3.143 \(\int x^4 C(b x)^2 \, dx\)

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

[Out]

4/15*x^3/b^2/Pi^2-1/10*x^3*cos(b^2*Pi*x^2)/b^2/Pi^2-8/5*x^2*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^3/Pi^2+1/5*x^5
*FresnelC(b*x)^2+16/5*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/b^5/Pi^3-2/5*x^4*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/b/P
i+11/20*x*sin(b^2*Pi*x^2)/b^4/Pi^3-43/40*FresnelS(b*x*2^(1/2))/b^5/Pi^3*2^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, 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 30

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

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.16, size = 137, normalized size = 0.77 \[ \frac {24 \pi ^3 b^5 x^5 C(b x)^2+32 \pi b^3 x^3+66 b x \sin \left (\pi b^2 x^2\right )-48 C(b x) \left (4 \pi b^2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )+\left (\pi ^2 b^4 x^4-8\right ) \sin \left (\frac {1}{2} \pi b^2 x^2\right )\right )-12 \pi b^3 x^3 \cos \left (\pi b^2 x^2\right )-129 \sqrt {2} S\left (\sqrt {2} b x\right )}{120 \pi ^3 b^5} \]

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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fricas [F]  time = 0.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} {\rm fresnelc}\left (b x\right )^{2}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} {\rm fresnelc}\left (b x\right )^{2}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 209, normalized size = 1.18 \[ \frac {\frac {b^{5} x^{5} \FresnelC \left (b x \right )^{2}}{5}-2 \FresnelC \left (b x \right ) \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }\right )+\frac {4 b^{3} x^{3}}{15 \pi ^{2}}+\frac {\frac {2 b x \sin \left (b^{2} \pi \,x^{2}\right )}{5 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi }}{\pi ^{2}}+\frac {-\frac {\pi \,b^{3} x^{3} \cos \left (b^{2} \pi \,x^{2}\right )}{2}+\frac {3 \pi \left (\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi ^{3}}}{b^{5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} {\rm fresnelc}\left (b x\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,{\mathrm {FresnelC}\left (b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4*FresnelC(b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} C^{2}\left (b x\right )\, dx \]

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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