\(\int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx\) [258]

Optimal result

Integrand size = 18, antiderivative size = 28 \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\frac {e^c \sqrt {\pi } \text {erfi}(b x)^{1+n}}{2 b (1+n)} \]

[Out]

1/2*exp(c)*erfi(b*x)^(1+n)*Pi^(1/2)/b/(1+n)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, 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, 30} \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\frac {\sqrt {\pi } e^c \text {erfi}(b x)^{n+1}}{2 b (n+1)} \]

[In]

Int[E^(c + b^2*x^2)*Erfi[b*x]^n,x]

[Out]

(E^c*Sqrt[Pi]*Erfi[b*x]^(1 + n))/(2*b*(1 + n))

Rule 30

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

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 x^n \, dx,x,\text {erfi}(b x)\right )}{2 b} \\ & = \frac {e^c \sqrt {\pi } \text {erfi}(b x)^{1+n}}{2 b (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\frac {e^c \sqrt {\pi } \text {erfi}(b x)^{1+n}}{2 b (1+n)} \]

[In]

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

[Out]

(E^c*Sqrt[Pi]*Erfi[b*x]^(1 + n))/(2*b*(1 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\frac {\sqrt {\pi } \operatorname {erfi}\left (b x\right )^{n} \operatorname {erfi}\left (b x\right ) e^{c}}{2 \, {\left (b n + b\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).

Time = 1.48 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\begin {cases} \tilde {\infty } x e^{c} & \text {for}\: b = 0 \wedge n = -1 \\0^{n} x e^{c} & \text {for}\: b = 0 \\\frac {\sqrt {\pi } e^{c} \log {\left (\operatorname {erfi}{\left (b x \right )} \right )}}{2 b} & \text {for}\: n = -1 \\\frac {\sqrt {\pi } e^{c} \operatorname {erfi}{\left (b x \right )} \operatorname {erfi}^{n}{\left (b x \right )}}{2 b n + 2 b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x*exp(c), Eq(b, 0) & Eq(n, -1)), (0**n*x*exp(c), Eq(b, 0)), (sqrt(pi)*exp(c)*log(erfi(b*x))/(2*
b), Eq(n, -1)), (sqrt(pi)*exp(c)*erfi(b*x)*erfi(b*x)**n/(2*b*n + 2*b), True))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.83 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int e^{c+b^2 x^2} \text {erfi}(b x)^n \, dx=\frac {\sqrt {\pi }\,{\mathrm {e}}^c\,{\mathrm {erfi}\left (b\,x\right )}^{n+1}}{2\,b\,\left (n+1\right )} \]

[In]

int(exp(c + b^2*x^2)*erfi(b*x)^n,x)

[Out]

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