\(\int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx\) [255]

Optimal result

Integrand size = 18, antiderivative size = 20 \[ \int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx=\frac {e^c \sqrt {\pi } \log (\text {erfi}(b x))}{2 b} \]

[Out]

1/2*exp(c)*ln(erfi(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6510, 29} \[ \int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx=\frac {\sqrt {\pi } e^c \log (\text {erfi}(b x))}{2 b} \]

[In]

Int[E^(c + b^2*x^2)/Erfi[b*x],x]

[Out]

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

Rule 29

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

Rule 6510

Int[E^((c_.) + (d_.)*(x_)^2)*Erfi[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[E^c*(Sqrt[Pi]/(2*b)), Subst[Int[x^n, x]
, x, Erfi[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 {erfi}(b x)\right )}{2 b} \\ & = \frac {e^c \sqrt {\pi } \log (\text {erfi}(b x))}{2 b} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[E^(c + b^2*x^2)/Erfi[b*x],x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx=\frac {\sqrt {\pi } e^{c} \log \left (\operatorname {erfi}\left (b x\right )\right )}{2 \, b} \]

[In]

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

[Out]

1/2*sqrt(pi)*e^c*log(erfi(b*x))/b

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx=\begin {cases} \frac {\sqrt {\pi } e^{c} \log {\left (\operatorname {erfi}{\left (b x \right )} \right )}}{2 b} & \text {for}\: b \neq 0 \\\tilde {\infty } x e^{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{c+b^2 x^2}}{\text {erfi}(b x)} \, dx=\frac {\sqrt {\pi }\,\ln \left (\mathrm {erfi}\left (b\,x\right )\right )\,{\mathrm {e}}^c}{2\,b} \]

[In]

int(exp(c + b^2*x^2)/erfi(b*x),x)

[Out]

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