3.52 \(\int e^{c-b^2 x^2} \text {erf}(b x)^n \, dx\)

Optimal. Leaf size=28 \[ \frac {\sqrt {\pi } e^c \text {erf}(b x)^{n+1}}{2 b (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6373, 30} \[ \frac {\sqrt {\pi } e^c \text {Erf}(b x)^{n+1}}{2 b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[E^(c - b^2*x^2)*Erf[b*x]^n,x]

[Out]

(E^c*Sqrt[Pi]*Erf[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 6373

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {\sqrt {\pi } e^c \text {erf}(b x)^{n+1}}{2 b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 24, normalized size = 0.86 \[ \frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{n} \operatorname {erf}\left (b x\right ) e^{c}}{2 \, {\left (b n + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {erf}\left (b x\right )^{n} e^{\left (-b^{2} x^{2} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{-b^{2} x^{2}+c} \erf \left (b x \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {erf}\left (b x\right )^{n} e^{\left (-b^{2} x^{2} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.16, size = 23, normalized size = 0.82 \[ \frac {\sqrt {\pi }\,{\mathrm {e}}^c\,{\mathrm {erf}\left (b\,x\right )}^{n+1}}{2\,b\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c - b^2*x^2)*erf(b*x)^n,x)

[Out]

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

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sympy [A]  time = 5.60, size = 63, normalized size = 2.25 \[ \begin {cases} \tilde {\infty } x e^{c} & \text {for}\: b = 0 \wedge n = -1 \\0^{n} x e^{c} & \text {for}\: b = 0 \\\frac {\sqrt {\pi } e^{c} \log {\left (\operatorname {erf}{\left (b x \right )} \right )}}{2 b} & \text {for}\: n = -1 \\\frac {\sqrt {\pi } e^{c} \operatorname {erf}{\left (b x \right )} \operatorname {erf}^{n}{\left (b x \right )}}{2 b n + 2 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-b**2*x**2+c)*erf(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(erf(b*x))/(2*b
), Eq(n, -1)), (sqrt(pi)*exp(c)*erf(b*x)*erf(b*x)**n/(2*b*n + 2*b), True))

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