3.6 \(\int x^2 S(b x) \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.052703, 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 = {6426, 3379, 3296, 2637} \[ -\frac{2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}+\frac{1}{3} x^3 S(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6426

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

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[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^2 S(b x) \, dx &=\frac{1}{3} x^3 S(b x)-\frac{1}{3} b \int x^3 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{3} x^3 S(b x)-\frac{1}{6} b \operatorname{Subst}\left (\int x \sin \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac{x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac{1}{3} x^3 S(b x)-\frac{\operatorname{Subst}\left (\int \cos \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{3 b \pi }\\ &=\frac{x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac{1}{3} x^3 S(b x)-\frac{2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.0115254, size = 59, normalized size = 1. \[ -\frac{2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}+\frac{1}{3} x^3 S(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 54, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it FresnelS} \left ( bx \right ) }{3}}+{\frac{{b}^{2}{x}^{2}}{3\,\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{2}{3\,{\pi }^{2}}\sin \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*FresnelS(b*x),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*fresnels(b*x), x)

________________________________________________________________________________________

Sympy [A]  time = 0.828279, size = 80, normalized size = 1.36 \begin{align*} \frac{x^{3} S\left (b x\right ) \Gamma \left (\frac{3}{4}\right )}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{x^{2} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{4 \pi b \Gamma \left (\frac{7}{4}\right )} - \frac{\sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{2 \pi ^{2} b^{3} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*fresnels(b*x)*gamma(3/4)/(4*gamma(7/4)) + x**2*cos(pi*b**2*x**2/2)*gamma(3/4)/(4*pi*b*gamma(7/4)) - sin(p
i*b**2*x**2/2)*gamma(3/4)/(2*pi**2*b**3*gamma(7/4))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x), x)