3.1.11 \(\int \frac {S(b x)}{x^3} \, dx\) [11]

Optimal. Leaf size=44 \[ \frac {1}{2} b^2 \pi \text {FresnelC}(b x)-\frac {S(b x)}{2 x^2}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x} \]

[Out]

1/2*b^2*Pi*FresnelC(b*x)-1/2*FresnelS(b*x)/x^2-1/2*b*sin(1/2*b^2*Pi*x^2)/x

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6561, 3468, 3433} \begin {gather*} \frac {1}{2} \pi b^2 \text {FresnelC}(b x)-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x}-\frac {S(b x)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^3,x]

[Out]

(b^2*Pi*FresnelC[b*x])/2 - FresnelS[b*x]/(2*x^2) - (b*Sin[(b^2*Pi*x^2)/2])/(2*x)

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3468

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x)^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1)
)), x] - Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

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^3} \, dx &=-\frac {S(b x)}{2 x^2}+\frac {1}{2} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac {S(b x)}{2 x^2}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}+\frac {1}{2} \left (b^3 \pi \right ) \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {1}{2} b^2 \pi C(b x)-\frac {S(b x)}{2 x^2}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 44, normalized size = 1.00 \begin {gather*} \frac {1}{2} b^2 \pi \text {FresnelC}(b x)-\frac {S(b x)}{2 x^2}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^3,x]

[Out]

(b^2*Pi*FresnelC[b*x])/2 - FresnelS[b*x]/(2*x^2) - (b*Sin[(b^2*Pi*x^2)/2])/(2*x)

________________________________________________________________________________________

Maple [A]
time = 0.33, size = 43, normalized size = 0.98

method	result	size
meijerg	\(\frac {\pi \,b^{3} x \hypergeom \left (\left [\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {5}{4}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{6}\)	\(27\)
derivativedivides	\(b^{2} \left (-\frac {\mathrm {S}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b x}+\frac {\pi \FresnelC \left (b x \right )}{2}\right )\)	\(43\)
default	\(b^{2} \left (-\frac {\mathrm {S}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b x}+\frac {\pi \FresnelC \left (b x \right )}{2}\right )\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

b^2*(-1/2*FresnelS(b*x)/b^2/x^2-1/2/b/x*sin(1/2*b^2*Pi*x^2)+1/2*Pi*FresnelC(b*x))

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.53, size = 61, normalized size = 1.39 \begin {gather*} -\frac {\sqrt {\frac {1}{2}} \sqrt {\pi x^{2}} {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{2}}{16 \, x} - \frac {\operatorname {S}\left (b x\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^3,x, algorithm="maxima")

[Out]

-1/16*sqrt(1/2)*sqrt(pi*x^2)*((I - 1)*sqrt(2)*gamma(-1/2, 1/2*I*pi*b^2*x^2) - (I + 1)*sqrt(2)*gamma(-1/2, -1/2
*I*pi*b^2*x^2))*b^2/x - 1/2*fresnel_sin(b*x)/x^2

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 45, normalized size = 1.02 \begin {gather*} \frac {\pi \sqrt {b^{2}} b x^{2} \operatorname {C}\left (\sqrt {b^{2}} x\right ) - b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \operatorname {S}\left (b x\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^3,x, algorithm="fricas")

[Out]

1/2*(pi*sqrt(b^2)*b*x^2*fresnel_cos(sqrt(b^2)*x) - b*x*sin(1/2*pi*b^2*x^2) - fresnel_sin(b*x))/x^2

________________________________________________________________________________________

Sympy [A]
time = 0.34, size = 51, normalized size = 1.16 \begin {gather*} \frac {\pi b^{3} x \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {5}{4}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**3,x)

[Out]

pi*b**3*x*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (5/4, 3/2, 7/4), -pi**2*b**4*x**4/16)/(32*gamma(5/4)*gamma(7
/4))

________________________________________________________________________________________

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^3,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)/x^3, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^3,x)

[Out]

int(FresnelS(b*x)/x^3, x)

________________________________________________________________________________________