\(\int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx\) [153]

Optimal result

Integrand size = 19, antiderivative size = 21 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx=\frac {e^c \sqrt {\pi }}{2 b \text {erfc}(b x)} \]

[Out]

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6509, 30} \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx=\frac {\sqrt {\pi } e^c}{2 b \text {erfc}(b x)} \]

[In]

Int[E^(c - b^2*x^2)/Erfc[b*x]^2,x]

[Out]

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

Rule 30

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

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]

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

Mathematica [A] (verified)

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

[In]

Integrate[E^(c - b^2*x^2)/Erfc[b*x]^2,x]

[Out]

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

Maple [F]

\[\int \frac {{\mathrm e}^{-b^{2} x^{2}+c}}{\operatorname {erfc}\left (b x \right )^{2}}d x\]

[In]

int(exp(-b^2*x^2+c)/erfc(b*x)^2,x)

[Out]

int(exp(-b^2*x^2+c)/erfc(b*x)^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx=-\frac {\sqrt {\pi } e^{c}}{2 \, {\left (b \operatorname {erf}\left (b x\right ) - b\right )}} \]

[In]

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

[Out]

-1/2*sqrt(pi)*e^c/(b*erf(b*x) - b)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx=\begin {cases} \frac {\sqrt {\pi } e^{c}}{2 b \operatorname {erfc}{\left (b x \right )}} & \text {for}\: b \neq 0 \\x e^{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)^2} \, dx=\frac {\sqrt {\pi }\,{\mathrm {e}}^c}{2\,b\,\mathrm {erfc}\left (b\,x\right )} \]

[In]

int(exp(c - b^2*x^2)/erfc(b*x)^2,x)

[Out]

(pi^(1/2)*exp(c))/(2*b*erfc(b*x))