\(\int e^{c+d x^2} x \text {erf}(b x) \, dx\) [55]

Optimal result

Integrand size = 15, antiderivative size = 57 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=\frac {e^{c+d x^2} \text {erf}(b x)}{2 d}-\frac {b e^c \text {erf}\left (\sqrt {b^2-d} x\right )}{2 \sqrt {b^2-d} d} \]

[Out]

1/2*exp(d*x^2+c)*erf(b*x)/d-1/2*b*exp(c)*erf(x*(b^2-d)^(1/2))/d/(b^2-d)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, 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 = {6517, 2236} \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=\frac {\text {erf}(b x) e^{c+d x^2}}{2 d}-\frac {b e^c \text {erf}\left (x \sqrt {b^2-d}\right )}{2 d \sqrt {b^2-d}} \]

[In]

Int[E^(c + d*x^2)*x*Erf[b*x],x]

[Out]

(E^(c + d*x^2)*Erf[b*x])/(2*d) - (b*E^c*Erf[Sqrt[b^2 - d]*x])/(2*Sqrt[b^2 - d]*d)

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 6517

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(a_.) + (b_.)*(x_)]*(x_), x_Symbol] :> Simp[E^(c + d*x^2)*(Erf[a + b*x]/(2*d)
), x] - Dist[b/(d*Sqrt[Pi]), Int[E^(-a^2 + c - 2*a*b*x - (b^2 - d)*x^2), x], x] /; FreeQ[{a, b, c, d}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=\frac {e^c \left (e^{d x^2} \text {erf}(b x)-\frac {b \text {erfi}\left (\sqrt {-b^2+d} x\right )}{\sqrt {-b^2+d}}\right )}{2 d} \]

[In]

Integrate[E^(c + d*x^2)*x*Erf[b*x],x]

[Out]

(E^c*(E^(d*x^2)*Erf[b*x] - (b*Erfi[Sqrt[-b^2 + d]*x])/Sqrt[-b^2 + d]))/(2*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\frac {\operatorname {erf}\left (b x \right ) b \,{\mathrm e}^{\frac {b^{2} d \,x^{2}+b^{2} c}{b^{2}}}}{2 d}-\frac {b \,{\mathrm e}^{c} \operatorname {erf}\left (\sqrt {1-\frac {d}{b^{2}}}\, b x \right )}{2 d \sqrt {1-\frac {d}{b^{2}}}}}{b}\)	\(67\)

[In]

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

[Out]

(1/2*erf(b*x)*b*exp((b^2*d*x^2+b^2*c)/b^2)/d-1/2*b/d*exp(c)/(1-d/b^2)^(1/2)*erf((1-d/b^2)^(1/2)*b*x))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=-\frac {\sqrt {b^{2} - d} b \operatorname {erf}\left (\sqrt {b^{2} - d} x\right ) e^{c} - {\left (b^{2} - d\right )} \operatorname {erf}\left (b x\right ) e^{\left (d x^{2} + c\right )}}{2 \, {\left (b^{2} d - d^{2}\right )}} \]

[In]

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

[Out]

-1/2*(sqrt(b^2 - d)*b*erf(sqrt(b^2 - d)*x)*e^c - (b^2 - d)*erf(b*x)*e^(d*x^2 + c))/(b^2*d - d^2)

Sympy [F]

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

[In]

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

[Out]

exp(c)*Integral(x*exp(d*x**2)*erf(b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=-\frac {b \operatorname {erf}\left (\sqrt {b^{2} - d} x\right ) e^{c}}{2 \, \sqrt {b^{2} - d} d} + \frac {\operatorname {erf}\left (b x\right ) e^{\left (d x^{2} + c\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*b*erf(sqrt(b^2 - d)*x)*e^c/(sqrt(b^2 - d)*d) + 1/2*erf(b*x)*e^(d*x^2 + c)/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=\frac {b \operatorname {erf}\left (-\sqrt {b^{2} - d} x\right ) e^{c}}{2 \, \sqrt {b^{2} - d} d} + \frac {\operatorname {erf}\left (b x\right ) e^{\left (d x^{2} + c\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*b*erf(-sqrt(b^2 - d)*x)*e^c/(sqrt(b^2 - d)*d) + 1/2*erf(b*x)*e^(d*x^2 + c)/d

Mupad [B] (verification not implemented)

Time = 5.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int e^{c+d x^2} x \text {erf}(b x) \, dx=\frac {{\mathrm {e}}^{d\,x^2}\,{\mathrm {e}}^c\,\mathrm {erf}\left (b\,x\right )}{2\,d}-\frac {b\,{\mathrm {e}}^c\,\mathrm {erf}\left (x\,\sqrt {b^2-d}\right )}{2\,d\,\sqrt {b^2-d}} \]

[In]

int(x*exp(c + d*x^2)*erf(b*x),x)

[Out]

(exp(d*x^2)*exp(c)*erf(b*x))/(2*d) - (b*exp(c)*erf(x*(b^2 - d)^(1/2)))/(2*d*(b^2 - d)^(1/2))