3.205 \(\int x^3 C(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=109 \[ -\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}-\frac {x^2 C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {2 C(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {x^3}{6 \pi b} \]

[Out]

1/6*x^3/b/Pi-x^2*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^2/Pi+2*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/b^4/Pi^2+1/4*x*s
in(b^2*Pi*x^2)/b^3/Pi^2-5/8*FresnelS(b*x*2^(1/2))/b^4/Pi^2*2^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6463, 6453, 3351, 3392, 30, 3386} \[ \frac {2 \text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \text {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}+\frac {x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {x^3}{6 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

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

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 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 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 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^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx &=-\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {2 \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx}{b^2 \pi }+\frac {\int x^2 \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=-\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}+\frac {\int x^2 \, dx}{2 b \pi }+\frac {\int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=\frac {x^3}{6 b \pi }-\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} b^4 \pi ^2}+\frac {2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}\\ &=\frac {x^3}{6 b \pi }-\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 90, normalized size = 0.83 \[ \frac {4 \pi b^3 x^3-24 C(b x) \left (\pi b^2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )-2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )\right )+6 b x \sin \left (\pi b^2 x^2\right )-15 \sqrt {2} S\left (\sqrt {2} b x\right )}{24 \pi ^2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(4*b^3*Pi*x^3 - 15*Sqrt[2]*FresnelS[Sqrt[2]*b*x] - 24*FresnelC[b*x]*(b^2*Pi*x^2*Cos[(b^2*Pi*x^2)/2] - 2*Sin[(b
^2*Pi*x^2)/2]) + 6*b*x*Sin[b^2*Pi*x^2])/(24*b^4*Pi^2)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")

[Out]

integral(x^3*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

integrate(x^3*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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maple [A]  time = 0.05, size = 120, normalized size = 1.10 \[ \frac {\frac {\FresnelC \left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{b^{3}}-\frac {\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{2 \pi ^{2}}-\frac {b^{3} x^{3}}{6 \pi }-\frac {\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{4 \pi }}{2 \pi }}{b^{3}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

integrate(x^3*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {FresnelC}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelC(b*x)*sin((Pi*b^2*x^2)/2),x)

[Out]

int(x^3*FresnelC(b*x)*sin((Pi*b^2*x^2)/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*fresnelc(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x**3*sin(pi*b**2*x**2/2)*fresnelc(b*x), x)

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