3.61 \(\int e^{c+\frac{1}{2} i b^2 \pi x^2} S(b x) \, dx\)

Optimal. Leaf size=64 \[ \frac{1}{4} i b e^c x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )-\frac{e^c \text{Erfi}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\pi } b x\right )^2}{8 b} \]

[Out]

-(E^c*Erfi[(1/2 + I/2)*b*Sqrt[Pi]*x]^2)/(8*b) + (I/4)*b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*
Pi*x^2]

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Rubi [A]  time = 0.0692265, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6436, 6376, 6375, 30} \[ \frac{1}{4} i b e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )-\frac{e^c \text{Erfi}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\pi } b x\right )^2}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(c + (I/2)*b^2*Pi*x^2)*FresnelS[b*x],x]

[Out]

-(E^c*Erfi[(1/2 + I/2)*b*Sqrt[Pi]*x]^2)/(8*b) + (I/4)*b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*
Pi*x^2]

Rule 6436

Int[E^((c_.) + (d_.)*(x_)^2)*FresnelS[(b_.)*(x_)], x_Symbol] :> Dist[(1 + I)/4, Int[E^(c + d*x^2)*Erf[(Sqrt[Pi
]*(1 + I)*b*x)/2], x], x] + Dist[(1 - I)/4, Int[E^(c + d*x^2)*Erf[(Sqrt[Pi]*(1 - I)*b*x)/2], x], x] /; FreeQ[{
b, c, d}, x] && EqQ[d^2, -((Pi^2*b^4)/4)]

Rule 6376

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(b_.)*(x_)], x_Symbol] :> Simp[(b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2},
b^2*x^2])/Sqrt[Pi], x] /; FreeQ[{b, c, d}, x] && EqQ[d, b^2]

Rule 6375

Int[E^((c_.) + (d_.)*(x_)^2)*Erfi[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[(E^c*Sqrt[Pi])/(2*b), Subst[Int[x^n, x]
, x, Erfi[b*x]], x] /; FreeQ[{b, c, d, n}, x] && EqQ[d, b^2]

Rule 30

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

Rubi steps

\begin{align*} \int e^{c+\frac{1}{2} i b^2 \pi x^2} S(b x) \, dx &=\left (-\frac{1}{4}-\frac{i}{4}\right ) \int e^{c+\frac{1}{2} i b^2 \pi x^2} \text{erfi}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) b \sqrt{\pi } x\right ) \, dx+\left (\frac{1}{4}+\frac{i}{4}\right ) \int e^{c+\frac{1}{2} i b^2 \pi x^2} \text{erf}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) b \sqrt{\pi } x\right ) \, dx\\ &=\frac{1}{4} i b e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )-\frac{e^c \operatorname{Subst}\left (\int x \, dx,x,\text{erfi}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) b \sqrt{\pi } x\right )\right )}{4 b}\\ &=-\frac{e^c \text{erfi}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) b \sqrt{\pi } x\right )^2}{8 b}+\frac{1}{4} i b e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )\\ \end{align*}

Mathematica [F]  time = 0.0304085, size = 0, normalized size = 0. \[ \int e^{c+\frac{1}{2} i b^2 \pi x^2} S(b x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[E^(c + (I/2)*b^2*Pi*x^2)*FresnelS[b*x],x]

[Out]

Integrate[E^(c + (I/2)*b^2*Pi*x^2)*FresnelS[b*x], x]

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Maple [F]  time = 0.114, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c+{\frac{i}{2}}{b}^{2}\pi \,{x}^{2}}}{\it FresnelS} \left ( bx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c+1/2*I*b^2*Pi*x^2)*FresnelS(b*x),x)

[Out]

int(exp(c+1/2*I*b^2*Pi*x^2)*FresnelS(b*x),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (\frac{1}{2} i \, \pi b^{2} x^{2} + c\right )}{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c+1/2*I*b^2*pi*x^2)*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(e^(1/2*I*pi*b^2*x^2 + c)*fresnels(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c+1/2*I*b^2*pi*x^2)*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(e^(1/2*I*pi*b^2*x^2 + c)*fresnels(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c+1/2*I*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

exp(c)*Integral(exp(I*pi*b**2*x**2/2)*fresnels(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (\frac{1}{2} i \, \pi b^{2} x^{2} + c\right )}{\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c+1/2*I*b^2*pi*x^2)*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(e^(1/2*I*pi*b^2*x^2 + c)*fresnels(b*x), x)