\(\int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx\) [152]

Optimal result

Integrand size = 19, antiderivative size = 20 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=-\frac {e^c \sqrt {\pi } \log (\text {erfc}(b x))}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, 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, 29} \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=-\frac {\sqrt {\pi } e^c \log (\text {erfc}(b x))}{2 b} \]

[In]

Int[E^(c - b^2*x^2)/Erfc[b*x],x]

[Out]

-1/2*(E^c*Sqrt[Pi]*Log[Erfc[b*x]])/b

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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} \, dx,x,\text {erfc}(b x)\right )}{2 b} \\ & = -\frac {e^c \sqrt {\pi } \log (\text {erfc}(b x))}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=-\frac {e^c \sqrt {\pi } \log (\text {erfc}(b x))}{2 b} \]

[In]

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

[Out]

-1/2*(E^c*Sqrt[Pi]*Log[Erfc[b*x]])/b

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=-\frac {\sqrt {\pi } e^{c} \log \left (\operatorname {erf}\left (b x\right ) - 1\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*sqrt(pi)*e^c*log(erf(b*x) - 1)/b

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=\begin {cases} - \frac {\sqrt {\pi } e^{c} \log {\left (\operatorname {erfc}{\left (b x \right )} \right )}}{2 b} & \text {for}\: b \neq 0 \\x e^{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.93 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{c-b^2 x^2}}{\text {erfc}(b x)} \, dx=-\frac {\sqrt {\pi }\,\ln \left (\mathrm {erfc}\left (b\,x\right )\right )\,{\mathrm {e}}^c}{2\,b} \]

[In]

int(exp(c - b^2*x^2)/erfc(b*x),x)

[Out]

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