\(\int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx\) [51]

Optimal result

Integrand size = 19, antiderivative size = 21 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {e^c \sqrt {\pi }}{4 b \text {erf}(b x)^2} \]

[Out]

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

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} \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {\sqrt {\pi } e^c}{4 b \text {erf}(b x)^2} \]

[In]

Int[E^(c - b^2*x^2)/Erf[b*x]^3,x]

[Out]

-1/4*(E^c*Sqrt[Pi])/(b*Erf[b*x]^2)

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^3} \, dx,x,\text {erf}(b x)\right )}{2 b} \\ & = -\frac {e^c \sqrt {\pi }}{4 b \text {erf}(b x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {e^c \sqrt {\pi }}{4 b \text {erf}(b x)^2} \]

[In]

Integrate[E^(c - b^2*x^2)/Erf[b*x]^3,x]

[Out]

-1/4*(E^c*Sqrt[Pi])/(b*Erf[b*x]^2)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

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

[In]

int(exp(-b^2*x^2+c)/erf(b*x)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {\sqrt {\pi } e^{c}}{4 \, b \operatorname {erf}\left (b x\right )^{2}} \]

[In]

integrate(exp(-b^2*x^2+c)/erf(b*x)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=- \frac {\sqrt {\pi } e^{c}}{4 b \operatorname {erf}^{2}{\left (b x \right )}} \]

[In]

integrate(exp(-b**2*x**2+c)/erf(b*x)**3,x)

[Out]

-sqrt(pi)*exp(c)/(4*b*erf(b*x)**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {\sqrt {\pi } e^{c}}{4 \, b \operatorname {erf}\left (b x\right )^{2}} \]

[In]

integrate(exp(-b^2*x^2+c)/erf(b*x)^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=\int { \frac {e^{\left (-b^{2} x^{2} + c\right )}}{\operatorname {erf}\left (b x\right )^{3}} \,d x } \]

[In]

integrate(exp(-b^2*x^2+c)/erf(b*x)^3,x, algorithm="giac")

[Out]

integrate(e^(-b^2*x^2 + c)/erf(b*x)^3, x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{c-b^2 x^2}}{\text {erf}(b x)^3} \, dx=-\frac {\sqrt {\pi }\,{\mathrm {e}}^c}{4\,b\,{\mathrm {erf}\left (b\,x\right )}^2} \]

[In]

int(exp(c - b^2*x^2)/erf(b*x)^3,x)

[Out]

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