\(\int x^4 \text {erf}(b x) \, dx\) [9]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 8, antiderivative size = 84 \[ \int x^4 \text {erf}(b x) \, dx=\frac {2 e^{-b^2 x^2}}{5 b^5 \sqrt {\pi }}+\frac {2 e^{-b^2 x^2} x^2}{5 b^3 \sqrt {\pi }}+\frac {e^{-b^2 x^2} x^4}{5 b \sqrt {\pi }}+\frac {1}{5} x^5 \text {erf}(b x) \]

[Out]

1/5*x^5*erf(b*x)+2/5/b^5/exp(b^2*x^2)/Pi^(1/2)+2/5*x^2/b^3/exp(b^2*x^2)/Pi^(1/2)+1/5*x^4/b/exp(b^2*x^2)/Pi^(1/
2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6496, 2243, 2240} \[ \int x^4 \text {erf}(b x) \, dx=\frac {x^4 e^{-b^2 x^2}}{5 \sqrt {\pi } b}+\frac {2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^5}+\frac {2 x^2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^3}+\frac {1}{5} x^5 \text {erf}(b x) \]

[In]

Int[x^4*Erf[b*x],x]

[Out]

2/(5*b^5*E^(b^2*x^2)*Sqrt[Pi]) + (2*x^2)/(5*b^3*E^(b^2*x^2)*Sqrt[Pi]) + x^4/(5*b*E^(b^2*x^2)*Sqrt[Pi]) + (x^5*
Erf[b*x])/5

Rule 2240

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 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 6496

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \text {erf}(b x)-\frac {(2 b) \int e^{-b^2 x^2} x^5 \, dx}{5 \sqrt {\pi }} \\ & = \frac {e^{-b^2 x^2} x^4}{5 b \sqrt {\pi }}+\frac {1}{5} x^5 \text {erf}(b x)-\frac {4 \int e^{-b^2 x^2} x^3 \, dx}{5 b \sqrt {\pi }} \\ & = \frac {2 e^{-b^2 x^2} x^2}{5 b^3 \sqrt {\pi }}+\frac {e^{-b^2 x^2} x^4}{5 b \sqrt {\pi }}+\frac {1}{5} x^5 \text {erf}(b x)-\frac {4 \int e^{-b^2 x^2} x \, dx}{5 b^3 \sqrt {\pi }} \\ & = \frac {2 e^{-b^2 x^2}}{5 b^5 \sqrt {\pi }}+\frac {2 e^{-b^2 x^2} x^2}{5 b^3 \sqrt {\pi }}+\frac {e^{-b^2 x^2} x^4}{5 b \sqrt {\pi }}+\frac {1}{5} x^5 \text {erf}(b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int x^4 \text {erf}(b x) \, dx=e^{-b^2 x^2} \left (\frac {2}{5 b^5 \sqrt {\pi }}+\frac {2 x^2}{5 b^3 \sqrt {\pi }}+\frac {x^4}{5 b \sqrt {\pi }}\right )+\frac {1}{5} x^5 \text {erf}(b x) \]

[In]

Integrate[x^4*Erf[b*x],x]

[Out]

(2/(5*b^5*Sqrt[Pi]) + (2*x^2)/(5*b^3*Sqrt[Pi]) + x^4/(5*b*Sqrt[Pi]))/E^(b^2*x^2) + (x^5*Erf[b*x])/5

