3.201 \(\int \cos (c+i b^2 x^2) \text{Erfc}(b x) \, dx\)

Optimal. Leaf size=85 \[ -\frac{b e^{-i c} x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{\sqrt{\pi } e^{i c} \text{Erfc}(b x)^2}{8 b}+\frac{\sqrt{\pi } e^{-i c} \text{Erfi}(b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0734826, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6408, 6377, 2204, 6376, 6374, 30} \[ -\frac{b e^{-i c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{\sqrt{\pi } e^{i c} \text{Erfc}(b x)^2}{8 b}+\frac{\sqrt{\pi } e^{-i c} \text{Erfi}(b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + I*b^2*x^2]*Erfc[b*x],x]

[Out]

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

Rule 6408

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

Rule 6377

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

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 6374

Int[E^((c_.) + (d_.)*(x_)^2)*Erfc[(b_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(E^c*Sqrt[Pi])/(2*b), Subst[Int[x^n, x
], x, Erfc[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 \cos \left (c+i b^2 x^2\right ) \text{erfc}(b x) \, dx &=\frac{1}{2} \int e^{i c-b^2 x^2} \text{erfc}(b x) \, dx+\frac{1}{2} \int e^{-i c+b^2 x^2} \text{erfc}(b x) \, dx\\ &=\frac{1}{2} \int e^{-i c+b^2 x^2} \, dx-\frac{1}{2} \int e^{-i c+b^2 x^2} \text{erf}(b x) \, dx-\frac{\left (e^{i c} \sqrt{\pi }\right ) \operatorname{Subst}(\int x \, dx,x,\text{erfc}(b x))}{4 b}\\ &=-\frac{e^{i c} \sqrt{\pi } \text{erfc}(b x)^2}{8 b}+\frac{e^{-i c} \sqrt{\pi } \text{erfi}(b x)}{4 b}-\frac{b e^{-i c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}\\ \end{align*}

Mathematica [F]  time = 1.61014, size = 0, normalized size = 0. \[ \int \cos \left (c+i b^2 x^2\right ) \text{Erfc}(b x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cos[c + I*b^2*x^2]*Erfc[b*x],x]

[Out]

Integrate[Cos[c + I*b^2*x^2]*Erfc[b*x], x]

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Maple [F]  time = 0.41, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( c+i{b}^{2}{x}^{2} \right ){\it erfc} \left ( bx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c+I*b^2*x^2)*erfc(b*x),x)

[Out]

int(cos(c+I*b^2*x^2)*erfc(b*x),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\sqrt{\pi } \cos \left (c\right ) \operatorname{erfc}\left (b x\right )^{2}}{8 \, b} - \frac{i \, \sqrt{\pi } \operatorname{erfc}\left (b x\right )^{2} \sin \left (c\right )}{8 \, b} + \frac{1}{2} \, \cos \left (c\right ) \int \operatorname{erfc}\left (b x\right ) e^{\left (b^{2} x^{2}\right )}\,{d x} - \frac{1}{2} i \, \int \operatorname{erfc}\left (b x\right ) e^{\left (b^{2} x^{2}\right )}\,{d x} \sin \left (c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(c+I*b^2*x^2)*erfc(b*x),x, algorithm="maxima")

[Out]

-1/8*sqrt(pi)*cos(c)*erfc(b*x)^2/b - 1/8*I*sqrt(pi)*erfc(b*x)^2*sin(c)/b + 1/2*cos(c)*integrate(erfc(b*x)*e^(b
^2*x^2), x) - 1/2*I*integrate(erfc(b*x)*e^(b^2*x^2), x)*sin(c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(c+I*b^2*x^2)*erfc(b*x),x, algorithm="fricas")

[Out]

integral(-1/2*((erf(b*x) - 1)*e^(-2*b^2*x^2 + 2*I*c) + erf(b*x) - 1)*e^(b^2*x^2 - I*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos{\left (i b^{2} x^{2} + c \right )} \operatorname{erfc}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(c+I*b**2*x**2)*erfc(b*x),x)

[Out]

Integral(cos(I*b**2*x**2 + c)*erfc(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (i \, b^{2} x^{2} + c\right ) \operatorname{erfc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(c+I*b^2*x^2)*erfc(b*x),x, algorithm="giac")

[Out]

integrate(cos(I*b^2*x^2 + c)*erfc(b*x), x)