3.110 \(\int x^7 \text{FresnelC}(b x) \, dx\)

Optimal. Leaf size=124 \[ -\frac{105 \text{FresnelC}(b x)}{8 \pi ^4 b^8}-\frac{x^7 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi b}+\frac{35 x^3 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{7 x^5 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^2 b^3}+\frac{105 x \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^7}+\frac{1}{8} x^8 \text{FresnelC}(b x) \]

[Out]

(105*x*Cos[(b^2*Pi*x^2)/2])/(8*b^7*Pi^4) - (7*x^5*Cos[(b^2*Pi*x^2)/2])/(8*b^3*Pi^2) - (105*FresnelC[b*x])/(8*b
^8*Pi^4) + (x^8*FresnelC[b*x])/8 + (35*x^3*Sin[(b^2*Pi*x^2)/2])/(8*b^5*Pi^3) - (x^7*Sin[(b^2*Pi*x^2)/2])/(8*b*
Pi)

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Rubi [A]  time = 0.0821382, antiderivative size = 124, 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, 3386, 3385, 3352} \[ -\frac{105 \text{FresnelC}(b x)}{8 \pi ^4 b^8}-\frac{x^7 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi b}+\frac{35 x^3 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^3 b^5}-\frac{7 x^5 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^2 b^3}+\frac{105 x \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^7}+\frac{1}{8} x^8 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^7*FresnelC[b*x],x]

[Out]

(105*x*Cos[(b^2*Pi*x^2)/2])/(8*b^7*Pi^4) - (7*x^5*Cos[(b^2*Pi*x^2)/2])/(8*b^3*Pi^2) - (105*FresnelC[b*x])/(8*b
^8*Pi^4) + (x^8*FresnelC[b*x])/8 + (35*x^3*Sin[(b^2*Pi*x^2)/2])/(8*b^5*Pi^3) - (x^7*Sin[(b^2*Pi*x^2)/2])/(8*b*
Pi)

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 3386

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

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, 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 x^7 C(b x) \, dx &=\frac{1}{8} x^8 C(b x)-\frac{1}{8} b \int x^8 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{8} x^8 C(b x)-\frac{x^7 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac{7 \int x^6 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{8 b \pi }\\ &=-\frac{7 x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{1}{8} x^8 C(b x)-\frac{x^7 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b \pi }+\frac{35 \int x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{8 b^3 \pi ^2}\\ &=-\frac{7 x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{1}{8} x^8 C(b x)+\frac{35 x^3 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{x^7 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b \pi }-\frac{105 \int x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}\\ &=\frac{105 x \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^7 \pi ^4}-\frac{7 x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{1}{8} x^8 C(b x)+\frac{35 x^3 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{x^7 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b \pi }-\frac{105 \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{8 b^7 \pi ^4}\\ &=\frac{105 x \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^7 \pi ^4}-\frac{7 x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2}-\frac{105 C(b x)}{8 b^8 \pi ^4}+\frac{1}{8} x^8 C(b x)+\frac{35 x^3 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac{x^7 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{8 b \pi }\\ \end{align*}

Mathematica [A]  time = 0.0712275, size = 89, normalized size = 0.72 \[ \frac{\left (\pi ^4 b^8 x^8-105\right ) \text{FresnelC}(b x)+\pi b^3 x^3 \left (35-\pi ^2 b^4 x^4\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )-7 b x \left (\pi ^2 b^4 x^4-15\right ) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 \pi ^4 b^8} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*FresnelC[b*x],x]

[Out]

(-7*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*(35 - b^4*
Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/(8*b^8*Pi^4)

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Maple [A]  time = 0.049, size = 123, normalized size = 1. \begin{align*}{\frac{1}{{b}^{8}} \left ({\frac{{\it FresnelC} \left ( bx \right ){b}^{8}{x}^{8}}{8}}-{\frac{{b}^{7}{x}^{7}}{8\,\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{7}{8\,\pi } \left ( -{\frac{{b}^{5}{x}^{5}}{\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+5\,{\frac{1}{\pi } \left ({\frac{{x}^{3}{b}^{3}\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}-3\,{\frac{1}{\pi } \left ( -{\frac{\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) bx}{\pi }}+{\frac{{\it FresnelC} \left ( bx \right ) }{\pi }} \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*FresnelC(b*x),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{7}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnelc(b*x),x, algorithm="maxima")

[Out]

integrate(x^7*fresnelc(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{7}{\rm fresnelc}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnelc(b*x),x, algorithm="fricas")

[Out]

integral(x^7*fresnelc(b*x), x)

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Sympy [A]  time = 2.60003, size = 184, normalized size = 1.48 \begin{align*} \frac{45 x^{8} C\left (b x\right ) \Gamma \left (\frac{1}{4}\right )}{512 \Gamma \left (\frac{13}{4}\right )} - \frac{45 x^{7} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{512 \pi b \Gamma \left (\frac{13}{4}\right )} - \frac{315 x^{5} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{512 \pi ^{2} b^{3} \Gamma \left (\frac{13}{4}\right )} + \frac{1575 x^{3} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{512 \pi ^{3} b^{5} \Gamma \left (\frac{13}{4}\right )} + \frac{4725 x \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{512 \pi ^{4} b^{7} \Gamma \left (\frac{13}{4}\right )} - \frac{4725 C\left (b x\right ) \Gamma \left (\frac{1}{4}\right )}{512 \pi ^{4} b^{8} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*fresnelc(b*x),x)

[Out]

45*x**8*fresnelc(b*x)*gamma(1/4)/(512*gamma(13/4)) - 45*x**7*sin(pi*b**2*x**2/2)*gamma(1/4)/(512*pi*b*gamma(13
/4)) - 315*x**5*cos(pi*b**2*x**2/2)*gamma(1/4)/(512*pi**2*b**3*gamma(13/4)) + 1575*x**3*sin(pi*b**2*x**2/2)*ga
mma(1/4)/(512*pi**3*b**5*gamma(13/4)) + 4725*x*cos(pi*b**2*x**2/2)*gamma(1/4)/(512*pi**4*b**7*gamma(13/4)) - 4
725*fresnelc(b*x)*gamma(1/4)/(512*pi**4*b**8*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{7}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*fresnelc(b*x),x, algorithm="giac")

[Out]

integrate(x^7*fresnelc(b*x), x)