3.1.43 \(\int \frac {S(b x)^2}{x^5} \, dx\) [43]

Optimal. Leaf size=127 \[ -\frac {b^2}{24 x^2}+\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{6 x}-\frac {1}{12} b^4 \pi ^2 S(b x)^2-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}+\frac {1}{12} b^4 \pi \text {Si}\left (b^2 \pi x^2\right ) \]

[Out]

-1/24*b^2/x^2+1/24*b^2*cos(b^2*Pi*x^2)/x^2-1/6*b^3*Pi*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x-1/12*b^4*Pi^2*Fresne
lS(b*x)^2-1/4*FresnelS(b*x)^2/x^4+1/12*b^4*Pi*Si(b^2*Pi*x^2)-1/6*b*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x^3

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Rubi [A]
time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6565, 6591, 6599, 6575, 30, 3456, 3461, 3378, 3380} \begin {gather*} -\frac {1}{12} \pi ^2 b^4 S(b x)^2-\frac {b S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^3}-\frac {b^2}{24 x^2}+\frac {b^2 \cos \left (\pi b^2 x^2\right )}{24 x^2}+\frac {1}{12} \pi b^4 \text {Si}\left (b^2 \pi x^2\right )-\frac {\pi b^3 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x}-\frac {S(b x)^2}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]^2/x^5,x]

[Out]

-1/24*b^2/x^2 + (b^2*Cos[b^2*Pi*x^2])/(24*x^2) - (b^3*Pi*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(6*x) - (b^4*Pi^2*
FresnelS[b*x]^2)/12 - FresnelS[b*x]^2/(4*x^4) - (b*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(6*x^3) + (b^4*Pi*SinInt
egral[b^2*Pi*x^2])/12

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 3456

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 6565

Int[FresnelS[(b_.)*(x_)]^2*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(FresnelS[b*x]^2/(m + 1)), x] - Dist[2*(b/(
m + 1)), Int[x^(m + 1)*Sin[(Pi/2)*b^2*x^2]*FresnelS[b*x], x], x] /; FreeQ[b, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6575

Int[FresnelS[(b_.)*(x_)]^(n_.)*Sin[(d_.)*(x_)^2], x_Symbol] :> Dist[Pi*(b/(2*d)), Subst[Int[x^n, x], x, Fresne
lS[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2/4)*b^4]

Rule 6591

Int[FresnelS[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[x^(m + 1)*Sin[d*x^2]*(FresnelS[b*x]/(m
+ 1)), x] + (-Dist[2*(d/(m + 1)), Int[x^(m + 2)*Cos[d*x^2]*FresnelS[b*x], x], x] + Dist[d/(Pi*b*(m + 1)), Int
[x^(m + 1)*Cos[2*d*x^2], x], x] - Simp[d*(x^(m + 2)/(Pi*b*(m + 1)*(m + 2))), x]) /; FreeQ[{b, d}, x] && EqQ[d^
2, (Pi^2/4)*b^4] && ILtQ[m, -2]

Rule 6599

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[x^(m + 1)*Cos[d*x^2]*(FresnelS[b*x]/(m
+ 1)), x] + (Dist[2*(d/(m + 1)), Int[x^(m + 2)*Sin[d*x^2]*FresnelS[b*x], x], x] - Dist[d/(Pi*b*(m + 1)), Int[
x^(m + 1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {S(b x)^2}{x^5} \, dx &=-\frac {S(b x)^2}{4 x^4}+\frac {1}{2} b \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac {b^2}{24 x^2}-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}-\frac {1}{12} b^2 \int \frac {\cos \left (b^2 \pi x^2\right )}{x^3} \, dx+\frac {1}{6} \left (b^3 \pi \right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{x^2} \, dx\\ &=-\frac {b^2}{24 x^2}-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{6 x}-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}-\frac {1}{24} b^2 \text {Subst}\left (\int \frac {\cos \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )+\frac {1}{12} \left (b^4 \pi \right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x} \, dx-\frac {1}{6} \left (b^5 \pi ^2\right ) \int S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac {b^2}{24 x^2}+\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{6 x}-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}+\frac {1}{24} b^4 \pi \text {Si}\left (b^2 \pi x^2\right )+\frac {1}{24} \left (b^4 \pi \right ) \text {Subst}\left (\int \frac {\sin \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right )-\frac {1}{6} \left (b^4 \pi ^2\right ) \text {Subst}(\int x \, dx,x,S(b x))\\ &=-\frac {b^2}{24 x^2}+\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{6 x}-\frac {1}{12} b^4 \pi ^2 S(b x)^2-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}+\frac {1}{12} b^4 \pi \text {Si}\left (b^2 \pi x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 127, normalized size = 1.00 \begin {gather*} -\frac {b^2}{24 x^2}+\frac {b^2 \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{6 x}-\frac {1}{12} b^4 \pi ^2 S(b x)^2-\frac {S(b x)^2}{4 x^4}-\frac {b S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^3}+\frac {1}{12} b^4 \pi \text {Si}\left (b^2 \pi x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]^2/x^5,x]

[Out]

-1/24*b^2/x^2 + (b^2*Cos[b^2*Pi*x^2])/(24*x^2) - (b^3*Pi*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(6*x) - (b^4*Pi^2*
FresnelS[b*x]^2)/12 - FresnelS[b*x]^2/(4*x^4) - (b*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(6*x^3) + (b^4*Pi*SinInt
egral[b^2*Pi*x^2])/12

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\mathrm {S}\left (b x \right )^{2}}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)^2/x^5,x)

[Out]

int(FresnelS(b*x)^2/x^5,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)^2/x^5,x, algorithm="maxima")

[Out]

integrate(fresnel_sin(b*x)^2/x^5, x)

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Fricas [A]
time = 0.36, size = 111, normalized size = 0.87 \begin {gather*} \frac {\pi b^{4} x^{4} \operatorname {Si}\left (\pi b^{2} x^{2}\right ) - 2 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} - b^{2} x^{2} - 2 \, b x \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {S}\left (b x\right )^{2}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)^2/x^5,x, algorithm="fricas")

[Out]

1/12*(pi*b^4*x^4*sin_integral(pi*b^2*x^2) - 2*pi*b^3*x^3*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) + b^2*x^2*cos(1/
2*pi*b^2*x^2)^2 - b^2*x^2 - 2*b*x*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2) - (pi^2*b^4*x^4 + 3)*fresnel_sin(b*x)^2
)/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {S^{2}\left (b x\right )}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)**2/x**5,x)

[Out]

Integral(fresnels(b*x)**2/x**5, x)

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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)^2/x^5,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)^2/x^5, 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 )}^2}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)^2/x^5,x)

[Out]

int(FresnelS(b*x)^2/x^5, x)

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