3.115 \(\int x^2 \text{FresnelC}(b x) \, dx\)

Optimal. Leaf size=59 \[ -\frac{x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac{2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{1}{3} x^3 \text{FresnelC}(b x) \]

[Out]

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

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Rubi [A]  time = 0.0543763, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6427, 3380, 3296, 2638} \[ -\frac{x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac{2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{1}{3} x^3 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*FresnelC[b*x],x]

[Out]

(-2*Cos[(b^2*Pi*x^2)/2])/(3*b^3*Pi^2) + (x^3*FresnelC[b*x])/3 - (x^2*Sin[(b^2*Pi*x^2)/2])/(3*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 2638

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

Rubi steps

\begin{align*} \int x^2 C(b x) \, dx &=\frac{1}{3} x^3 C(b x)-\frac{1}{3} b \int x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{3} x^3 C(b x)-\frac{1}{6} b \operatorname{Subst}\left (\int x \cos \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 C(b x)-\frac{x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac{\operatorname{Subst}\left (\int \sin \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{3 b \pi }\\ &=-\frac{2 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac{1}{3} x^3 C(b x)-\frac{x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b \pi }\\ \end{align*}

Mathematica [A]  time = 0.0136692, size = 59, normalized size = 1. \[ -\frac{x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac{2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{1}{3} x^3 \text{FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*FresnelC[b*x],x]

[Out]

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

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Maple [A]  time = 0.05, size = 54, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it FresnelC} \left ( bx \right ) }{3}}-{\frac{{b}^{2}{x}^{2}}{3\,\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{2}{3\,{\pi }^{2}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelC(b*x),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnelc(b*x),x, algorithm="maxima")

[Out]

integrate(x^2*fresnelc(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2}{\rm fresnelc}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnelc(b*x),x, algorithm="fricas")

[Out]

integral(x^2*fresnelc(b*x), x)

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Sympy [A]  time = 1.03035, size = 80, normalized size = 1.36 \begin{align*} \frac{x^{3} C\left (b x\right ) \Gamma \left (\frac{1}{4}\right )}{12 \Gamma \left (\frac{5}{4}\right )} - \frac{x^{2} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{12 \pi b \Gamma \left (\frac{5}{4}\right )} - \frac{\cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{1}{4}\right )}{6 \pi ^{2} b^{3} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*fresnelc(b*x),x)

[Out]

x**3*fresnelc(b*x)*gamma(1/4)/(12*gamma(5/4)) - x**2*sin(pi*b**2*x**2/2)*gamma(1/4)/(12*pi*b*gamma(5/4)) - cos
(pi*b**2*x**2/2)*gamma(1/4)/(6*pi**2*b**3*gamma(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnelc(b*x),x, algorithm="giac")

[Out]

integrate(x^2*fresnelc(b*x), x)