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

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

[Out]

-1/40*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^2-1/5*FresnelS(b*x)/x^5-1/80*b^5*Pi^2*Si(1/2*b^2*Pi*x^2)-1/20*b*sin(1/2*b^2
*Pi*x^2)/x^4

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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]

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 = 0.04, size = 77, normalized size = 1.00 \[ -\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{20 x^4}-\frac {1}{80} \pi ^2 b^5 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{40 x^2}-\frac {S(b x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/40*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/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

________________________________________________________________________________________

fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right )}{x^{6}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{6}}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 71, normalized size = 0.92 \[ b^{5} \left (-\frac {\mathrm {S}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \Si \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right ) \]

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)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{6}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelS(b*x)/x^6, x)

________________________________________________________________________________________

sympy [A]  time = 0.91, size = 46, normalized size = 0.60 \[ - \frac {\pi b^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x^{2} \Gamma \left (\frac {7}{4}\right )} \]

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))

________________________________________________________________________________________