3.112 \(\int x^4 \text {erfc}(b x) \, dx\)

Optimal. Leaf size=84 \[ -\frac {x^4 e^{-b^2 x^2}}{5 \sqrt {\pi } b}-\frac {2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^5}-\frac {2 x^2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^3}+\frac {1}{5} x^5 \text {erfc}(b x) \]

[Out]

1/5*x^5*erfc(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)

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6362, 2212, 2209} \[ -\frac {x^4 e^{-b^2 x^2}}{5 \sqrt {\pi } b}-\frac {2 x^2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^3}-\frac {2 e^{-b^2 x^2}}{5 \sqrt {\pi } b^5}+\frac {1}{5} x^5 \text {Erfc}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*Erfc[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
*Erfc[b*x])/5

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 2212

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 6362

Int[Erfc[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Erfc[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*} \int x^4 \text {erfc}(b x) \, dx &=\frac {1}{5} x^5 \text {erfc}(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 {erfc}(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 {erfc}(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 {erfc}(b x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.79 \[ e^{-b^2 x^2} \left (-\frac {2}{5 \sqrt {\pi } b^5}-\frac {2 x^2}{5 \sqrt {\pi } b^3}-\frac {x^4}{5 \sqrt {\pi } b}\right )+\frac {1}{5} x^5 \text {erfc}(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Erfc[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*Erfc[b*x])/5

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fricas [A]  time = 0.56, size = 60, normalized size = 0.71 \[ -\frac {\pi b^{5} x^{5} \operatorname {erf}\left (b x\right ) - \pi b^{5} x^{5} + \sqrt {\pi } {\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \pi b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 49, normalized size = 0.58 \[ -\frac {1}{5} \, x^{5} \operatorname {erf}\left (b x\right ) + \frac {1}{5} \, x^{5} - \frac {{\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \sqrt {\pi } b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 72, normalized size = 0.86 \[ \frac {\frac {b^{5} x^{5} \mathrm {erfc}\left (b x \right )}{5}+\frac {-\frac {{\mathrm e}^{-b^{2} x^{2}} b^{4} x^{4}}{5}-\frac {2 \,{\mathrm e}^{-b^{2} x^{2}} b^{2} x^{2}}{5}-\frac {2 \,{\mathrm e}^{-b^{2} x^{2}}}{5}}{\sqrt {\pi }}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^5*(1/5*b^5*x^5*erfc(b*x)+2/5/Pi^(1/2)*(-1/2/exp(b^2*x^2)*b^4*x^4-1/exp(b^2*x^2)*b^2*x^2-1/exp(b^2*x^2)))

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maxima [A]  time = 0.62, size = 44, normalized size = 0.52 \[ \frac {1}{5} \, x^{5} \operatorname {erfc}\left (b x\right ) - \frac {{\left (b^{4} x^{4} + 2 \, b^{2} x^{2} + 2\right )} e^{\left (-b^{2} x^{2}\right )}}{5 \, \sqrt {\pi } b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.23, size = 70, normalized size = 0.83 \[ \frac {x^5\,\mathrm {erfc}\left (b\,x\right )}{5}-\frac {\frac {2\,{\mathrm {e}}^{-b^2\,x^2}}{5\,\sqrt {\pi }}+\frac {2\,b^2\,x^2\,{\mathrm {e}}^{-b^2\,x^2}}{5\,\sqrt {\pi }}+\frac {b^4\,x^4\,{\mathrm {e}}^{-b^2\,x^2}}{5\,\sqrt {\pi }}}{b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.74, size = 78, normalized size = 0.93 \[ \begin {cases} \frac {x^{5} \operatorname {erfc}{\left (b x \right )}}{5} - \frac {x^{4} e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b} - \frac {2 x^{2} e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b^{3}} - \frac {2 e^{- b^{2} x^{2}}}{5 \sqrt {\pi } b^{5}} & \text {for}\: b \neq 0 \\\frac {x^{5}}{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**5*erfc(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)), (x**5/5, True))

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