3.14 \(\int \frac{S(b x)}{x^6} \, dx\)

Optimal. Leaf size=77 \[ -\frac{1}{80} \pi ^2 b^5 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{20 x^4}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{40 x^2}-\frac{S(b x)}{5 x^5} \]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(40*x^2) - FresnelS[b*x]/(5*x^5) - (b*Sin[(b^2*Pi*x^2)/2])/(20*x^4) - (b^5*Pi^2*
SinIntegral[(b^2*Pi*x^2)/2])/80

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Rubi [A]  time = 0.0892369, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6426, 3379, 3297, 3299} \[ -\frac{1}{80} \pi ^2 b^5 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{20 x^4}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{40 x^2}-\frac{S(b x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^6,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(40*x^2) - FresnelS[b*x]/(5*x^5) - (b*Sin[(b^2*Pi*x^2)/2])/(20*x^4) - (b^5*Pi^2*
SinIntegral[(b^2*Pi*x^2)/2])/80

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 3299

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

Rubi steps

\begin{align*} \int \frac{S(b x)}{x^6} \, dx &=-\frac{S(b x)}{5 x^5}+\frac{1}{5} b \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^5} \, dx\\ &=-\frac{S(b x)}{5 x^5}+\frac{1}{10} b \operatorname{Subst}\left (\int \frac{\sin \left (\frac{1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right )\\ &=-\frac{S(b x)}{5 x^5}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{20 x^4}+\frac{1}{40} \left (b^3 \pi \right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac{S(b x)}{5 x^5}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac{1}{80} \left (b^5 \pi ^2\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right )\\ &=-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac{S(b x)}{5 x^5}-\frac{b \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac{1}{80} b^5 \pi ^2 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0184752, size = 77, normalized size = 1. \[ -\frac{1}{80} \pi ^2 b^5 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{20 x^4}-\frac{\pi b^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{40 x^2}-\frac{S(b x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^6,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(40*x^2) - FresnelS[b*x]/(5*x^5) - (b*Sin[(b^2*Pi*x^2)/2])/(20*x^4) - (b^5*Pi^2*
SinIntegral[(b^2*Pi*x^2)/2])/80

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Maple [A]  time = 0.049, size = 71, normalized size = 0.9 \begin{align*}{b}^{5} \left ( -{\frac{{\it FresnelS} \left ( bx \right ) }{5\,{b}^{5}{x}^{5}}}-{\frac{1}{20\,{x}^{4}{b}^{4}}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi }{20} \left ( -{\frac{1}{2\,{b}^{2}{x}^{2}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{\pi }{4}{\it Si} \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^6,x)

[Out]

b^5*(-1/5*FresnelS(b*x)/b^5/x^5-1/20*sin(1/2*b^2*Pi*x^2)/b^4/x^4+1/20*Pi*(-1/2/b^2/x^2*cos(1/2*b^2*Pi*x^2)-1/4
*Pi*Si(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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^6,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^6, 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^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^6,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^6, x)

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Sympy [A]  time = 1.21138, size = 46, normalized size = 0.6 \begin{align*} - \frac{\pi b^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{1}{2}, \frac{3}{2}, \frac{7}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x^{2} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**6,x)

[Out]

-pi*b**3*gamma(3/4)*hyper((-1/2, 3/4), (1/2, 3/2, 7/4), -pi**2*b**4*x**4/16)/(16*x**2*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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^6,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^6, x)