\(\int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^2} \, dx\) [50]

Optimal result

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

[Out]

-1/2*exp(c)*Pi^(1/2)/b/erf(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 = {6508, 30} \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^2} \, dx=-\frac {\sqrt {\pi } e^c}{2 b \text {erf}(b x)} \]

[In]

Int[E^(c - b^2*x^2)/Erf[b*x]^2,x]

[Out]

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

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}\left (\int \frac {1}{x^2} \, dx,x,\text {erf}(b x)\right )}{2 b} \\ & = -\frac {e^c \sqrt {\pi }}{2 b \text {erf}(b x)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^2} \, dx=- \frac {\sqrt {\pi } e^{c}}{2 b \operatorname {erf}{\left (b x \right )}} \]

[In]

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

[Out]

-sqrt(pi)*exp(c)/(2*b*erf(b*x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^2} \, dx=-\frac {\sqrt {\pi } e^{c}}{2 \, b \operatorname {erf}\left (b x\right )} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^2} \, dx=-\frac {\sqrt {\pi }\,{\mathrm {e}}^c}{2\,b\,\mathrm {erf}\left (b\,x\right )} \]

[In]

int(exp(c - b^2*x^2)/erf(b*x)^2,x)

[Out]

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