3.1.13 \(\int \frac {S(b x)}{x^5} \, dx\) [13]

Optimal. Leaf size=69 \[ -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 S(b x)-\frac {S(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \]

[Out]

-1/12*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x-1/12*b^4*Pi^2*FresnelS(b*x)-1/4*FresnelS(b*x)/x^4-1/12*b*sin(1/2*b^2*Pi*x^2
)/x^3

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Rubi [A]
time = 0.03, antiderivative size = 69, 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, 3468, 3469, 3432} \begin {gather*} -\frac {1}{12} \pi ^2 b^4 S(b x)-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x^3}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{12 x}-\frac {S(b x)}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x - (b^4*Pi^2*FresnelS[b*x])/12 - FresnelS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^
2)/2])/(12*x^3)

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[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 3469

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x)^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1)
)), x] + Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Sin[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^5} \, dx &=-\frac {S(b x)}{4 x^4}+\frac {1}{4} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac {S(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3}+\frac {1}{12} \left (b^3 \pi \right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {S(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3}-\frac {1}{12} \left (b^5 \pi ^2\right ) \int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 S(b x)-\frac {S(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 69, normalized size = 1.00 \begin {gather*} -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 S(b x)-\frac {S(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x - (b^4*Pi^2*FresnelS[b*x])/12 - FresnelS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^
2)/2])/(12*x^3)

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Maple [A]
time = 0.33, size = 65, normalized size = 0.94

method	result	size
derivativedivides	\(b^{4} \left (-\frac {\mathrm {S}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\mathrm {S}\left (b x \right )\right )}{12}\right )\)	\(65\)
default	\(b^{4} \left (-\frac {\mathrm {S}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\mathrm {S}\left (b x \right )\right )}{12}\right )\)	\(65\)
meijerg	\(\frac {\pi ^{2} b^{4} \left (-\frac {32 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi x b}-\frac {32 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2} x^{3} b^{3}}-\frac {32 \left (x^{4} \pi ^{2} b^{4}+3\right ) \mathrm {S}\left (b x \right )}{3 \pi ^{2} x^{4} b^{4}}\right )}{128}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.52, size = 61, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {3}{2}} {\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{4}}{64 \, x^{3}} - \frac {\operatorname {S}\left (b x\right )}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*sqrt(1/2)*(pi*x^2)^(3/2)*(-(I + 1)*sqrt(2)*gamma(-3/2, 1/2*I*pi*b^2*x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -
1/2*I*pi*b^2*x^2))*b^4/x^3 - 1/4*fresnel_sin(b*x)/x^4

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Fricas [A]
time = 0.34, size = 54, normalized size = 0.78 \begin {gather*} -\frac {\pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + b x \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 )}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.61, size = 110, normalized size = 1.59 \begin {gather*} \frac {\pi ^{2} b^{4} S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 \Gamma \left (\frac {7}{4}\right )} + \frac {\pi b^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x \Gamma \left (\frac {7}{4}\right )} + \frac {b \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x^{3} \Gamma \left (\frac {7}{4}\right )} + \frac {3 S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 x^{4} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

pi**2*b**4*fresnels(b*x)*gamma(-1/4)/(64*gamma(7/4)) + pi*b**3*cos(pi*b**2*x**2/2)*gamma(-1/4)/(64*x*gamma(7/4
)) + b*sin(pi*b**2*x**2/2)*gamma(-1/4)/(64*x**3*gamma(7/4)) + 3*fresnels(b*x)*gamma(-1/4)/(64*x**4*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^5,x, algorithm="giac")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelS(b*x)/x^5, x)

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