\(\int e^{c-b^2 x^2} \text {erf}(b x) \, dx\) [48]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {e^c \sqrt {\pi } \text {erf}(b x)^2}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6508, 30} \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } e^c \text {erf}(b x)^2}{4 b} \]

[In]

Int[E^(c - b^2*x^2)*Erf[b*x],x]

[Out]

(E^c*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 {\left (e^c \sqrt {\pi }\right ) \text {Subst}(\int x \, dx,x,\text {erf}(b x))}{2 b} \\ & = \frac {e^c \sqrt {\pi } \text {erf}(b x)^2}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {e^c \sqrt {\pi } \text {erf}(b x)^2}{4 b} \]

[In]

Integrate[E^(c - b^2*x^2)*Erf[b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {{\mathrm e}^{c} \operatorname {erf}\left (b x \right )^{2} \sqrt {\pi }}{4 b}\)	\(17\)

[In]

int(exp(-b^2*x^2+c)*erf(b*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2} e^{c}}{4 \, b} \]

[In]

integrate(exp(-b^2*x^2+c)*erf(b*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\begin {cases} \frac {\sqrt {\pi } e^{c} \operatorname {erf}^{2}{\left (b x \right )}}{4 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(exp(-b**2*x**2+c)*erf(b*x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2} e^{c}}{4 \, b} \]

[In]

integrate(exp(-b^2*x^2+c)*erf(b*x),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(exp(-b^2*x^2+c)*erf(b*x),x, algorithm="giac")

[Out]

integrate(erf(b*x)*e^(-b^2*x^2 + c), x)

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 4.43 \[ \int e^{c-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi }\,{\mathrm {erf}\left (x\,\sqrt {b^2}\right )}^2\,{\mathrm {e}}^c}{4\,b}-\frac {\sqrt {\pi }\,{\mathrm {e}}^c\,\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {-b^2}}\right )\,\mathrm {erf}\left (b\,x\right )}{2\,\sqrt {-b^2}}+\frac {b\,\sqrt {\pi }\,\mathrm {erf}\left (x\,\sqrt {b^2}\right )\,{\mathrm {e}}^c\,\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {-b^2}}\right )}{2\,\sqrt {b^2}\,\sqrt {-b^2}} \]

[In]

int(exp(c - b^2*x^2)*erf(b*x),x)

[Out]

(pi^(1/2)*erf(x*(b^2)^(1/2))^2*exp(c))/(4*b) - (pi^(1/2)*exp(c)*erfi((b^2*x)/(-b^2)^(1/2))*erf(b*x))/(2*(-b^2)
^(1/2)) + (b*pi^(1/2)*erf(x*(b^2)^(1/2))*exp(c)*erfi((b^2*x)/(-b^2)^(1/2)))/(2*(b^2)^(1/2)*(-b^2)^(1/2))