3.16 \(\int \frac {S(b x)}{x^8} \, dx\)

Optimal. Leaf size=102 \[ -\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{42 x^6}-\frac {1}{672} \pi ^3 b^7 \text {Ci}\left (\frac {1}{2} b^2 \pi x^2\right )+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{336 x^2}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{168 x^4}-\frac {S(b x)}{7 x^7} \]

[Out]

-1/672*b^7*Pi^3*Ci(1/2*b^2*Pi*x^2)-1/168*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^4-1/7*FresnelS(b*x)/x^7-1/42*b*sin(1/2*b
^2*Pi*x^2)/x^6+1/336*b^5*Pi^2*sin(1/2*b^2*Pi*x^2)/x^2

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Rubi [A]  time = 0.12, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6426, 3379, 3297, 3302} \[ -\frac {1}{672} \pi ^3 b^7 \text {CosIntegral}\left (\frac {1}{2} \pi b^2 x^2\right )+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{336 x^2}-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{42 x^6}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{168 x^4}-\frac {S(b x)}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^8,x]

[Out]

-(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/(168*x^4) - (b^7*Pi^3*CosIntegral[(b^2*Pi*x^2)/2])/672 - FresnelS[b*x]/(7*x^7) -
(b*Sin[(b^2*Pi*x^2)/2])/(42*x^6) + (b^5*Pi^2*Sin[(b^2*Pi*x^2)/2])/(336*x^2)

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]

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^8} \, dx &=-\frac {S(b x)}{7 x^7}+\frac {1}{7} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^7} \, dx\\ &=-\frac {S(b x)}{7 x^7}+\frac {1}{14} b \operatorname {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^4} \, dx,x,x^2\right )\\ &=-\frac {S(b x)}{7 x^7}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {1}{84} \left (b^3 \pi \right ) \operatorname {Subst}\left (\int \frac {\cos \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 )}{168 x^4}-\frac {S(b x)}{7 x^7}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}-\frac {1}{336} \left (b^5 \pi ^2\right ) \operatorname {Subst}\left (\int \frac {\sin \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 )}{168 x^4}-\frac {S(b x)}{7 x^7}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2}-\frac {1}{672} \left (b^7 \pi ^3\right ) \operatorname {Subst}\left (\int \frac {\cos \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 )}{168 x^4}-\frac {1}{672} b^7 \pi ^3 \text {Ci}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {S(b x)}{7 x^7}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 85, normalized size = 0.83 \[ \frac {1}{672} \left (-\pi ^3 b^7 \text {Ci}\left (\frac {1}{2} b^2 \pi x^2\right )+\frac {2 b \left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^6}-\frac {4 \pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^4}-\frac {96 S(b x)}{x^7}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^8,x]

[Out]

((-4*b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x^4 - b^7*Pi^3*CosIntegral[(b^2*Pi*x^2)/2] - (96*FresnelS[b*x])/x^7 + (2*b*(-
8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/x^6)/672

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right )}{x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^8,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)/x^8, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^8,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)/x^8, x)

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maple [A]  time = 0.02, size = 93, normalized size = 0.91 \[ b^{7} \left (-\frac {\mathrm {S}\left (b x \right )}{7 b^{7} x^{7}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{42 b^{6} x^{6}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}+\frac {\pi \Ci \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{42}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^8,x)

[Out]

b^7*(-1/7*FresnelS(b*x)/b^7/x^7-1/42*sin(1/2*b^2*Pi*x^2)/b^6/x^6+1/42*Pi*(-1/4/b^4/x^4*cos(1/2*b^2*Pi*x^2)-1/4
*Pi*(-1/2*sin(1/2*b^2*Pi*x^2)/b^2/x^2+1/4*Pi*Ci(1/2*b^2*Pi*x^2))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x^8,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)/x^8, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^8,x)

[Out]

int(FresnelS(b*x)/x^8, x)

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sympy [A]  time = 2.05, size = 68, normalized size = 0.67 \[ \frac {\pi ^{5} b^{11} x^{4} \Gamma \left (\frac {11}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {11}{4} \\ 2, 3, \frac {7}{2}, \frac {15}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{61440 \Gamma \left (\frac {15}{4}\right )} - \frac {\pi ^{3} b^{7} \log {\left (b^{4} x^{4} \right )}}{1344} - \frac {\pi b^{3}}{24 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**8,x)

[Out]

pi**5*b**11*x**4*gamma(11/4)*hyper((1, 1, 11/4), (2, 3, 7/2, 15/4), -pi**2*b**4*x**4/16)/(61440*gamma(15/4)) -
pi**3*b**7*log(b**4*x**4)/1344 - pi*b**3/(24*x**4)

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