3.2.43 \(\int x^4 \text {FresnelC}(b x)^2 \, dx\) [143]

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

[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 = 0.14, 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 = {6566, 6590, 6598, 6588, 3432, 3473, 30, 3467, 3466} \begin {gather*} -\frac {43 S\left (\sqrt {2} b x\right )}{20 \sqrt {2} \pi ^3 b^5}-\frac {2 x^4 \text {FresnelC}(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 \text {FresnelC}(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 \text {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 \text {FresnelC}(b x)^2 \end {gather*}

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 3432

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

Rule 3466

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 3467

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 3473

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 6566

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/2)*b^2*x^2]*FresnelC[b*x], x], x] /; FreeQ[b, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6588

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^2/4)*b^4]

Rule 6590

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^2/4)*b^4] && IGtQ[m, 1]

Rule 6598

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^2/4)*b^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 = 0.10, size = 137, normalized size = 0.77 \begin {gather*} \frac {32 b^3 \pi x^3-12 b^3 \pi x^3 \cos \left (b^2 \pi x^2\right )+24 b^5 \pi ^3 x^5 \text {FresnelC}(b x)^2-129 \sqrt {2} S\left (\sqrt {2} b x\right )-48 \text {FresnelC}(b x) \left (4 b^2 \pi x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )+\left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )+66 b x \sin \left (b^2 \pi x^2\right )}{120 b^5 \pi ^3} \end {gather*}

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 = 0.80, size = 209, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {\frac {\FresnelC \left (b x \right )^{2} b^{5} x^{5}}{5}-2 \FresnelC \left (b x \right ) \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }\right )+\frac {4 b^{3} x^{3}}{15 \pi ^{2}}+\frac {\frac {2 b x \sin \left (b^{2} \pi \,x^{2}\right )}{5 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi }}{\pi ^{2}}+\frac {-\frac {\pi \,b^{3} x^{3} \cos \left (b^{2} \pi \,x^{2}\right )}{2}+\frac {3 \pi \left (\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi ^{3}}}{b^{5}}\)	\(209\)
default	\(\frac {\frac {\FresnelC \left (b x \right )^{2} b^{5} x^{5}}{5}-2 \FresnelC \left (b x \right ) \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }\right )+\frac {4 b^{3} x^{3}}{15 \pi ^{2}}+\frac {\frac {2 b x \sin \left (b^{2} \pi \,x^{2}\right )}{5 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi }}{\pi ^{2}}+\frac {-\frac {\pi \,b^{3} x^{3} \cos \left (b^{2} \pi \,x^{2}\right )}{2}+\frac {3 \pi \left (\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{5 \pi ^{3}}}{b^{5}}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*FresnelC(b*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^5*(1/5*FresnelC(b*x)^2*b^5*x^5-2*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnel_cos(b*x)^2, x)

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Fricas [A]
time = 0.38, size = 149, normalized size = 0.84 \begin {gather*} \frac {24 \, \pi ^{3} b^{6} x^{5} \operatorname {C}\left (b x\right )^{2} - 24 \, \pi b^{4} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} + 44 \, \pi b^{4} x^{3} - 192 \, \pi b^{3} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) - 129 \, \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right ) + 12 \, {\left (11 \, b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, {\left (\pi ^{2} b^{5} x^{4} - 8 \, b\right )} \operatorname {C}\left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{120 \, \pi ^{3} b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*pi^3*b^6*x^5*fresnel_cos(b*x)^2 - 24*pi*b^4*x^3*cos(1/2*pi*b^2*x^2)^2 + 44*pi*b^4*x^3 - 192*pi*b^3*x
^2*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x) - 129*sqrt(2)*sqrt(b^2)*fresnel_sin(sqrt(2)*sqrt(b^2)*x) + 12*(11*b^2*
x*cos(1/2*pi*b^2*x^2) - 4*(pi^2*b^5*x^4 - 8*b)*fresnel_cos(b*x))*sin(1/2*pi*b^2*x^2))/(pi^3*b^6)

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnel_cos(b*x)^2, x)

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

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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