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

Optimal. Leaf size=124 \[ -\frac {5 C\left (\sqrt {2} b x\right )}{6 \sqrt {2} \pi ^2 b^3}+\frac {2 x^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi b}+\frac {x \cos \left (\pi b^2 x^2\right )}{6 \pi ^2 b^2}+\frac {2 x}{3 \pi ^2 b^2}-\frac {4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 S(b x)^2 \]

[Out]

2/3*x/b^2/Pi^2+1/6*x*cos(b^2*Pi*x^2)/b^2/Pi^2+2/3*x^2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b/Pi+1/3*x^3*FresnelS(
b*x)^2-4/3*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2-5/12*FresnelC(b*x*2^(1/2))/b^3/Pi^2*2^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6430, 6454, 6460, 3357, 3352, 3385} \[ -\frac {5 \text {FresnelC}\left (\sqrt {2} b x\right )}{6 \sqrt {2} \pi ^2 b^3}-\frac {4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac {2 x^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi b}+\frac {x \cos \left (\pi b^2 x^2\right )}{6 \pi ^2 b^2}+\frac {2 x}{3 \pi ^2 b^2}+\frac {1}{3} x^3 S(b x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/(3*b^2*Pi^2) + (x*Cos[b^2*Pi*x^2])/(6*b^2*Pi^2) - (5*FresnelC[Sqrt[2]*b*x])/(6*Sqrt[2]*b^3*Pi^2) + (2*x^
2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(3*b*Pi) + (x^3*FresnelS[b*x]^2)/3 - (4*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2]
)/(3*b^3*Pi^2)

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3357

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 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 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 6460

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

Rubi steps

\begin {align*} \int x^2 S(b x)^2 \, dx &=\frac {1}{3} x^3 S(b x)^2-\frac {1}{3} (2 b) \int x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {\int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{3 \pi }-\frac {4 \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{3 b \pi }\\ &=\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}-\frac {\int \cos \left (b^2 \pi x^2\right ) \, dx}{6 b^2 \pi ^2}+\frac {4 \int \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {4 \int \left (\frac {1}{2}-\frac {1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {2 x}{3 b^2 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}-\frac {2 \int \cos \left (b^2 \pi x^2\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {2 x}{3 b^2 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}-\frac {\sqrt {2} C\left (\sqrt {2} b x\right )}{3 b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 100, normalized size = 0.81 \[ \frac {4 \pi ^2 b^3 x^3 S(b x)^2+8 S(b x) \left (\pi b^2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )-2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )\right )+2 b x \left (\cos \left (\pi b^2 x^2\right )+4\right )-5 \sqrt {2} C\left (\sqrt {2} b x\right )}{12 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*(4 + Cos[b^2*Pi*x^2]) - 5*Sqrt[2]*FresnelC[Sqrt[2]*b*x] + 4*b^3*Pi^2*x^3*FresnelS[b*x]^2 + 8*FresnelS[b
*x]*(b^2*Pi*x^2*Cos[(b^2*Pi*x^2)/2] - 2*Sin[(b^2*Pi*x^2)/2]))/(12*b^3*Pi^2)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} {\rm fresnels}\left (b x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*fresnels(b*x)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm fresnels}\left (b x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x)^2, x)

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maple [A]  time = 0.05, size = 122, normalized size = 0.98 \[ \frac {\frac {b^{3} x^{3} \mathrm {S}\left (b x \right )^{2}}{3}-2 \,\mathrm {S}\left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}\right )+\frac {2 b x}{3 \pi ^{2}}-\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{3 \pi ^{2}}-\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4 \pi }}{3 \pi }}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm fresnels}\left (b x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnels(b*x)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {FresnelS}\left (b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*FresnelS(b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} S^{2}\left (b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*fresnels(b*x)**2, x)

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