\(\int \frac {\text {erf}(b x)}{x^6} \, dx\) [14]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (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 = 81 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=-\frac {b e^{-b^2 x^2}}{10 \sqrt {\pi } x^4}+\frac {b^3 e^{-b^2 x^2}}{10 \sqrt {\pi } x^2}-\frac {\text {erf}(b x)}{5 x^5}+\frac {b^5 \operatorname {ExpIntegralEi}\left (-b^2 x^2\right )}{10 \sqrt {\pi }} \]

[Out]

-1/5*erf(b*x)/x^5-1/10*b/exp(b^2*x^2)/x^4/Pi^(1/2)+1/10*b^3/exp(b^2*x^2)/x^2/Pi^(1/2)+1/10*b^5*Ei(-b^2*x^2)/Pi
^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, 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, 2245, 2241} \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=-\frac {b e^{-b^2 x^2}}{10 \sqrt {\pi } x^4}+\frac {b^5 \operatorname {ExpIntegralEi}\left (-b^2 x^2\right )}{10 \sqrt {\pi }}+\frac {b^3 e^{-b^2 x^2}}{10 \sqrt {\pi } x^2}-\frac {\text {erf}(b x)}{5 x^5} \]

[In]

Int[Erf[b*x]/x^6,x]

[Out]

-1/10*b/(E^(b^2*x^2)*Sqrt[Pi]*x^4) + b^3/(10*E^(b^2*x^2)*Sqrt[Pi]*x^2) - Erf[b*x]/(5*x^5) + (b^5*ExpIntegralEi
[-(b^2*x^2)])/(10*Sqrt[Pi])

Rule 2241

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

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

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 {\text {erf}(b x)}{5 x^5}+\frac {(2 b) \int \frac {e^{-b^2 x^2}}{x^5} \, dx}{5 \sqrt {\pi }} \\ & = -\frac {b e^{-b^2 x^2}}{10 \sqrt {\pi } x^4}-\frac {\text {erf}(b x)}{5 x^5}-\frac {b^3 \int \frac {e^{-b^2 x^2}}{x^3} \, dx}{5 \sqrt {\pi }} \\ & = -\frac {b e^{-b^2 x^2}}{10 \sqrt {\pi } x^4}+\frac {b^3 e^{-b^2 x^2}}{10 \sqrt {\pi } x^2}-\frac {\text {erf}(b x)}{5 x^5}+\frac {b^5 \int \frac {e^{-b^2 x^2}}{x} \, dx}{5 \sqrt {\pi }} \\ & = -\frac {b e^{-b^2 x^2}}{10 \sqrt {\pi } x^4}+\frac {b^3 e^{-b^2 x^2}}{10 \sqrt {\pi } x^2}-\frac {\text {erf}(b x)}{5 x^5}+\frac {b^5 \operatorname {ExpIntegralEi}\left (-b^2 x^2\right )}{10 \sqrt {\pi }} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=\frac {b e^{-b^2 x^2} x \left (-1+b^2 x^2\right )-2 \sqrt {\pi } \text {erf}(b x)+b^5 x^5 \operatorname {ExpIntegralEi}\left (-b^2 x^2\right )}{10 \sqrt {\pi } x^5} \]

[In]

Integrate[Erf[b*x]/x^6,x]

[Out]

((b*x*(-1 + b^2*x^2))/E^(b^2*x^2) - 2*Sqrt[Pi]*Erf[b*x] + b^5*x^5*ExpIntegralEi[-(b^2*x^2)])/(10*Sqrt[Pi]*x^5)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.66 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78

method result size
meijerg \(\frac {b^{5} \left (-\frac {b^{2} x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{2}\right ], \left [2, 4, \frac {9}{2}\right ], -b^{2} x^{2}\right )}{21}-\frac {19}{50}+\frac {\gamma }{5}+\frac {2 \ln \left (x \right )}{5}+\frac {2 \ln \left (b \right )}{5}-\frac {1}{b^{4} x^{4}}+\frac {2}{3 b^{2} x^{2}}\right )}{2 \sqrt {\pi }}\) \(63\)
parts \(-\frac {\operatorname {erf}\left (b x \right )}{5 x^{5}}+\frac {2 b \left (-\frac {{\mathrm e}^{-b^{2} x^{2}}}{4 x^{4}}-\frac {b^{2} \left (-\frac {{\mathrm e}^{-b^{2} x^{2}}}{2 x^{2}}+\frac {b^{2} \operatorname {Ei}_{1}\left (b^{2} x^{2}\right )}{2}\right )}{2}\right )}{5 \sqrt {\pi }}\) \(66\)
derivativedivides \(b^{5} \left (-\frac {\operatorname {erf}\left (b x \right )}{5 b^{5} x^{5}}+\frac {-\frac {{\mathrm e}^{-b^{2} x^{2}}}{10 b^{4} x^{4}}+\frac {{\mathrm e}^{-b^{2} x^{2}}}{10 x^{2} b^{2}}-\frac {\operatorname {Ei}_{1}\left (b^{2} x^{2}\right )}{10}}{\sqrt {\pi }}\right )\) \(71\)
default \(b^{5} \left (-\frac {\operatorname {erf}\left (b x \right )}{5 b^{5} x^{5}}+\frac {-\frac {{\mathrm e}^{-b^{2} x^{2}}}{10 b^{4} x^{4}}+\frac {{\mathrm e}^{-b^{2} x^{2}}}{10 x^{2} b^{2}}-\frac {\operatorname {Ei}_{1}\left (b^{2} x^{2}\right )}{10}}{\sqrt {\pi }}\right )\) \(71\)

