3.18 \(\int \frac {S(b x)}{x^{10}} \, dx\)

Optimal. Leaf size=127 \[ -\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{72 x^8}+\frac {\pi ^4 b^9 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}+\frac {\pi ^3 b^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3456 x^2}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{1728 x^4}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{432 x^6}-\frac {S(b x)}{9 x^9} \]

[Out]

-1/432*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^6+1/3456*b^7*Pi^3*cos(1/2*b^2*Pi*x^2)/x^2-1/9*FresnelS(b*x)/x^9+1/6912*b^9
*Pi^4*Si(1/2*b^2*Pi*x^2)-1/72*b*sin(1/2*b^2*Pi*x^2)/x^8+1/1728*b^5*Pi^2*sin(1/2*b^2*Pi*x^2)/x^4

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Rubi [A]  time = 0.15, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6426, 3379, 3297, 3299} \[ \frac {\pi ^4 b^9 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{1728 x^4}-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{72 x^8}+\frac {\pi ^3 b^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3456 x^2}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{432 x^6}-\frac {S(b x)}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^10,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(432*x^6) + (b^7*Pi^3*Cos[(b^2*Pi*x^2)/2])/(3456*x^2) - FresnelS[b*x]/(9*x^9) -
(b*Sin[(b^2*Pi*x^2)/2])/(72*x^8) + (b^5*Pi^2*Sin[(b^2*Pi*x^2)/2])/(1728*x^4) + (b^9*Pi^4*SinIntegral[(b^2*Pi*x
^2)/2])/6912

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^{10}} \, dx &=-\frac {S(b x)}{9 x^9}+\frac {1}{9} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^9} \, dx\\ &=-\frac {S(b x)}{9 x^9}+\frac {1}{18} b \operatorname {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^5} \, dx,x,x^2\right )\\ &=-\frac {S(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {1}{144} \left (b^3 \pi \right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^4} \, dx,x,x^2\right )\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {S(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}-\frac {1}{864} \left (b^5 \pi ^2\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right )\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {S(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}-\frac {\left (b^7 \pi ^3\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )}{3456}\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}-\frac {S(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {\left (b^9 \pi ^4\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right )}{6912}\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}-\frac {S(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {b^9 \pi ^4 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 96, normalized size = 0.76 \[ \frac {\pi ^4 b^9 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )+\frac {4 b \left (\pi ^2 b^4 x^4-24\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^8}+\frac {2 \pi b^3 \left (\pi ^2 b^4 x^4-8\right ) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^6}-\frac {768 S(b x)}{x^9}}{6912} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^10,x]

[Out]

((2*b^3*Pi*(-8 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/x^6 - (768*FresnelS[b*x])/x^9 + (4*b*(-24 + b^4*Pi^2*x^4)*
Sin[(b^2*Pi*x^2)/2])/x^8 + b^9*Pi^4*SinIntegral[(b^2*Pi*x^2)/2])/6912

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right )}{x^{10}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^10,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^10, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^10,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^10, x)

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maple [A]  time = 0.02, size = 115, normalized size = 0.91 \[ b^{9} \left (-\frac {\mathrm {S}\left (b x \right )}{9 b^{9} x^{9}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{72 b^{8} x^{8}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{6} x^{6}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \Si \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{72}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^10,x)

[Out]

b^9*(-1/9*FresnelS(b*x)/b^9/x^9-1/72*sin(1/2*b^2*Pi*x^2)/b^8/x^8+1/72*Pi*(-1/6/b^6/x^6*cos(1/2*b^2*Pi*x^2)-1/6
*Pi*(-1/4*sin(1/2*b^2*Pi*x^2)/b^4/x^4+1/4*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^10,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^10, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^{10}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^10,x)

[Out]

int(FresnelS(b*x)/x^10, x)

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sympy [A]  time = 2.76, size = 48, normalized size = 0.38 \[ - \frac {\pi b^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{48 x^{6} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**10,x)

[Out]

-pi*b**3*gamma(3/4)*hyper((-3/2, 3/4), (-1/2, 3/2, 7/4), -pi**2*b**4*x**4/16)/(48*x**6*gamma(7/4))

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