\(\int e^{x^2} \sin (b x) \, dx\) [57]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 69 \[ \int e^{x^2} \sin (b x) \, dx=\frac {1}{4} i e^{\frac {b^2}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-i b+2 x)\right )-\frac {1}{4} i e^{\frac {b^2}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (i b+2 x)\right ) \]

[Out]

-1/4*I*exp(1/4*b^2)*erfi(1/2*I*b-x)*Pi^(1/2)-1/4*I*exp(1/4*b^2)*erfi(1/2*I*b+x)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4560, 2266, 2235} \[ \int e^{x^2} \sin (b x) \, dx=\frac {1}{4} i \sqrt {\pi } e^{\frac {b^2}{4}} \text {erfi}\left (\frac {1}{2} (2 x-i b)\right )-\frac {1}{4} i \sqrt {\pi } e^{\frac {b^2}{4}} \text {erfi}\left (\frac {1}{2} (2 x+i b)\right ) \]

[In]

Int[E^x^2*Sin[b*x],x]

[Out]

(I/4)*E^(b^2/4)*Sqrt[Pi]*Erfi[((-I)*b + 2*x)/2] - (I/4)*E^(b^2/4)*Sqrt[Pi]*Erfi[(I*b + 2*x)/2]

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 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 4560

Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^n, x], x] /; FreeQ[F, x] && (LinearQ
[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} i e^{-i b x+x^2}-\frac {1}{2} i e^{i b x+x^2}\right ) \, dx \\ & = \frac {1}{2} i \int e^{-i b x+x^2} \, dx-\frac {1}{2} i \int e^{i b x+x^2} \, dx \\ & = \frac {1}{2} \left (i e^{\frac {b^2}{4}}\right ) \int e^{\frac {1}{4} (-i b+2 x)^2} \, dx-\frac {1}{2} \left (i e^{\frac {b^2}{4}}\right ) \int e^{\frac {1}{4} (i b+2 x)^2} \, dx \\ & = \frac {1}{4} i e^{\frac {b^2}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-i b+2 x)\right )-\frac {1}{4} i e^{\frac {b^2}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (i b+2 x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.62 \[ \int e^{x^2} \sin (b x) \, dx=\frac {1}{4} e^{\frac {b^2}{4}} \sqrt {\pi } \left (\text {erf}\left (\frac {b}{2}-i x\right )+\text {erf}\left (\frac {b}{2}+i x\right )\right ) \]

[In]

Integrate[E^x^2*Sin[b*x],x]

[Out]

(E^(b^2/4)*Sqrt[Pi]*(Erf[b/2 - I*x] + Erf[b/2 + I*x]))/4

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61

method result size
risch \(\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {b^{2}}{4}} \operatorname {erf}\left (-i x +\frac {b}{2}\right )}{4}+\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {b^{2}}{4}} \operatorname {erf}\left (i x +\frac {b}{2}\right )}{4}\) \(42\)

[In]

int(exp(x^2)*sin(x*b),x,method=_RETURNVERBOSE)

[Out]

1/4*Pi^(1/2)*exp(1/4*b^2)*erf(-I*x+1/2*b)+1/4*Pi^(1/2)*exp(1/4*b^2)*erf(I*x+1/2*b)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.43 \[ \int e^{x^2} \sin (b x) \, dx=\frac {1}{4} \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, b + i \, x\right ) - \operatorname {erf}\left (-\frac {1}{2} \, b + i \, x\right )\right )} e^{\left (\frac {1}{4} \, b^{2}\right )} \]

[In]

integrate(exp(x^2)*sin(b*x),x, algorithm="fricas")

[Out]

1/4*sqrt(pi)*(erf(1/2*b + I*x) - erf(-1/2*b + I*x))*e^(1/4*b^2)

Sympy [F]

\[ \int e^{x^2} \sin (b x) \, dx=\int e^{x^{2}} \sin {\left (b x \right )}\, dx \]

[In]

integrate(exp(x**2)*sin(b*x),x)

[Out]

Integral(exp(x**2)*sin(b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.54 \[ \int e^{x^2} \sin (b x) \, dx=\frac {1}{4} \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, b + i \, x\right ) e^{\left (\frac {1}{4} \, b^{2}\right )} - \operatorname {erf}\left (-\frac {1}{2} \, b + i \, x\right ) e^{\left (\frac {1}{4} \, b^{2}\right )}\right )} \]

[In]

integrate(exp(x^2)*sin(b*x),x, algorithm="maxima")

[Out]

1/4*sqrt(pi)*(erf(1/2*b + I*x)*e^(1/4*b^2) - erf(-1/2*b + I*x)*e^(1/4*b^2))

Giac [F]

\[ \int e^{x^2} \sin (b x) \, dx=\int { e^{\left (x^{2}\right )} \sin \left (b x\right ) \,d x } \]

[In]

integrate(exp(x^2)*sin(b*x),x, algorithm="giac")

[Out]

integrate(e^(x^2)*sin(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int e^{x^2} \sin (b x) \, dx=\int {\mathrm {e}}^{x^2}\,\sin \left (b\,x\right ) \,d x \]

[In]

int(exp(x^2)*sin(b*x),x)

[Out]

int(exp(x^2)*sin(b*x), x)