3.186 \(\int x^2 \cos (\frac{1}{2} b^2 \pi x^2) \text{FresnelC}(b x) \, dx\)

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

[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}, (-I/2)*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*P
i) + (x*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

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Rubi [A]  time = 0.0595254, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6455, 6447, 3379, 2638} \[ -\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi b}+\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi b}-\frac{\text{FresnelC}(b x) S(b x)}{2 \pi b^3}+\frac{x \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

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}, (-I/2)*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*P
i) + (x*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

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 6447

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

Rule 3379

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 2638

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

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{\operatorname{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*}

Mathematica [F]  time = 0.208928, size = 0, normalized size = 0. \[ \int x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) \text{FresnelC}(b x) \, dx \]

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 = 0.079, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ){\it FresnelC} \left ( bx \right ) \, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

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, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right ), x\right ) \end{align*}

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, algorithm="fricas")

[Out]

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

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

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right )\,{d x} \end{align*}

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, algorithm="giac")

[Out]

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