3.12 \(\int \frac{S(b x)}{x^4} \, dx\)

Optimal. Leaf size=52 \[ \frac{1}{12} \pi b^3 \text{CosIntegral}\left (\frac{1}{2} \pi b^2 x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{S(b x)}{3 x^3} \]

[Out]

(b^3*Pi*CosIntegral[(b^2*Pi*x^2)/2])/12 - FresnelS[b*x]/(3*x^3) - (b*Sin[(b^2*Pi*x^2)/2])/(6*x^2)

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Rubi [A]  time = 0.0637342, antiderivative size = 52, 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, 3379, 3297, 3302} \[ \frac{1}{12} \pi b^3 \text{CosIntegral}\left (\frac{1}{2} \pi b^2 x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{S(b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^4,x]

[Out]

(b^3*Pi*CosIntegral[(b^2*Pi*x^2)/2])/12 - FresnelS[b*x]/(3*x^3) - (b*Sin[(b^2*Pi*x^2)/2])/(6*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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

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

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{S(b x)}{x^4} \, dx &=-\frac{S(b x)}{3 x^3}+\frac{1}{3} b \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^3} \, dx\\ &=-\frac{S(b x)}{3 x^3}+\frac{1}{6} b \operatorname{Subst}\left (\int \frac{\sin \left (\frac{1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )\\ &=-\frac{S(b x)}{3 x^3}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 x^2}+\frac{1}{12} \left (b^3 \pi \right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right )\\ &=\frac{1}{12} b^3 \pi \text{Ci}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{S(b x)}{3 x^3}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{6 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0143616, size = 52, normalized size = 1. \[ \frac{1}{12} \pi b^3 \text{CosIntegral}\left (\frac{1}{2} \pi b^2 x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{S(b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^4,x]

[Out]

(b^3*Pi*CosIntegral[(b^2*Pi*x^2)/2])/12 - FresnelS[b*x]/(3*x^3) - (b*Sin[(b^2*Pi*x^2)/2])/(6*x^2)

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Maple [A]  time = 0.052, size = 49, normalized size = 0.9 \begin{align*}{b}^{3} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{3\,{x}^{3}{b}^{3}}}-{\frac{1}{6\,{b}^{2}{x}^{2}}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi }{12}{\it Ci} \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^4,x)

[Out]

b^3*(-1/3*FresnelS(b*x)/b^3/x^3-1/6*sin(1/2*b^2*Pi*x^2)/b^2/x^2+1/12*Pi*Ci(1/2*b^2*Pi*x^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^4,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^4, 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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^4,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^4, x)

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Sympy [A]  time = 1.38716, size = 56, normalized size = 1.08 \begin{align*} - \frac{\pi ^{3} b^{7} x^{4} \Gamma \left (\frac{7}{4}\right ){{}_{3}F_{4}\left (\begin{matrix} 1, 1, \frac{7}{4} \\ 2, 2, \frac{5}{2}, \frac{11}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{768 \Gamma \left (\frac{11}{4}\right )} + \frac{\pi b^{3} \log{\left (b^{4} x^{4} \right )}}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**4,x)

[Out]

-pi**3*b**7*x**4*gamma(7/4)*hyper((1, 1, 7/4), (2, 2, 5/2, 11/4), -pi**2*b**4*x**4/16)/(768*gamma(11/4)) + pi*
b**3*log(b**4*x**4)/24

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^4,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^4, x)