Integrand size = 8, antiderivative size = 71 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\frac {b e^{-b^2 x^2}}{6 \sqrt {\pi } x^3}-\frac {b^3 e^{-b^2 x^2}}{3 \sqrt {\pi } x}-\frac {1}{3} b^4 \text {erf}(b x)-\frac {\text {erfc}(b x)}{4 x^4} \] Output:
1/6*b/exp(b^2*x^2)/Pi^(1/2)/x^3-1/3*b^3/exp(b^2*x^2)/Pi^(1/2)/x-1/3*b^4*er f(b*x)-1/4*erfc(b*x)/x^4
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\frac {1}{12} \left (\frac {2 e^{-b^2 x^2} \left (b-2 b^3 x^2\right )}{\sqrt {\pi } x^3}-4 b^4 \text {erf}(b x)-\frac {3 \text {erfc}(b x)}{x^4}\right ) \] Input:
Integrate[Erfc[b*x]/x^5,x]
Output:
((2*(b - 2*b^3*x^2))/(E^(b^2*x^2)*Sqrt[Pi]*x^3) - 4*b^4*Erf[b*x] - (3*Erfc [b*x])/x^4)/12
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6916, 2643, 2643, 2634}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {erfc}(b x)}{x^5} \, dx\) |
\(\Big \downarrow \) 6916 |
\(\displaystyle -\frac {b \int \frac {e^{-b^2 x^2}}{x^4}dx}{2 \sqrt {\pi }}-\frac {\text {erfc}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle -\frac {b \left (-\frac {2}{3} b^2 \int \frac {e^{-b^2 x^2}}{x^2}dx-\frac {e^{-b^2 x^2}}{3 x^3}\right )}{2 \sqrt {\pi }}-\frac {\text {erfc}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle -\frac {b \left (-\frac {2}{3} b^2 \left (-2 b^2 \int e^{-b^2 x^2}dx-\frac {e^{-b^2 x^2}}{x}\right )-\frac {e^{-b^2 x^2}}{3 x^3}\right )}{2 \sqrt {\pi }}-\frac {\text {erfc}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 2634 |
\(\displaystyle -\frac {b \left (-\frac {2}{3} b^2 \left (\sqrt {\pi } (-b) \text {erf}(b x)-\frac {e^{-b^2 x^2}}{x}\right )-\frac {e^{-b^2 x^2}}{3 x^3}\right )}{2 \sqrt {\pi }}-\frac {\text {erfc}(b x)}{4 x^4}\) |
Input:
Int[Erfc[b*x]/x^5,x]
Output:
-1/2*(b*(-1/3*1/(E^(b^2*x^2)*x^3) - (2*b^2*(-(1/(E^(b^2*x^2)*x)) - b*Sqrt[ Pi]*Erf[b*x]))/3))/Sqrt[Pi] - Erfc[b*x]/(4*x^4)
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr eeQ[{F, a, b, c, d}, x] && NegQ[b]
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] - Simp[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]))
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] + Simp[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]
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87
method | result | size |
parts | \(-\frac {\operatorname {erfc}\left (b x \right )}{4 x^{4}}-\frac {b \left (-\frac {{\mathrm e}^{-b^{2} x^{2}}}{3 x^{3}}-\frac {2 b^{2} \left (-\frac {{\mathrm e}^{-b^{2} x^{2}}}{x}-b \sqrt {\pi }\, \operatorname {erf}\left (b x \right )\right )}{3}\right )}{2 \sqrt {\pi }}\) | \(62\) |
parallelrisch | \(\frac {4 \,\operatorname {erfc}\left (b x \right ) x^{4} \sqrt {\pi }\, b^{4}-4 \,{\mathrm e}^{-b^{2} x^{2}} x^{3} b^{3}+2 \,{\mathrm e}^{-b^{2} x^{2}} b x -3 \,\operatorname {erfc}\left (b x \right ) \sqrt {\pi }}{12 \sqrt {\pi }\, x^{4}}\) | \(64\) |
derivativedivides | \(b^{4} \left (-\frac {\operatorname {erfc}\left (b x \right )}{4 b^{4} x^{4}}-\frac {-\frac {{\mathrm e}^{-b^{2} x^{2}}}{3 b^{3} x^{3}}+\frac {2 \,{\mathrm e}^{-b^{2} x^{2}}}{3 x b}+\frac {2 \,\operatorname {erf}\left (b x \right ) \sqrt {\pi }}{3}}{2 \sqrt {\pi }}\right )\) | \(69\) |
