3.143 \(\int x^4 \text{FresnelC}(b x)^2 \, dx\)

Optimal. Leaf size=177 \[ -\frac{2 x^4 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac{16 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac{8 x^2 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}-\frac{43 S\left (\sqrt{2} b x\right )}{20 \sqrt{2} \pi ^3 b^5}+\frac{4 x^3}{15 \pi ^2 b^2}+\frac{11 x \sin \left (\pi b^2 x^2\right )}{20 \pi ^3 b^4}-\frac{x^3 \cos \left (\pi b^2 x^2\right )}{10 \pi ^2 b^2}+\frac{1}{5} x^5 \text{FresnelC}(b x)^2 \]

[Out]

(4*x^3)/(15*b^2*Pi^2) - (x^3*Cos[b^2*Pi*x^2])/(10*b^2*Pi^2) - (8*x^2*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(5*b^3
*Pi^2) + (x^5*FresnelC[b*x]^2)/5 - (43*FresnelS[Sqrt[2]*b*x])/(20*Sqrt[2]*b^5*Pi^3) + (16*FresnelC[b*x]*Sin[(b
^2*Pi*x^2)/2])/(5*b^5*Pi^3) - (2*x^4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(5*b*Pi) + (11*x*Sin[b^2*Pi*x^2])/(20*
b^4*Pi^3)

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Rubi [A]  time = 0.187988, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6431, 6455, 6463, 6453, 3351, 3392, 30, 3386, 3385} \[ -\frac{2 x^4 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac{16 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac{8 x^2 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}-\frac{43 S\left (\sqrt{2} b x\right )}{20 \sqrt{2} \pi ^3 b^5}+\frac{4 x^3}{15 \pi ^2 b^2}+\frac{11 x \sin \left (\pi b^2 x^2\right )}{20 \pi ^3 b^4}-\frac{x^3 \cos \left (\pi b^2 x^2\right )}{10 \pi ^2 b^2}+\frac{1}{5} x^5 \text{FresnelC}(b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^4*FresnelC[b*x]^2,x]

[Out]

(4*x^3)/(15*b^2*Pi^2) - (x^3*Cos[b^2*Pi*x^2])/(10*b^2*Pi^2) - (8*x^2*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(5*b^3
*Pi^2) + (x^5*FresnelC[b*x]^2)/5 - (43*FresnelS[Sqrt[2]*b*x])/(20*Sqrt[2]*b^5*Pi^3) + (16*FresnelC[b*x]*Sin[(b
^2*Pi*x^2)/2])/(5*b^5*Pi^3) - (2*x^4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(5*b*Pi) + (11*x*Sin[b^2*Pi*x^2])/(20*
b^4*Pi^3)

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 6463

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

Rule 6453

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

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]

Rule 3392

Int[Cos[(a_.) + ((b_.)*(x_)^(n_))/2]^2*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m, x], x] + Dist[1/2, Int[x^m*
Cos[2*a + b*x^n], x], x] /; FreeQ[{a, b, m, n}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[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 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^4 C(b x)^2 \, dx &=\frac{1}{5} x^5 C(b x)^2-\frac{1}{5} (2 b) \int x^5 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=\frac{1}{5} x^5 C(b x)^2-\frac{2 x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{\int x^4 \sin \left (b^2 \pi x^2\right ) \, dx}{5 \pi }+\frac{8 \int x^3 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{5 b \pi }\\ &=-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{10 b^2 \pi ^2}-\frac{8 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)^2-\frac{2 x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{16 \int x \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x) \, dx}{5 b^3 \pi ^2}+\frac{3 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{10 b^2 \pi ^2}+\frac{8 \int x^2 \cos ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{5 b^2 \pi ^2}\\ &=-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{10 b^2 \pi ^2}-\frac{8 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)^2+\frac{16 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac{2 x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{3 x \sin \left (b^2 \pi x^2\right )}{20 b^4 \pi ^3}-\frac{3 \int \sin \left (b^2 \pi x^2\right ) \, dx}{20 b^4 \pi ^3}-\frac{8 \int \sin \left (b^2 \pi x^2\right ) \, dx}{5 b^4 \pi ^3}+\frac{4 \int x^2 \, dx}{5 b^2 \pi ^2}+\frac{4 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{5 b^2 \pi ^2}\\ &=\frac{4 x^3}{15 b^2 \pi ^2}-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{10 b^2 \pi ^2}-\frac{8 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)^2-\frac{3 S\left (\sqrt{2} b x\right )}{20 \sqrt{2} b^5 \pi ^3}-\frac{4 \sqrt{2} S\left (\sqrt{2} b x\right )}{5 b^5 \pi ^3}+\frac{16 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac{2 x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{11 x \sin \left (b^2 \pi x^2\right )}{20 b^4 \pi ^3}-\frac{2 \int \sin \left (b^2 \pi x^2\right ) \, dx}{5 b^4 \pi ^3}\\ &=\frac{4 x^3}{15 b^2 \pi ^2}-\frac{x^3 \cos \left (b^2 \pi x^2\right )}{10 b^2 \pi ^2}-\frac{8 x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{5 b^3 \pi ^2}+\frac{1}{5} x^5 C(b x)^2-\frac{3 S\left (\sqrt{2} b x\right )}{20 \sqrt{2} b^5 \pi ^3}-\frac{\sqrt{2} S\left (\sqrt{2} b x\right )}{b^5 \pi ^3}+\frac{16 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac{2 x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac{11 x \sin \left (b^2 \pi x^2\right )}{20 b^4 \pi ^3}\\ \end{align*}

