3.2.26 \(\int \frac {\text {FresnelC}(b x)}{x^9} \, dx\) [126]

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

[Out]

-1/56*b*cos(1/2*b^2*Pi*x^2)/x^7+1/840*b^5*Pi^2*cos(1/2*b^2*Pi*x^2)/x^3+1/840*b^8*Pi^4*FresnelC(b*x)-1/8*Fresne
lC(b*x)/x^8+1/280*b^3*Pi*sin(1/2*b^2*Pi*x^2)/x^5-1/840*b^7*Pi^3*sin(1/2*b^2*Pi*x^2)/x

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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 = {6562, 3469, 3468, 3433} \begin {gather*} \frac {1}{840} \pi ^4 b^8 \text {FresnelC}(b x)-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{56 x^7}-\frac {\pi ^3 b^7 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x}+\frac {\pi ^2 b^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x^3}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{280 x^5}-\frac {\text {FresnelC}(b x)}{8 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/56*(b*Cos[(b^2*Pi*x^2)/2])/x^7 + (b^5*Pi^2*Cos[(b^2*Pi*x^2)/2])/(840*x^3) + (b^8*Pi^4*FresnelC[b*x])/840 -
FresnelC[b*x]/(8*x^8) + (b^3*Pi*Sin[(b^2*Pi*x^2)/2])/(280*x^5) - (b^7*Pi^3*Sin[(b^2*Pi*x^2)/2])/(840*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 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 6562

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

Rubi steps

\begin {align*} \int \frac {C(b x)}{x^9} \, dx &=-\frac {C(b x)}{8 x^8}+\frac {1}{8} b \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8} \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}-\frac {C(b x)}{8 x^8}-\frac {1}{56} \left (b^3 \pi \right ) \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}-\frac {C(b x)}{8 x^8}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {1}{280} \left (b^5 \pi ^2\right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}-\frac {C(b x)}{8 x^8}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}+\frac {1}{840} \left (b^7 \pi ^3\right ) \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}-\frac {C(b x)}{8 x^8}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {b^7 \pi ^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}+\frac {1}{840} \left (b^9 \pi ^4\right ) \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}+\frac {1}{840} b^8 \pi ^4 C(b x)-\frac {C(b x)}{8 x^8}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {b^7 \pi ^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 85, normalized size = 0.71 \begin {gather*} \frac {b x \left (-15+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 ) \text {FresnelC}(b x)+b^3 \pi x^3 \left (3-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[FresnelC[b*x]/x^9,x]

[Out]

(b*x*(-15 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + (-105 + b^8*Pi^4*x^8)*FresnelC[b*x] + b^3*Pi*x^3*(3 - 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 = 108, normalized size = 0.91

method	result	size
meijerg	\(-\frac {b \hypergeom \left (\left [-\frac {7}{4}, \frac {1}{4}\right ], \left [-\frac {3}{4}, \frac {1}{2}, \frac {5}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{7 x^{7}}\)	\(26\)
derivativedivides	\(b^{8} \left (-\frac {\FresnelC \left (b x \right )}{8 b^{8} x^{8}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}+\pi \FresnelC \left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\)	\(108\)
default	\(b^{8} \left (-\frac {\FresnelC \left (b x \right )}{8 b^{8} x^{8}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}+\pi \FresnelC \left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.54, 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 {C}\left (b x\right )}{8 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(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_cos(b*x)/x^8

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Fricas [A]
time = 0.36, size = 81, normalized size = 0.68 \begin {gather*} \frac {{\left (\pi ^{2} b^{5} x^{5} - 15 \, b x\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{4} b^{8} x^{8} - 105\right )} \operatorname {C}\left (b x\right ) - {\left (\pi ^{3} b^{7} x^{7} - 3 \, \pi b^{3} x^{3}\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_cos(b*x)/x^9,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 1.84, size = 185, normalized size = 1.55 \begin {gather*} \frac {\pi ^{4} b^{8} C\left (b x\right ) \Gamma \left (- \frac {7}{4}\right )}{2560 \Gamma \left (\frac {5}{4}\right )} - \frac {\pi ^{3} b^{7} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {7}{4}\right )}{2560 x \Gamma \left (\frac {5}{4}\right )} + \frac {\pi ^{2} b^{5} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {7}{4}\right )}{2560 x^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {3 \pi b^{3} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {7}{4}\right )}{2560 x^{5} \Gamma \left (\frac {5}{4}\right )} - \frac {3 b \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {7}{4}\right )}{512 x^{7} \Gamma \left (\frac {5}{4}\right )} - \frac {21 C\left (b x\right ) \Gamma \left (- \frac {7}{4}\right )}{512 x^{8} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

pi**4*b**8*fresnelc(b*x)*gamma(-7/4)/(2560*gamma(5/4)) - pi**3*b**7*sin(pi*b**2*x**2/2)*gamma(-7/4)/(2560*x*ga
mma(5/4)) + pi**2*b**5*cos(pi*b**2*x**2/2)*gamma(-7/4)/(2560*x**3*gamma(5/4)) + 3*pi*b**3*sin(pi*b**2*x**2/2)*
gamma(-7/4)/(2560*x**5*gamma(5/4)) - 3*b*cos(pi*b**2*x**2/2)*gamma(-7/4)/(512*x**7*gamma(5/4)) - 21*fresnelc(b
*x)*gamma(-7/4)/(512*x**8*gamma(5/4))

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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_cos(b*x)/x^9,x, algorithm="giac")

[Out]

integrate(fresnel_cos(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 {FresnelC}\left (b\,x\right )}{x^9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelC(b*x)/x^9, x)

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