3.200 \(\int \text{Erfc}(b x) \sin (c-i b^2 x^2) \, dx\)

Optimal. Leaf size=91 \[ \frac{i b e^{i c} x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{i \sqrt{\pi } e^{-i c} \text{Erfc}(b x)^2}{8 b}-\frac{i \sqrt{\pi } e^{i c} \text{Erfi}(b x)}{4 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0754201, antiderivative size = 91, 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 = {6405, 6374, 30, 6377, 2204, 6376} \[ \frac{i b e^{i c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{i \sqrt{\pi } e^{-i c} \text{Erfc}(b x)^2}{8 b}-\frac{i \sqrt{\pi } e^{i c} \text{Erfi}(b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Erfc[b*x]*Sin[c - I*b^2*x^2],x]

[Out]

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

Rule 6405

Int[Erfc[(b_.)*(x_)]*Sin[(c_.) + (d_.)*(x_)^2], x_Symbol] :> Dist[I/2, Int[E^(-(I*c) - I*d*x^2)*Erfc[b*x], x],
 x] - Dist[I/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 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]

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]

Rubi steps

\begin{align*} \int \text{erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx &=\frac{1}{2} i \int e^{-i c-b^2 x^2} \text{erfc}(b x) \, dx-\frac{1}{2} i \int e^{i c+b^2 x^2} \text{erfc}(b x) \, dx\\ &=-\left (\frac{1}{2} i \int e^{i c+b^2 x^2} \, dx\right )+\frac{1}{2} i \int e^{i c+b^2 x^2} \text{erf}(b x) \, dx-\frac{\left (i e^{-i c} \sqrt{\pi }\right ) \operatorname{Subst}(\int x \, dx,x,\text{erfc}(b x))}{4 b}\\ &=-\frac{i e^{-i c} \sqrt{\pi } \text{erfc}(b x)^2}{8 b}-\frac{i e^{i c} \sqrt{\pi } \text{erfi}(b x)}{4 b}+\frac{i 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 [A]  time = 0.437127, size = 101, normalized size = 1.11 \[ \frac{1}{2} i \left (\frac{b x^2 (\cos (c)+i \sin (c)) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{\sqrt{\pi }}-\frac{\sqrt{\pi } \left (\text{Erf}(b x)^2 (\cos (c)-i \sin (c))-2 \text{Erf}(b x) (\cos (c)-i \sin (c))+2 \text{Erfi}(b x) (\cos (c)+i \sin (c))\right )}{4 b}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Erfc[b*x]*Sin[c - I*b^2*x^2],x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.19, size = 0, normalized size = 0. \begin{align*} \int -{\it erfc} \left ( bx \right ) \sin \left ( -c+i{b}^{2}{x}^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \sin{\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(-erfc(b*x)*sin(-c+I*b**2*x**2),x)

[Out]

-Integral(sin(I*b**2*x**2 - c)*erfc(b*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-erfc(b*x)*sin(I*b^2*x^2 - c), x)