3.1.17 \(\int \frac {S(b x)}{x^9} \, dx\) [17]

Optimal. Leaf size=119 \[ -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}+\frac {1}{840} b^8 \pi ^4 S(b x)-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3} \]

[Out]

-1/280*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^5+1/840*b^7*Pi^3*cos(1/2*b^2*Pi*x^2)/x+1/840*b^8*Pi^4*FresnelS(b*x)-1/8*Fr
esnelS(b*x)/x^8-1/56*b*sin(1/2*b^2*Pi*x^2)/x^7+1/840*b^5*Pi^2*sin(1/2*b^2*Pi*x^2)/x^3

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Rubi [A]
time = 0.05, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, 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}{840} \pi ^4 b^8 S(b x)-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{56 x^7}+\frac {\pi ^3 b^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x^3}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{280 x^5}-\frac {S(b x)}{8 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x]/x^9,x]

[Out]

-1/280*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x^5 + (b^7*Pi^3*Cos[(b^2*Pi*x^2)/2])/(840*x) + (b^8*Pi^4*FresnelS[b*x])/84
0 - FresnelS[b*x]/(8*x^8) - (b*Sin[(b^2*Pi*x^2)/2])/(56*x^7) + (b^5*Pi^2*Sin[(b^2*Pi*x^2)/2])/(840*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^9} \, dx &=-\frac {S(b x)}{8 x^8}+\frac {1}{8} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8} \, dx\\ &=-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {1}{56} \left (b^3 \pi \right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}-\frac {1}{280} \left (b^5 \pi ^2\right ) \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}-\frac {1}{840} \left (b^7 \pi ^3\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 )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}+\frac {1}{840} \left (b^9 \pi ^4\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 )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}+\frac {1}{840} b^8 \pi ^4 S(b x)-\frac {S(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 84, normalized size = 0.71 \begin {gather*} \frac {b^3 \pi x^3 \left (-3+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )+\left (-105+b^8 \pi ^4 x^8\right ) S(b x)+b x \left (-15+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x]/x^9,x]

[Out]

(b^3*Pi*x^3*(-3 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + (-105 + b^8*Pi^4*x^8)*FresnelS[b*x] + b*x*(-15 + b^4*Pi^
2*x^4)*Sin[(b^2*Pi*x^2)/2])/(840*x^8)

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Maple [A]
time = 0.32, size = 109, normalized size = 0.92

method	result	size
meijerg	\(-\frac {\pi \,b^{3} \hypergeom \left (\left [-\frac {5}{4}, \frac {3}{4}\right ], \left [-\frac {1}{4}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{30 x^{5}}\)	\(29\)
derivativedivides	\(b^{8} \left (-\frac {\mathrm {S}\left (b x \right )}{8 b^{8} x^{8}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\mathrm {S}\left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\)	\(109\)
default	\(b^{8} \left (-\frac {\mathrm {S}\left (b x \right )}{8 b^{8} x^{8}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\mathrm {S}\left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^9,x,method=_RETURNVERBOSE)

[Out]

b^8*(-1/8*FresnelS(b*x)/b^8/x^8-1/56/b^7/x^7*sin(1/2*b^2*Pi*x^2)+1/56*Pi*(-1/5/b^5/x^5*cos(1/2*b^2*Pi*x^2)-1/5
*Pi*(-1/3/b^3/x^3*sin(1/2*b^2*Pi*x^2)+1/3*Pi*(-1/b/x*cos(1/2*b^2*Pi*x^2)-Pi*FresnelS(b*x)))))

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Maxima [C] Result contains complex when optimal does not.
time = 0.52, size = 61, normalized size = 0.51 \begin {gather*} -\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {7}{2}} {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{8}}{512 \, x^{7}} - \frac {\operatorname {S}\left (b x\right )}{8 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^9,x, algorithm="maxima")

[Out]

-1/512*sqrt(1/2)*(pi*x^2)^(7/2)*((I + 1)*sqrt(2)*gamma(-7/2, 1/2*I*pi*b^2*x^2) - (I - 1)*sqrt(2)*gamma(-7/2, -
1/2*I*pi*b^2*x^2))*b^8/x^7 - 1/8*fresnel_sin(b*x)/x^8

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Fricas [A]
time = 0.35, size = 80, normalized size = 0.67 \begin {gather*} \frac {{\left (\pi ^{3} b^{7} x^{7} - 3 \, \pi b^{3} x^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{4} b^{8} x^{8} - 105\right )} \operatorname {S}\left (b x\right ) + {\left (\pi ^{2} b^{5} x^{5} - 15 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{840 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x)/x^9,x, algorithm="fricas")

[Out]

1/840*((pi^3*b^7*x^7 - 3*pi*b^3*x^3)*cos(1/2*pi*b^2*x^2) + (pi^4*b^8*x^8 - 105)*fresnel_sin(b*x) + (pi^2*b^5*x
^5 - 15*b*x)*sin(1/2*pi*b^2*x^2))/x^8

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Sympy [A]
time = 1.82, size = 185, normalized size = 1.55 \begin {gather*} \frac {\pi ^{4} b^{8} S\left (b x\right ) \Gamma \left (- \frac {5}{4}\right )}{3584 \Gamma \left (\frac {7}{4}\right )} + \frac {\pi ^{3} b^{7} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x \Gamma \left (\frac {7}{4}\right )} + \frac {\pi ^{2} b^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {3 \pi b^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{5} \Gamma \left (\frac {7}{4}\right )} - \frac {15 b \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{7} \Gamma \left (\frac {7}{4}\right )} - \frac {15 S\left (b x\right ) \Gamma \left (- \frac {5}{4}\right )}{512 x^{8} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)/x**9,x)

[Out]

pi**4*b**8*fresnels(b*x)*gamma(-5/4)/(3584*gamma(7/4)) + pi**3*b**7*cos(pi*b**2*x**2/2)*gamma(-5/4)/(3584*x*ga
mma(7/4)) + pi**2*b**5*sin(pi*b**2*x**2/2)*gamma(-5/4)/(3584*x**3*gamma(7/4)) - 3*pi*b**3*cos(pi*b**2*x**2/2)*
gamma(-5/4)/(3584*x**5*gamma(7/4)) - 15*b*sin(pi*b**2*x**2/2)*gamma(-5/4)/(3584*x**7*gamma(7/4)) - 15*fresnels
(b*x)*gamma(-5/4)/(512*x**8*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^9,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)/x^9, 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^9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)/x^9,x)

[Out]

int(FresnelS(b*x)/x^9, x)

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