\(\int e^{-b^2 x^2} \text {erf}(b x) \, dx\) [83]

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

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]

[Out]

1/4*erf(b*x)^2*Pi^(1/2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6508, 30} \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]

[In]

Int[Erf[b*x]/E^(b^2*x^2),x]

[Out]

(Sqrt[Pi]*Erf[b*x]^2)/(4*b)

Rule 30

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

Rule 6508

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\pi } \text {Subst}(\int x \, dx,x,\text {erf}(b x))}{2 b} \\ & = \frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]

[In]

Integrate[Erf[b*x]/E^(b^2*x^2),x]

[Out]

(Sqrt[Pi]*Erf[b*x]^2)/(4*b)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method result size
default \(\frac {\operatorname {erf}\left (b x \right )^{2} \sqrt {\pi }}{4 b}\) \(15\)

[In]

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

[Out]

1/4*erf(b*x)^2*Pi^(1/2)/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2}}{4 \, b} \]

[In]

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

[Out]

1/4*sqrt(pi)*erf(b*x)^2/b

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\begin {cases} \frac {\sqrt {\pi } \operatorname {erf}^{2}{\left (b x \right )}}{4 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(erf(b*x)/exp(b**2*x**2),x)

[Out]

Piecewise((sqrt(pi)*erf(b*x)**2/(4*b), Ne(b, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2}}{4 \, b} \]

[In]

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

[Out]

1/4*sqrt(pi)*erf(b*x)^2/b

Giac [F]

\[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\int { \operatorname {erf}\left (b x\right ) e^{\left (-b^{2} x^{2}\right )} \,d x } \]

[In]

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

[Out]

integrate(erf(b*x)*e^(-b^2*x^2), x)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.28 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi }\,\mathrm {erf}\left (x\,\sqrt {b^2}\right )\,\mathrm {erf}\left (b\,x\right )}{2\,\sqrt {b^2}}-\frac {\sqrt {\pi }\,{\mathrm {erf}\left (x\,\sqrt {b^2}\right )}^2}{4\,b} \]

[In]

int(exp(-b^2*x^2)*erf(b*x),x)

[Out]

(pi^(1/2)*erf(x*(b^2)^(1/2))*erf(b*x))/(2*(b^2)^(1/2)) - (pi^(1/2)*erf(x*(b^2)^(1/2))^2)/(4*b)