3.4 \(\int x^4 S(b x) \, dx\)

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

[Out]

(-8*Cos[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) + (x^4*Cos[(b^2*Pi*x^2)/2])/(5*b*Pi) + (x^5*FresnelS[b*x])/5 - (4*x^2*Si
n[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2)

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Rubi [A]  time = 0.0801435, 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 = {6426, 3379, 3296, 2638} \[ -\frac{4 x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac{x^4 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi b}-\frac{8 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}+\frac{1}{5} x^5 S(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*Cos[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) + (x^4*Cos[(b^2*Pi*x^2)/2])/(5*b*Pi) + (x^5*FresnelS[b*x])/5 - (4*x^2*Si
n[(b^2*Pi*x^2)/2])/(5*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 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^4 S(b x) \, dx &=\frac{1}{5} x^5 S(b x)-\frac{1}{5} b \int x^5 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{1}{5} x^5 S(b x)-\frac{1}{10} b \operatorname{Subst}\left (\int x^2 \sin \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )\\ &=\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{1}{5} x^5 S(b x)-\frac{2 \operatorname{Subst}\left (\int x \cos \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi }\\ &=\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{1}{5} x^5 S(b x)-\frac{4 x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac{4 \operatorname{Subst}\left (\int \sin \left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2}\\ &=-\frac{8 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac{x^4 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{1}{5} x^5 S(b x)-\frac{4 x^2 \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.0407404, size = 71, normalized size = 0.85 \[ -\frac{4 x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac{\left (\pi ^2 b^4 x^4-8\right ) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}+\frac{1}{5} x^5 S(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 80, normalized size = 1. \begin{align*}{\frac{1}{{b}^{5}} \left ({\frac{{b}^{5}{x}^{5}{\it FresnelS} \left ( bx \right ) }{5}}+{\frac{{x}^{4}{b}^{4}}{5\,\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{4}{5\,\pi } \left ({\frac{{b}^{2}{x}^{2}}{\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+2\,{\frac{\cos \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*FresnelS(b*x),x)

[Out]

1/b^5*(1/5*b^5*x^5*FresnelS(b*x)+1/5/Pi*b^4*x^4*cos(1/2*b^2*Pi*x^2)-4/5/Pi*(1/Pi*b^2*x^2*sin(1/2*b^2*Pi*x^2)+2
/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^{4}{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnels(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4*fresnels(b*x), x)

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Sympy [A]  time = 1.62895, size = 121, normalized size = 1.44 \begin{align*} \frac{3 x^{5} S\left (b x\right ) \Gamma \left (\frac{3}{4}\right )}{20 \Gamma \left (\frac{7}{4}\right )} + \frac{3 x^{4} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{20 \pi b \Gamma \left (\frac{7}{4}\right )} - \frac{3 x^{2} \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac{7}{4}\right )} - \frac{6 \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**5*fresnels(b*x)*gamma(3/4)/(20*gamma(7/4)) + 3*x**4*cos(pi*b**2*x**2/2)*gamma(3/4)/(20*pi*b*gamma(7/4)) -
3*x**2*sin(pi*b**2*x**2/2)*gamma(3/4)/(5*pi**2*b**3*gamma(7/4)) - 6*cos(pi*b**2*x**2/2)*gamma(3/4)/(5*pi**3*b
**5*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnels(b*x), x)