3.1.14 \(\int \frac {S(b x)}{x^6} \, dx\) [14]

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

[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

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Rubi [A]
time = 0.06, 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 = {6561, 3460, 3378, 3380} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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

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 3380

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 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^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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 77, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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

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Maple [A]
time = 0.38, size = 71, normalized size = 0.92

method	result	size
meijerg	\(-\frac {\pi \,b^{3} \hypergeom \left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{12 x^{2}}\)	\(29\)
derivativedivides	\(b^{5} \left (-\frac {\mathrm {S}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \sinIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\)	\(71\)
default	\(b^{5} \left (-\frac {\mathrm {S}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \sinIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^6,x,method=_RETURNVERBOSE)

[Out]

b^5*(-1/5*FresnelS(b*x)/b^5/x^5-1/20/b^4/x^4*sin(1/2*b^2*Pi*x^2)+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)))

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Maxima [C] Result contains complex when optimal does not.
time = 0.32, size = 48, normalized size = 0.62 \begin {gather*} -\frac {1}{80} \, {\left (-i \, \pi ^{2} \Gamma \left (-2, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + i \, \pi ^{2} \Gamma \left (-2, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{5} - \frac {\operatorname {S}\left (b x\right )}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^6,x, algorithm="maxima")

[Out]

-1/80*(-I*pi^2*gamma(-2, 1/2*I*pi*b^2*x^2) + I*pi^2*gamma(-2, -1/2*I*pi*b^2*x^2))*b^5 - 1/5*fresnel_sin(b*x)/x
^5

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Fricas [A]
time = 0.37, size = 65, normalized size = 0.84 \begin {gather*} -\frac {\pi ^{2} b^{5} x^{5} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 16 \, \operatorname {S}\left (b x\right )}{80 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^6,x, algorithm="fricas")

[Out]

-1/80*(pi^2*b^5*x^5*sin_integral(1/2*pi*b^2*x^2) + 2*pi*b^3*x^3*cos(1/2*pi*b^2*x^2) + 4*b*x*sin(1/2*pi*b^2*x^2
) + 16*fresnel_sin(b*x))/x^5

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Sympy [A]
time = 0.60, size = 46, normalized size = 0.60 \begin {gather*} - \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 )} \end {gather*}

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))

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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^6,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)/x^6, x)

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

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)

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