3.126 \(\int \frac{\text{FresnelC}(b x)}{x^9} \, dx\)

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

[Out]

-(b*Cos[(b^2*Pi*x^2)/2])/(56*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)

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Rubi [A]  time = 0.0784687, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6427, 3388, 3387, 3352} \[ \frac{1}{840} \pi ^4 b^8 \text{FresnelC}(b x)-\frac{\pi ^3 b^7 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{840 x}+\frac{\pi b^3 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{280 x^5}+\frac{\pi ^2 b^5 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{840 x^3}-\frac{b \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{56 x^7}-\frac{\text{FresnelC}(b x)}{8 x^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[(b^2*Pi*x^2)/2])/(56*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 6427

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*b^2*x^2)/2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rule 3388

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 3387

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 3352

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

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*}

Mathematica [A]  time = 0.0577638, size = 85, normalized size = 0.71 \[ \frac{\left (\pi ^4 b^8 x^8-105\right ) \text{FresnelC}(b x)+\pi b^3 x^3 \left (3-\pi ^2 b^4 x^4\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )+b x \left (\pi ^2 b^4 x^4-15\right ) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{840 x^8} \]

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 = 0.048, size = 108, normalized size = 0.9 \begin{align*}{b}^{8} \left ( -{\frac{{\it FresnelC} \left ( bx \right ) }{8\,{b}^{8}{x}^{8}}}-{\frac{1}{56\,{b}^{7}{x}^{7}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{\pi }{56} \left ( -{\frac{1}{5\,{b}^{5}{x}^{5}}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{\pi }{5} \left ( -{\frac{1}{3\,{x}^{3}{b}^{3}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{\pi }{3} \left ( -{\frac{1}{bx}\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+\pi \,{\it FresnelC} \left ( bx \right ) \right ) } \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnelc}\left (b x\right )}{x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x)/x^9, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm fresnelc}\left (b x\right )}{x^{9}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnelc(b*x)/x^9, x)

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Sympy [A]  time = 3.70601, size = 185, normalized size = 1.55 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnelc}\left (b x\right )}{x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x^9,x, algorithm="giac")

[Out]

integrate(fresnelc(b*x)/x^9, x)