3.145 \(\int x^2 \text{FresnelC}(b x)^2 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.105312, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6431, 6455, 6461, 3358, 3352, 3385} \[ -\frac{2 x^2 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac{4 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac{5 \text{FresnelC}\left (\sqrt{2} b x\right )}{6 \sqrt{2} \pi ^2 b^3}-\frac{x \cos \left (\pi b^2 x^2\right )}{6 \pi ^2 b^2}+\frac{2 x}{3 \pi ^2 b^2}+\frac{1}{3} x^3 \text{FresnelC}(b x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6431

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

Rule 6455

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelC[b*x])/(
2*d), x] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(4*d), 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 6461

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

Rule 3358

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0882343, size = 100, normalized size = 0.81 \[ \frac{4 \pi ^2 b^3 x^3 \text{FresnelC}(b x)^2-8 \text{FresnelC}(b x) \left (\pi b^2 x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )+2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )\right )-2 b x \left (\cos \left (\pi b^2 x^2\right )-4\right )+5 \sqrt{2} \text{FresnelC}\left (\sqrt{2} b x\right )}{12 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.075, size = 122, normalized size = 1. \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3} \left ({\it FresnelC} \left ( bx \right ) \right ) ^{2}}{3}}-2\,{\it FresnelC} \left ( bx \right ) \left ( 1/3\,{\frac{{b}^{2}{x}^{2}\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+2/3\,{\frac{\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) +{\frac{2\,bx}{3\,{\pi }^{2}}}+{\frac{\sqrt{2}{\it FresnelC} \left ( bx\sqrt{2} \right ) }{3\,{\pi }^{2}}}+{\frac{1}{3\,\pi } \left ( -{\frac{bx\cos \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}+{\frac{\sqrt{2}{\it FresnelC} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*b^3*x^3*FresnelC(b*x)^2-2*FresnelC(b*x)*(1/3/Pi*b^2*x^2*sin(1/2*b^2*Pi*x^2)+2/3/Pi^2*cos(1/2*b^2*Pi
*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., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm fresnelc}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnelc(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*fresnelc(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*fresnelc(b*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnelc(b*x)^2, x)