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65

method result size
meijerg \(\frac {-\frac {4}{5}+\frac {2 \left (3 b^{4} x^{4}+6 b^{2} x^{2}+6\right ) {\mathrm e}^{-b^{2} x^{2}}}{15}+\frac {2 b^{5} x^{5} \operatorname {erf}\left (b x \right ) \sqrt {\pi }}{5}}{2 b^{5} \sqrt {\pi }}\) \(55\)
parallelrisch \(\frac {b^{5} x^{5} \operatorname {erf}\left (b x \right ) \sqrt {\pi }+{\mathrm e}^{-b^{2} x^{2}} x^{4} b^{4}+2 x^{2} {\mathrm e}^{-b^{2} x^{2}} b^{2}+2 \,{\mathrm e}^{-b^{2} x^{2}}}{5 b^{5} \sqrt {\pi }}\) \(68\)
derivativedivides \(\frac {\frac {\operatorname {erf}\left (b x \right ) b^{5} x^{5}}{5}-\frac {2 \left (-\frac {{\mathrm e}^{-b^{2} x^{2}} x^{4} b^{4}}{2}-x^{2} {\mathrm e}^{-b^{2} x^{2}} b^{2}-{\mathrm e}^{-b^{2} x^{2}}\right )}{5 \sqrt {\pi }}}{b^{5}}\) \(72\)
default \(\frac {\frac {\operatorname {erf}\left (b x \right ) b^{5} x^{5}}{5}-\frac {2 \left (-\frac {{\mathrm e}^{-b^{2} x^{2}} x^{4} b^{4}}{2}-x^{2} {\mathrm e}^{-b^{2} x^{2}} b^{2}-{\mathrm e}^{-b^{2} x^{2}}\right )}{5 \sqrt {\pi }}}{b^{5}}\) \(72\)
parts \(\frac {x^{5} \operatorname {erf}\left (b x \right )}{5}-\frac {2 b \left (-\frac {x^{4} {\mathrm e}^{-b^{2} x^{2}}}{2 b^{2}}+\frac {-\frac {x^{2} {\mathrm e}^{-b^{2} x^{2}}}{b^{2}}-\frac {{\mathrm e}^{-b^{2} x^{2}}}{b^{4}}}{b^{2}}\right )}{5 \sqrt {\pi }}\) \(72\)

[In]

int(x^4*erf(b*x),x,method=_RETURNVERBOSE)

[Out]

1/2/b^5/Pi^(1/2)*(-4/5+2/15*(3*b^4*x^4+6*b^2*x^2+6)*exp(-b^2*x^2)+2/5*b^5*x^5*erf(b*x)*Pi^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int x^4 \text {erf}(b x) \, dx=\frac {\pi b^{5} x^{5} \operatorname {erf}\left (b x\right ) + \sqrt {\pi } {\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \pi b^{5}} \]

[In]

integrate(x^4*erf(b*x),x, algorithm="fricas")

[Out]

1/5*(pi*b^5*x^5*erf(b*x) + sqrt(pi)*(b^4*x^4 + 2*b^2*x^2 + 2)*e^(-b^2*x^2))/(pi*b^5)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int x^4 \text {erf}(b x) \, dx=\begin {cases} \frac {x^{5} \operatorname {erf}{\left (b x \right )}}{5} + \frac {x^{4} e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b} + \frac {2 x^{2} e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b^{3}} + \frac {2 e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b^{5}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(x**4*erf(b*x),x)

[Out]

Piecewise((x**5*erf(b*x)/5 + x**4*exp(-b**2*x**2)/(5*sqrt(pi)*b) + 2*x**2*exp(-b**2*x**2)/(5*sqrt(pi)*b**3) +
2*exp(-b**2*x**2)/(5*sqrt(pi)*b**5), Ne(b, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52 \[ \int x^4 \text {erf}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {erf}\left (b x\right ) + \frac {{\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \sqrt {\pi } b^{5}} \]

[In]

integrate(x^4*erf(b*x),x, algorithm="maxima")

[Out]

1/5*x^5*erf(b*x) + 1/5*(b^4*x^4 + 2*b^2*x^2 + 2)*e^(-b^2*x^2)/(sqrt(pi)*b^5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52 \[ \int x^4 \text {erf}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {erf}\left (b x\right ) + \frac {{\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \sqrt {\pi } b^{5}} \]

[In]

integrate(x^4*erf(b*x),x, algorithm="giac")

[Out]

1/5*x^5*erf(b*x) + 1/5*(b^4*x^4 + 2*b^2*x^2 + 2)*e^(-b^2*x^2)/(sqrt(pi)*b^5)

Mupad [B] (verification not implemented)

Time = 5.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52 \[ \int x^4 \text {erf}(b x) \, dx=\frac {x^5\,\mathrm {erf}\left (b\,x\right )}{5}+\frac {{\mathrm {e}}^{-b^2\,x^2}\,\left (b^4\,x^4+2\,b^2\,x^2+2\right )}{5\,b^5\,\sqrt {\pi }} \]

[In]

int(x^4*erf(b*x),x)

[Out]

(x^5*erf(b*x))/5 + (exp(-b^2*x^2)*(2*b^2*x^2 + b^4*x^4 + 2))/(5*b^5*pi^(1/2))