[In]

int(erf(b*x)/x^6,x,method=_RETURNVERBOSE)

[Out]

1/2/Pi^(1/2)*b^5*(-1/21*b^2*x^2*hypergeom([1,1,7/2],[2,4,9/2],-b^2*x^2)-19/50+1/5*gamma+2/5*ln(x)+2/5*ln(b)-1/
b^4/x^4+2/3/b^2/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=-\frac {2 \, \pi \operatorname {erf}\left (b x\right ) - \sqrt {\pi } {\left (b^{5} x^{5} {\rm Ei}\left (-b^{2} x^{2}\right ) + {\left (b^{3} x^{3} - b x\right )} e^{\left (-b^{2} x^{2}\right )}\right )}}{10 \, \pi x^{5}} \]

[In]

integrate(erf(b*x)/x^6,x, algorithm="fricas")

[Out]

-1/10*(2*pi*erf(b*x) - sqrt(pi)*(b^5*x^5*Ei(-b^2*x^2) + (b^3*x^3 - b*x)*e^(-b^2*x^2)))/(pi*x^5)

Sympy [A] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=- \frac {b^{5} \operatorname {E}_{1}\left (b^{2} x^{2}\right )}{10 \sqrt {\pi }} + \frac {b^{3} e^{- b^{2} x^{2}}}{10 \sqrt {\pi } x^{2}} - \frac {b e^{- b^{2} x^{2}}}{10 \sqrt {\pi } x^{4}} + \frac {\operatorname {erfc}{\left (b x \right )}}{5 x^{5}} - \frac {1}{5 x^{5}} \]

[In]

integrate(erf(b*x)/x**6,x)

[Out]

-b**5*expint(1, b**2*x**2)/(10*sqrt(pi)) + b**3*exp(-b**2*x**2)/(10*sqrt(pi)*x**2) - b*exp(-b**2*x**2)/(10*sqr
t(pi)*x**4) + erfc(b*x)/(5*x**5) - 1/(5*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.33 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=-\frac {b^{5} \Gamma \left (-2, b^{2} x^{2}\right )}{5 \, \sqrt {\pi }} - \frac {\operatorname {erf}\left (b x\right )}{5 \, x^{5}} \]

[In]

integrate(erf(b*x)/x^6,x, algorithm="maxima")

[Out]

-1/5*b^5*gamma(-2, b^2*x^2)/sqrt(pi) - 1/5*erf(b*x)/x^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=-\frac {\operatorname {erf}\left (b x\right )}{5 \, x^{5}} + \frac {b^{10} x^{4} {\rm Ei}\left (-b^{2} x^{2}\right ) + b^{8} x^{2} e^{\left (-b^{2} x^{2}\right )} - b^{6} e^{\left (-b^{2} x^{2}\right )}}{10 \, \sqrt {\pi } b^{5} x^{4}} \]

[In]

integrate(erf(b*x)/x^6,x, algorithm="giac")

[Out]

-1/5*erf(b*x)/x^5 + 1/10*(b^10*x^4*Ei(-b^2*x^2) + b^8*x^2*e^(-b^2*x^2) - b^6*e^(-b^2*x^2))/(sqrt(pi)*b^5*x^4)

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int \frac {\text {erf}(b x)}{x^6} \, dx=\frac {b^5\,\mathrm {ei}\left (-b^2\,x^2\right )}{10\,\sqrt {\pi }}-\frac {\mathrm {erf}\left (b\,x\right )}{5\,x^5}-\frac {\frac {b\,{\mathrm {e}}^{-b^2\,x^2}}{2}-\frac {b^3\,x^2\,{\mathrm {e}}^{-b^2\,x^2}}{2}}{5\,x^4\,\sqrt {\pi }} \]

[In]

int(erf(b*x)/x^6,x)

[Out]

(b^5*ei(-b^2*x^2))/(10*pi^(1/2)) - erf(b*x)/(5*x^5) - ((b*exp(-b^2*x^2))/2 - (b^3*x^2*exp(-b^2*x^2))/2)/(5*x^4
*pi^(1/2))