\(\int \text {erfc}(b x) \sin (c-i b^2 x^2) \, dx\) [200]

Optimal result

Integrand size = 18, antiderivative size = 91 \[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=-\frac {i e^{-i c} \sqrt {\pi } \text {erfc}(b x)^2}{8 b}-\frac {i e^{i c} \sqrt {\pi } \text {erfi}(b x)}{4 b}+\frac {i b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }} \]

[Out]

1/2*I*b*exp(I*c)*x^2*hypergeom([1, 1],[3/2, 2],b^2*x^2)/Pi^(1/2)-1/8*I*erfc(b*x)^2*Pi^(1/2)/b/exp(I*c)-1/4*I*e
xp(I*c)*erfi(b*x)*Pi^(1/2)/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6540, 6509, 30, 6512, 2235, 6511} \[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\frac {i b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}-\frac {i \sqrt {\pi } e^{-i c} \text {erfc}(b x)^2}{8 b}-\frac {i \sqrt {\pi } e^{i c} \text {erfi}(b x)}{4 b} \]

[In]

Int[Erfc[b*x]*Sin[c - I*b^2*x^2],x]

[Out]

((-1/8*I)*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*Hype
rgeometricPFQ[{1, 1}, {3/2, 2}, b^2*x^2])/Sqrt[Pi]

Rule 30

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

Rule 2235

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 6509

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 6511

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

Rule 6512

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 6540

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 2.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\frac {1}{2} i \left (-\frac {\sqrt {\pi } \left (-2 \text {erf}(b x) (\cos (c)-i \sin (c))+\text {erf}(b x)^2 (\cos (c)-i \sin (c))+2 \text {erfi}(b x) (\cos (c)+i \sin (c))\right )}{4 b}+\frac {b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right ) (\cos (c)+i \sin (c))}{\sqrt {\pi }}\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int -\operatorname {erfc}\left (b x \right ) \sin \left (i b^{2} x^{2}-c \right )d x\]

[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)

Fricas [F]

\[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\int { -\operatorname {erfc}\left (b x\right ) \sin \left (i \, b^{2} x^{2} - c\right ) \,d x } \]

[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]

\[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=- \int \sin {\left (i b^{2} x^{2} - c \right )} \operatorname {erfc}{\left (b x \right )}\, dx \]

[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)

Maxima [F]

\[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\int { -\operatorname {erfc}\left (b x\right ) \sin \left (i \, b^{2} x^{2} - c\right ) \,d x } \]

[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)

Giac [F]

\[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\int { -\operatorname {erfc}\left (b x\right ) \sin \left (i \, b^{2} x^{2} - c\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \text {erfc}(b x) \sin \left (c-i b^2 x^2\right ) \, dx=\int \sin \left (c-b^2\,x^2\,1{}\mathrm {i}\right )\,\mathrm {erfc}\left (b\,x\right ) \,d x \]

[In]

int(sin(c - b^2*x^2*1i)*erfc(b*x),x)

[Out]

int(sin(c - b^2*x^2*1i)*erfc(b*x), x)