Mathematica [A]  time = 0.133759, size = 137, normalized size = 0.77 \[ \frac{24 \pi ^3 b^5 x^5 \text{FresnelC}(b x)^2-48 \text{FresnelC}(b x) \left (\left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac{1}{2} \pi b^2 x^2\right )+4 \pi b^2 x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )\right )+32 \pi b^3 x^3+66 b x \sin \left (\pi b^2 x^2\right )-12 \pi b^3 x^3 \cos \left (\pi b^2 x^2\right )-129 \sqrt{2} S\left (\sqrt{2} b x\right )}{120 \pi ^3 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*FresnelC[b*x]^2,x]

[Out]

(32*b^3*Pi*x^3 - 12*b^3*Pi*x^3*Cos[b^2*Pi*x^2] + 24*b^5*Pi^3*x^5*FresnelC[b*x]^2 - 129*Sqrt[2]*FresnelS[Sqrt[2
]*b*x] - 48*FresnelC[b*x]*(4*b^2*Pi*x^2*Cos[(b^2*Pi*x^2)/2] + (-8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2]) + 66*b*
x*Sin[b^2*Pi*x^2])/(120*b^5*Pi^3)

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Maple [A]  time = 0.076, size = 209, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{5}} \left ({\frac{{b}^{5}{x}^{5} \left ({\it FresnelC} \left ( bx \right ) \right ) ^{2}}{5}}-2\,{\it FresnelC} \left ( bx \right ) \left ( 1/5\,{\frac{{x}^{4}{b}^{4}\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}-4/5\,{\frac{1}{\pi } \left ( -{\frac{{b}^{2}{x}^{2}\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{\pi }}+2\,{\frac{\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) } \right ) +{\frac{4\,{x}^{3}{b}^{3}}{15\,{\pi }^{2}}}+{\frac{4}{5\,{\pi }^{2}} \left ({\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}-{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) }+{\frac{1}{5\,{\pi }^{3}} \left ( -{\frac{\pi \,{b}^{3}{x}^{3}\cos \left ({b}^{2}\pi \,{x}^{2} \right ) }{2}}+{\frac{3\,\pi }{2} \left ({\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}-{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) }-4\,\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*FresnelC(b*x)^2,x)

[Out]

1/b^5*(1/5*b^5*x^5*FresnelC(b*x)^2-2*FresnelC(b*x)*(1/5/Pi*b^4*x^4*sin(1/2*b^2*Pi*x^2)-4/5/Pi*(-1/Pi*b^2*x^2*c
os(1/2*b^2*Pi*x^2)+2/Pi^2*sin(1/2*b^2*Pi*x^2)))+4/15/Pi^2*b^3*x^3+4/5/Pi^2*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/4/Pi*
2^(1/2)*FresnelS(b*x*2^(1/2)))+1/5/Pi^3*(-1/2*Pi*b^3*x^3*cos(b^2*Pi*x^2)+3/2*Pi*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/
4/Pi*2^(1/2)*FresnelS(b*x*2^(1/2)))-4*2^(1/2)*FresnelS(b*x*2^(1/2))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^4*fresnelc(b*x)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{4}{\rm fresnelc}\left (b x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x)^2,x, algorithm="fricas")

[Out]

integral(x^4*fresnelc(b*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} C^{2}\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*fresnelc(b*x)**2,x)

[Out]

Integral(x**4*fresnelc(b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*fresnelc(b*x)^2,x, algorithm="giac")

[Out]

integrate(x^4*fresnelc(b*x)^2, x)