3.13 \(\int \frac{S(b x)}{x^5} \, dx\)

Optimal. Leaf size=69 \[ -\frac{1}{12} \pi ^2 b^4 S(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x^3}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x}-\frac{S(b x)}{4 x^4} \]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(12*x) - (b^4*Pi^2*FresnelS[b*x])/12 - FresnelS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^
2)/2])/(12*x^3)

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Rubi [A]  time = 0.0433676, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6426, 3387, 3388, 3351} \[ -\frac{1}{12} \pi ^2 b^4 S(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x^3}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x}-\frac{S(b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^5,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(12*x) - (b^4*Pi^2*FresnelS[b*x])/12 - FresnelS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^
2)/2])/(12*x^3)

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 3388

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[((e*x)^(m + 1)*Cos[c + d*x^n])/(e*(m + 1
)), x] + Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

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 \frac{S(b x)}{x^5} \, dx &=-\frac{S(b x)}{4 x^4}+\frac{1}{4} b \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac{S(b x)}{4 x^4}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{12 x^3}+\frac{1}{12} \left (b^3 \pi \right ) \int \frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{12 x}-\frac{S(b x)}{4 x^4}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{12 x^3}-\frac{1}{12} \left (b^5 \pi ^2\right ) \int \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{12 x}-\frac{1}{12} b^4 \pi ^2 S(b x)-\frac{S(b x)}{4 x^4}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{12 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0145281, size = 69, normalized size = 1. \[ -\frac{1}{12} \pi ^2 b^4 S(b x)-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x^3}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{12 x}-\frac{S(b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^5,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(12*x) - (b^4*Pi^2*FresnelS[b*x])/12 - FresnelS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^
2)/2])/(12*x^3)

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Maple [A]  time = 0.05, size = 65, normalized size = 0.9 \begin{align*}{b}^{4} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{4\,{x}^{4}{b}^{4}}}-{\frac{1}{12\,{x}^{3}{b}^{3}}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi }{12} \left ( -{\frac{1}{bx}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-\pi \,{\it FresnelS} \left ( bx \right ) \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^5,x)

[Out]

b^4*(-1/4*FresnelS(b*x)/b^4/x^4-1/12*sin(1/2*b^2*Pi*x^2)/b^3/x^3+1/12*Pi*(-1/b/x*cos(1/2*b^2*Pi*x^2)-Pi*Fresne
lS(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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^5,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^5, 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^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^5,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^5, x)

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Sympy [A]  time = 1.11749, size = 110, normalized size = 1.59 \begin{align*} \frac{\pi ^{2} b^{4} S\left (b x\right ) \Gamma \left (- \frac{1}{4}\right )}{64 \Gamma \left (\frac{7}{4}\right )} + \frac{\pi b^{3} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac{1}{4}\right )}{64 x \Gamma \left (\frac{7}{4}\right )} + \frac{b \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac{1}{4}\right )}{64 x^{3} \Gamma \left (\frac{7}{4}\right )} + \frac{3 S\left (b x\right ) \Gamma \left (- \frac{1}{4}\right )}{64 x^{4} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**5,x)

[Out]

pi**2*b**4*fresnels(b*x)*gamma(-1/4)/(64*gamma(7/4)) + pi*b**3*cos(pi*b**2*x**2/2)*gamma(-1/4)/(64*x*gamma(7/4
)) + b*sin(pi*b**2*x**2/2)*gamma(-1/4)/(64*x**3*gamma(7/4)) + 3*fresnels(b*x)*gamma(-1/4)/(64*x**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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^5,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^5, x)