = 0.623655, size = 51, normalized size = 1.16 \begin{align*} \frac{\pi b^{3} x \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{5}{4}, \frac{3}{2}, \frac{7}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**3,x)

[Out]

pi*b**3*x*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (5/4, 3/2, 7/4), -pi**2*b**4*x**4/16)/(32*gamma(5/4)*gamma(7
/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^3,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^3, x)