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

Optimal. Leaf size=64 \[ \frac {e^c \text {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\pi } b x\right )^2}{8 b}-\frac {1}{4} i b e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right ) \]

[Out]

1/8*exp(c)*erf((1/2+1/2*I)*b*x*Pi^(1/2))^2/b-1/4*I*b*exp(c)*x^2*HypergeometricPFQ([1, 1],[3/2, 2],-1/2*I*b^2*P
i*x^2)

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Rubi [A]  time = 0.07, antiderivative size = 64, normalized size of antiderivative = 1.00, 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, 6373, 30, 6378} \[ \frac {e^c \text {Erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\pi } b x\right )^2}{8 b}-\frac {1}{4} i b e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^c*Erf[(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*P
i*x^2]

Rule 30

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

Rule 6373

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

Rule 6378

Int[E^((c_.) + (d_.)*(x_)^2)*Erfi[(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 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)]

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 {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )\right )}{4 b}\\ &=\frac {e^c \text {erf}\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*}

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Mathematica [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int e^{c-\frac {1}{2} i b^2 \pi x^2} S(b x) \, dx \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} + c\right )} {\rm fresnels}\left (b x\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} + c\right )} {\rm fresnels}\left (b x\right )\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{c -\frac {i b^{2} \pi \,x^{2}}{2}} \mathrm {S}\left (b x \right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} + c\right )} {\rm fresnels}\left (b x\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {e}}^{c-\frac {\Pi \,b^2\,x^2\,1{}\mathrm {i}}{2}}\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c - (Pi*b^2*x^2*1i)/2)*FresnelS(b*x),x)

[Out]

int(exp(c - (Pi*b^2*x^2*1i)/2)*FresnelS(b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{c} \int e^{- \frac {i \pi b^{2} x^{2}}{2}} S\left (b x\right )\, dx \]

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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