3.2.86 \(\int x^2 \cos (\frac {1}{2} b^2 \pi x^2) \text {FresnelC}(b x) \, dx\) [186]

Optimal. Leaf size=136 \[ \frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\text {FresnelC}(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {x \text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \]

[Out]

1/4*cos(b^2*Pi*x^2)/b^3/Pi^2-1/2*FresnelC(b*x)*FresnelS(b*x)/b^3/Pi-1/8*I*x^2*hypergeom([1, 1],[3/2, 2],-1/2*I
*b^2*Pi*x^2)/b/Pi+1/8*I*x^2*hypergeom([1, 1],[3/2, 2],1/2*I*b^2*Pi*x^2)/b/Pi+x*FresnelC(b*x)*sin(1/2*b^2*Pi*x^
2)/b^2/Pi

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Rubi [A]
time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6590, 6582, 3460, 2718} \begin {gather*} -\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 \pi b}-\frac {\text {FresnelC}(b x) S(b x)}{2 \pi b^3}+\frac {x \text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[b^2*Pi*x^2]/(4*b^3*Pi^2) - (FresnelC[b*x]*FresnelS[b*x])/(2*b^3*Pi) - ((I/8)*x^2*HypergeometricPFQ[{1, 1},
{3/2, 2}, (-1/2*I)*b^2*Pi*x^2])/(b*Pi) + ((I/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b
*Pi) + (x*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 6582

Int[FresnelC[(b_.)*(x_)]*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[b*Pi*FresnelC[b*x]*(FresnelS[b*x]/(4*d)), x] + (
Simp[(1/8)*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I)*d*x^2], x] - Simp[(1/8)*I*b*x^2*HypergeometricPFQ[
{1, 1}, {3/2, 2}, I*d*x^2], 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]

Rubi steps

\begin {align*} \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx &=\frac {x C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac {\int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {C(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {x C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\text {Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=\frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {C(b x) S(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {x C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }\\ \end {align*}

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Mathematica [F]
time = 0.15, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \text {FresnelC}(b x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \FresnelC \left (b x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

integrate(x^2*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x), x)

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Fricas [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^2*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

integral(x^2*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cos(pi*b**2*x**2/2)*fresnelc(b*x), 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^2*cos(1/2*b^2*pi*x^2)*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^2*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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