3.35 \(\int x^3 S(b x)^2 \, dx\)

Optimal. Leaf size=140 \[ -\frac{3 x S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi ^2 b^3}+\frac{x^3 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac{3 S(b x)^2}{4 \pi ^2 b^4}+\frac{3 x^2}{8 \pi ^2 b^2}-\frac{\sin \left (\pi b^2 x^2\right )}{2 \pi ^3 b^4}+\frac{x^2 \cos \left (\pi b^2 x^2\right )}{8 \pi ^2 b^2}+\frac{1}{4} x^4 S(b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.150878, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6430, 6454, 6462, 3379, 2634, 6440, 30, 3296, 2637} \[ -\frac{3 x S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi ^2 b^3}+\frac{x^3 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac{3 S(b x)^2}{4 \pi ^2 b^4}+\frac{3 x^2}{8 \pi ^2 b^2}-\frac{\sin \left (\pi b^2 x^2\right )}{2 \pi ^3 b^4}+\frac{x^2 \cos \left (\pi b^2 x^2\right )}{8 \pi ^2 b^2}+\frac{1}{4} x^4 S(b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelS[b*x]^2,x]

[Out]

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

Rule 6430

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

Rule 6454

Int[FresnelS[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(x^(m - 1)*Cos[d*x^2]*FresnelS[b*x])/
(2*d), x] + (Dist[(m - 1)/(2*d), Int[x^(m - 2)*Cos[d*x^2]*FresnelS[b*x], x], x] + Dist[1/(2*b*Pi), Int[x^(m -
1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 6462

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelS[b*x])/(
2*d), x] + (-Dist[1/(Pi*b), Int[x^(m - 1)*Sin[d*x^2]^2, x], x] - Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*
FresnelS[b*x], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[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 2634

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

Rule 6440

Int[FresnelS[(b_.)*(x_)]^(n_.)*Sin[(d_.)*(x_)^2], x_Symbol] :> Dist[(Pi*b)/(2*d), Subst[Int[x^n, x], x, Fresne
lS[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A]  time = 0.0055243, size = 140, normalized size = 1. \[ -\frac{3 x S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi ^2 b^3}+\frac{x^3 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac{3 S(b x)^2}{4 \pi ^2 b^4}+\frac{3 x^2}{8 \pi ^2 b^2}-\frac{\sin \left (\pi b^2 x^2\right )}{2 \pi ^3 b^4}+\frac{x^2 \cos \left (\pi b^2 x^2\right )}{8 \pi ^2 b^2}+\frac{1}{4} x^4 S(b x)^2 \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*FresnelS[b*x]^2,x]

[Out]

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

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ({\it FresnelS} \left ( bx \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(b*x)^2,x)

[Out]

int(x^3*FresnelS(b*x)^2,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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^3*fresnels(b*x)^2, 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 )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x)^2,x, algorithm="fricas")

[Out]

integral(x^3*fresnels(b*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} S^{2}\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*fresnels(b*x)**2,x)

[Out]

Integral(x**3*fresnels(b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm fresnels}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnels(b*x)^2,x, algorithm="giac")

[Out]

integrate(x^3*fresnels(b*x)^2, x)