3.24 \(\int x \text {erf}(b x)^2 \, dx\)

Optimal. Leaf size=71 \[ \frac {x e^{-b^2 x^2} \text {erf}(b x)}{\sqrt {\pi } b}-\frac {\text {erf}(b x)^2}{4 b^2}+\frac {e^{-2 b^2 x^2}}{2 \pi b^2}+\frac {1}{2} x^2 \text {erf}(b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6364, 6385, 6373, 30, 2209} \[ \frac {x e^{-b^2 x^2} \text {Erf}(b x)}{\sqrt {\pi } b}-\frac {\text {Erf}(b x)^2}{4 b^2}+\frac {e^{-2 b^2 x^2}}{2 \pi b^2}+\frac {1}{2} x^2 \text {Erf}(b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x*Erf[b*x]^2,x]

[Out]

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

Rule 30

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6364

Int[Erf[(b_.)*(x_)]^2*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*Erf[b*x]^2)/(m + 1), x] - Dist[(4*b)/(Sqrt[Pi]*
(m + 1)), Int[(x^(m + 1)*Erf[b*x])/E^(b^2*x^2), x], x] /; FreeQ[b, x] && (IGtQ[m, 0] || ILtQ[(m + 1)/2, 0])

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]

Rule 6385

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(a_.) + (b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*E^(c + d*x^2)*Erf
[a + b*x])/(2*d), x] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*E^(c + d*x^2)*Erf[a + b*x], x], x] - Dist[b/(d*Sqrt
[Pi]), Int[x^(m - 1)*E^(-a^2 + c - 2*a*b*x - (b^2 - d)*x^2), x], x]) /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int x \text {erf}(b x)^2 \, dx &=\frac {1}{2} x^2 \text {erf}(b x)^2-\frac {(2 b) \int e^{-b^2 x^2} x^2 \text {erf}(b x) \, dx}{\sqrt {\pi }}\\ &=\frac {e^{-b^2 x^2} x \text {erf}(b x)}{b \sqrt {\pi }}+\frac {1}{2} x^2 \text {erf}(b x)^2-\frac {2 \int e^{-2 b^2 x^2} x \, dx}{\pi }-\frac {\int e^{-b^2 x^2} \text {erf}(b x) \, dx}{b \sqrt {\pi }}\\ &=\frac {e^{-2 b^2 x^2}}{2 b^2 \pi }+\frac {e^{-b^2 x^2} x \text {erf}(b x)}{b \sqrt {\pi }}+\frac {1}{2} x^2 \text {erf}(b x)^2-\frac {\operatorname {Subst}(\int x \, dx,x,\text {erf}(b x))}{2 b^2}\\ &=\frac {e^{-2 b^2 x^2}}{2 b^2 \pi }+\frac {e^{-b^2 x^2} x \text {erf}(b x)}{b \sqrt {\pi }}-\frac {\text {erf}(b x)^2}{4 b^2}+\frac {1}{2} x^2 \text {erf}(b x)^2\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.90 \[ \frac {\pi \left (2 b^2 x^2-1\right ) \text {erf}(b x)^2+4 \sqrt {\pi } b x e^{-b^2 x^2} \text {erf}(b x)+2 e^{-2 b^2 x^2}}{4 \pi b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Erf[b*x]^2,x]

[Out]

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

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fricas [A]  time = 0.45, size = 59, normalized size = 0.83 \[ \frac {4 \, \sqrt {\pi } b x \operatorname {erf}\left (b x\right ) e^{\left (-b^{2} x^{2}\right )} - {\left (\pi - 2 \, \pi b^{2} x^{2}\right )} \operatorname {erf}\left (b x\right )^{2} + 2 \, e^{\left (-2 \, b^{2} x^{2}\right )}}{4 \, \pi b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*erf(b*x)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*erf(b*x)^2,x, algorithm="giac")

[Out]

integrate(x*erf(b*x)^2, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int x \erf \left (b x \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*erf(b*x)^2,x)

[Out]

int(x*erf(b*x)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {-\frac {e^{\left (-2 \, b^{2} x^{2}\right )}}{2 \, b^{2}}}{\pi } + \frac {4 \, b x \operatorname {erf}\left (b x\right ) e^{\left (-b^{2} x^{2}\right )} + {\left (2 \, \sqrt {\pi } b^{2} x^{2} - \sqrt {\pi }\right )} \operatorname {erf}\left (b x\right )^{2}}{4 \, \sqrt {\pi } b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*erf(b*x)^2,x, algorithm="maxima")

[Out]

-2*integrate(x*e^(-2*b^2*x^2), x)/pi + 1/4*(4*b*x*erf(b*x)*e^(-b^2*x^2) + (2*sqrt(pi)*b^2*x^2 - sqrt(pi))*erf(
b*x)^2)/(sqrt(pi)*b^2)

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mupad [B]  time = 0.17, size = 67, normalized size = 0.94 \[ \frac {\frac {{\mathrm {e}}^{-2\,b^2\,x^2}}{2}+b\,x\,\sqrt {\pi }\,{\mathrm {e}}^{-b^2\,x^2}\,\mathrm {erf}\left (b\,x\right )}{b^2\,\pi }-\frac {\frac {{\mathrm {erf}\left (b\,x\right )}^2}{4}-\frac {b^2\,x^2\,{\mathrm {erf}\left (b\,x\right )}^2}{2}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*erf(b*x)^2,x)

[Out]

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

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sympy [A]  time = 0.68, size = 65, normalized size = 0.92 \[ \begin {cases} \frac {x^{2} \operatorname {erf}^{2}{\left (b x \right )}}{2} + \frac {x e^{- b^{2} x^{2}} \operatorname {erf}{\left (b x \right )}}{\sqrt {\pi } b} - \frac {\operatorname {erf}^{2}{\left (b x \right )}}{4 b^{2}} + \frac {e^{- 2 b^{2} x^{2}}}{2 \pi b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*erf(b*x)**2,x)

[Out]

Piecewise((x**2*erf(b*x)**2/2 + x*exp(-b**2*x**2)*erf(b*x)/(sqrt(pi)*b) - erf(b*x)**2/(4*b**2) + exp(-2*b**2*x
**2)/(2*pi*b**2), Ne(b, 0)), (0, True))

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