default | \(b^{4} \left (-\frac {\operatorname {erfc}\left (b x \right )}{4 b^{4} x^{4}}-\frac {-\frac {{\mathrm e}^{-b^{2} x^{2}}}{3 b^{3} x^{3}}+\frac {2 \,{\mathrm e}^{-b^{2} x^{2}}}{3 x b}+\frac {2 \,\operatorname {erf}\left (b x \right ) \sqrt {\pi }}{3}}{2 \sqrt {\pi }}\right )\) | \(69\) |
Input:
int(erfc(b*x)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*erfc(b*x)/x^4-1/2/Pi^(1/2)*b*(-1/3/x^3*exp(-b^2*x^2)-2/3*b^2*(-1/x*ex p(-b^2*x^2)-b*Pi^(1/2)*erf(b*x)))
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=-\frac {3 \, \pi + 2 \, \sqrt {\pi } {\left (2 \, b^{3} x^{3} - b x\right )} e^{\left (-b^{2} x^{2}\right )} - {\left (3 \, \pi - 4 \, \pi b^{4} x^{4}\right )} \operatorname {erf}\left (b x\right )}{12 \, \pi x^{4}} \] Input:
integrate(erfc(b*x)/x^5,x, algorithm="fricas")
Output:
-1/12*(3*pi + 2*sqrt(pi)*(2*b^3*x^3 - b*x)*e^(-b^2*x^2) - (3*pi - 4*pi*b^4 *x^4)*erf(b*x))/(pi*x^4)
Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\frac {b^{4} \operatorname {erfc}{\left (b x \right )}}{3} - \frac {b^{3} e^{- b^{2} x^{2}}}{3 \sqrt {\pi } x} + \frac {b e^{- b^{2} x^{2}}}{6 \sqrt {\pi } x^{3}} - \frac {\operatorname {erfc}{\left (b x \right )}}{4 x^{4}} \] Input:
integrate(erfc(b*x)/x**5,x)
Output:
b**4*erfc(b*x)/3 - b**3*exp(-b**2*x**2)/(3*sqrt(pi)*x) + b*exp(-b**2*x**2) /(6*sqrt(pi)*x**3) - erfc(b*x)/(4*x**4)
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\frac {b^{4} {\left (x^{2}\right )}^{\frac {3}{2}} \Gamma \left (-\frac {3}{2}, b^{2} x^{2}\right )}{4 \, \sqrt {\pi } x^{3}} - \frac {\operatorname {erfc}\left (b x\right )}{4 \, x^{4}} \] Input:
integrate(erfc(b*x)/x^5,x, algorithm="maxima")
Output:
1/4*b^4*(x^2)^(3/2)*gamma(-3/2, b^2*x^2)/(sqrt(pi)*x^3) - 1/4*erfc(b*x)/x^ 4
\[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\int { \frac {\operatorname {erfc}\left (b x\right )}{x^{5}} \,d x } \] Input:
integrate(erfc(b*x)/x^5,x, algorithm="giac")
Output:
integrate(erfc(b*x)/x^5, x)
Time = 3.88 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=-\frac {\frac {\mathrm {erfc}\left (b\,x\right )}{4}+\frac {b^3\,x^3\,{\mathrm {e}}^{-b^2\,x^2}}{3\,\sqrt {\pi }}-\frac {b\,x\,{\mathrm {e}}^{-b^2\,x^2}}{6\,\sqrt {\pi }}}{x^4}-\frac {b^5\,\mathrm {erfi}\left (x\,\sqrt {-b^2}\right )}{3\,\sqrt {-b^2}} \] Input:
int(erfc(b*x)/x^5,x)
Output:
- (erfc(b*x)/4 + (b^3*x^3*exp(-b^2*x^2))/(3*pi^(1/2)) - (b*x*exp(-b^2*x^2) )/(6*pi^(1/2)))/x^4 - (b^5*erfi(x*(-b^2)^(1/2)))/(3*(-b^2)^(1/2))
\[ \int \frac {\text {erfc}(b x)}{x^5} \, dx=\frac {3 e^{b^{2} x^{2}} \mathrm {erf}\left (b x \right ) \pi +4 \sqrt {\pi }\, e^{b^{2} x^{2}} \left (\int \frac {1}{e^{b^{2} x^{2}} x^{2}}d x \right ) b^{3} x^{4}-3 e^{b^{2} x^{2}} \pi +2 \sqrt {\pi }\, b x}{12 e^{b^{2} x^{2}} \pi \,x^{4}} \] Input:
int(erfc(b*x)/x^5,x)
Output:
(3*e**(b**2*x**2)*erf(b*x)*pi + 4*sqrt(pi)*e**(b**2*x**2)*int(1/(e**(b**2* x**2)*x**2),x)*b**3*x**4 - 3*e**(b**2*x**2)*pi + 2*sqrt(pi)*b*x)/(12*e**(b **2*x**2)*pi*x**4)