3.5 \(\int x^3 S(b x) \, dx\)

Optimal. Leaf size=74 \[ \frac{3 S(b x)}{4 \pi ^2 b^4}-\frac{3 x \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi b}+\frac{1}{4} x^4 S(b x) \]

[Out]

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

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Rubi [A]  time = 0.0474889, antiderivative size = 74, 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, 3385, 3386, 3351} \[ \frac{3 S(b x)}{4 \pi ^2 b^4}-\frac{3 x \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi b}+\frac{1}{4} x^4 S(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelS[b*x],x]

[Out]

(x^3*Cos[(b^2*Pi*x^2)/2])/(4*b*Pi) + (3*FresnelS[b*x])/(4*b^4*Pi^2) + (x^4*FresnelS[b*x])/4 - (3*x*Sin[(b^2*Pi
*x^2)/2])/(4*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 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 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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[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^3 S(b x) \, dx &=\frac{1}{4} x^4 S(b x)-\frac{1}{4} b \int x^4 \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac{1}{4} x^4 S(b x)-\frac{3 \int x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{4 b \pi }\\ &=\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac{1}{4} x^4 S(b x)-\frac{3 x \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac{3 \int \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}\\ &=\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac{3 S(b x)}{4 b^4 \pi ^2}+\frac{1}{4} x^4 S(b x)-\frac{3 x \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.0159426, size = 74, normalized size = 1. \[ \frac{3 S(b x)}{4 \pi ^2 b^4}-\frac{3 x \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{x^3 \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{4 \pi b}+\frac{1}{4} x^4 S(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*FresnelS[b*x],x]

[Out]

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

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Maple [A]  time = 0.046, size = 70, normalized size = 1. \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{b}^{4}{x}^{4}{\it FresnelS} \left ( bx \right ) }{4}}+{\frac{{x}^{3}{b}^{3}}{4\,\pi }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{3}{4\,\pi } \left ({\frac{bx}{\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{{\it FresnelS} \left ( bx \right ) }{\pi }} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(b*x),x)

[Out]

1/b^4*(1/4*b^4*x^4*FresnelS(b*x)+1/4/Pi*b^3*x^3*cos(1/2*b^2*Pi*x^2)-3/4/Pi*(1/Pi*b*x*sin(1/2*b^2*Pi*x^2)-1/Pi*
FresnelS(b*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*fresnels(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3}{\rm fresnels}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^3*fresnels(b*x), x)

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Sympy [A]  time = 0.93364, size = 112, normalized size = 1.51 \begin{align*} \frac{21 x^{4} S\left (b x\right ) \Gamma \left (\frac{3}{4}\right )}{64 \Gamma \left (\frac{11}{4}\right )} + \frac{21 x^{3} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{64 \pi b \Gamma \left (\frac{11}{4}\right )} - \frac{63 x \sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac{3}{4}\right )}{64 \pi ^{2} b^{3} \Gamma \left (\frac{11}{4}\right )} + \frac{63 S\left (b x\right ) \Gamma \left (\frac{3}{4}\right )}{64 \pi ^{2} b^{4} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*fresnels(b*x),x)

[Out]

21*x**4*fresnels(b*x)*gamma(3/4)/(64*gamma(11/4)) + 21*x**3*cos(pi*b**2*x**2/2)*gamma(3/4)/(64*pi*b*gamma(11/4
)) - 63*x*sin(pi*b**2*x**2/2)*gamma(3/4)/(64*pi**2*b**3*gamma(11/4)) + 63*fresnels(b*x)*gamma(3/4)/(64*pi**2*b
**4*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^3*fresnels(b*x), x)