3.113 \(\int x^4 \text{FresnelC}(b x) \, dx\)

Optimal. Leaf size=84 \[ -\frac{x^4 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac{8 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac{4 x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac{1}{5} x^5 \text{FresnelC}(b x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0766497, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6427, 3380, 3296, 2637} \[ -\frac{x^4 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac{8 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac{4 x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac{1}{5} x^5 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*FresnelC[b*x],x]

[Out]

(-4*x^2*Cos[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2) + (x^5*FresnelC[b*x])/5 + (8*Sin[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) - (x^
4*Sin[(b^2*Pi*x^2)/2])/(5*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 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^4 C(b x) \, dx &=\frac{1}{5} x^5 C(b x)-\frac{1}{5} b \int x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{5} x^5 C(b x)-\frac{1}{10} b \operatorname{Subst}\left (\int x^2 \cos \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac{1}{5} x^5 C(b x)-\frac{x^4 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{2 \operatorname{Subst}\left (\int x \sin \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi }\\ &=-\frac{4 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)-\frac{x^4 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{4 \operatorname{Subst}\left (\int \cos \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2}\\ &=-\frac{4 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)+\frac{8 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac{x^4 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }\\ \end{align*}

Mathematica [A]  time = 0.0434825, size = 71, normalized size = 0.85 \[ -\frac{\left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac{4 x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac{1}{5} x^5 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*FresnelC[b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 81, normalized size = 1. \begin{align*}{\frac{1}{{b}^{5}} \left ({\frac{{b}^{5}{x}^{5}{\it FresnelC} \left ( bx \right ) }{5}}-{\frac{{x}^{4}{b}^{4}}{5\,\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+{\frac{4}{5\,\pi } \left ( -{\frac{{b}^{2}{x}^{2}}{\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+2\,{\frac{\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*FresnelC(b*x),x)

[Out]

1/b^5*(1/5*b^5*x^5*FresnelC(b*x)-1/5/Pi*b^4*x^4*sin(1/2*b^2*Pi*x^2)+4/5/Pi*(-1/Pi*b^2*x^2*cos(1/2*b^2*Pi*x^2)+
2/Pi^2*sin(1/2*b^2*Pi*x^2)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x),x, algorithm="maxima")

[Out]

integrate(x^4*fresnelc(b*x), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{4}{\rm fresnelc}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x),x, algorithm="fricas")

[Out]

integral(x^4*fresnelc(b*x), x)

________________________________________________________________________________________

Sympy [A]  time = 1.223, size = 116, normalized size = 1.38 \begin{align*} \frac{x^{5} C\left (b x\right ) \Gamma \left (\frac{1}{4}\right )}{20 \Gamma \left (\frac{5}{4}\right )} - \frac{x^{4} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{20 \pi b \Gamma \left (\frac{5}{4}\right )} - \frac{x^{2} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac{5}{4}\right )} + \frac{2 \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*fresnelc(b*x),x)

[Out]

x**5*fresnelc(b*x)*gamma(1/4)/(20*gamma(5/4)) - x**4*sin(pi*b**2*x**2/2)*gamma(1/4)/(20*pi*b*gamma(5/4)) - x**
2*cos(pi*b**2*x**2/2)*gamma(1/4)/(5*pi**2*b**3*gamma(5/4)) + 2*sin(pi*b**2*x**2/2)*gamma(1/4)/(5*pi**3*b**5*ga
mma(5/4))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x),x, algorithm="giac")

[Out]

integrate(x^4*fresnelc(b*x), x)