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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6426, 3379, 3297, 3302} \[ \frac {1}{12} \pi b^3 \text {CosIntegral}\left (\frac {1}{2} \pi b^2 x^2\right )-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac {S(b x)}{3 x^3} \]

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 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^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*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 1.00 \[ -\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^2}+\frac {1}{12} \pi b^3 \text {Ci}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {S(b x)}{3 x^3} \]

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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right )}{x^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 49, normalized size = 0.94 \[ b^{3} \left (-\frac {\mathrm {S}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \Ci \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right ) \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{4}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelS(b*x)/x^4, x)

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sympy [A]  time = 1.01, size = 56, normalized size = 1.08 \[ - \frac {\pi ^{3} b^{7} x^{4} \Gamma \left (\frac {7}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {7}{4} \\ 2, 2, \frac {5}{2}, \frac {11}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{768 \Gamma \left (\frac {11}{4}\right )} + \frac {\pi b^{3} \log {\left (b^{4} x^{4} \right )}}{24} \]

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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