3.1.12 \(\int \frac {S(b x)}{x^4} \, dx\) [12]

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

[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 = 0.05, 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 = {6561, 3460, 3378, 3383} \begin {gather*} -\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^2}+\frac {1}{12} \pi b^3 \text {CosIntegral}\left (\frac {1}{2} \pi b^2 x^2\right )-\frac {S(b x)}{3 x^3} \end {gather*}

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 3378

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 3383

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 3460

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 6561

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/2)*b^2*x^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 \text {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 ) \text {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 = 0.01, size = 52, normalized size = 1.00 \begin {gather*} \frac {1}{12} b^3 \pi \text {CosIntegral}\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 {gather*}

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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Maple [A]
time = 0.35, size = 49, normalized size = 0.94

method	result	size
derivativedivides	\(b^{3} \left (-\frac {\mathrm {S}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \cosineIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\)	\(49\)
default	\(b^{3} \left (-\frac {\mathrm {S}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \cosineIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\)	\(49\)
meijerg	\(\frac {\pi ^{\frac {3}{2}} b^{3} \left (-\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2, \frac {5}{2}, \frac {11}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{21}+\frac {\frac {16 \gamma }{3}-\frac {16 \ln \left (2\right )}{3}-\frac {80}{9}+\frac {32 \ln \left (x \right )}{3}+\frac {16 \ln \left (\pi \right )}{3}+\frac {32 \ln \left (b \right )}{3}}{\sqrt {\pi }}\right )}{64}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.32, size = 42, normalized size = 0.81 \begin {gather*} \frac {1}{24} \, {\left (\pi \Gamma \left (-1, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \pi \Gamma \left (-1, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{3} - \frac {\operatorname {S}\left (b x\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^4,x, algorithm="maxima")

[Out]

1/24*(pi*gamma(-1, 1/2*I*pi*b^2*x^2) + pi*gamma(-1, -1/2*I*pi*b^2*x^2))*b^3 - 1/3*fresnel_sin(b*x)/x^3

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Fricas [A]
time = 0.35, size = 62, normalized size = 1.19 \begin {gather*} \frac {\pi b^{3} x^{3} \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \pi b^{3} x^{3} \operatorname {Ci}\left (-\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 8 \, \operatorname {S}\left (b x\right )}{24 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^4,x, algorithm="fricas")

[Out]

1/24*(pi*b^3*x^3*cos_integral(1/2*pi*b^2*x^2) + pi*b^3*x^3*cos_integral(-1/2*pi*b^2*x^2) - 4*b*x*sin(1/2*pi*b^
2*x^2) - 8*fresnel_sin(b*x))/x^3

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Sympy [A]
time = 0.72, size = 56, normalized size = 1.08 \begin {gather*} - \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} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^4,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)/x^4, x)

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

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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