3.55.58 \(\int \frac {e^{-1/x} (-2 x-x^3-2 x^4+(1-x) (i \pi +\log (-2+2 e)))}{x^3} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{-1/x} \left (i \pi -x \left (2+x^2\right )+\log (-2 (1-e))\right )}{x} \]

________________________________________________________________________________________

Rubi [B]  time = 0.53, antiderivative size = 84, normalized size of antiderivative = 2.55, number of steps used = 11, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6741, 6742, 2206, 2210, 2214, 2212, 2209} \begin {gather*} -e^{-1/x} x^2-e^{-1/x} (2+i \pi +\log (-2 (1-e)))+e^{-1/x} (\log (-2 (1-e))+i \pi )+\frac {e^{-1/x} (\log (-2 (1-e))+i \pi )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x - x^3 - 2*x^4 + (1 - x)*(I*Pi + Log[-2 + 2*E]))/(E^x^(-1)*x^3),x]

[Out]

-(x^2/E^x^(-1)) + (I*Pi + Log[-2*(1 - E)])/E^x^(-1) + (I*Pi + Log[-2*(1 - E)])/(E^x^(-1)*x) - (2 + I*Pi + Log[
-2*(1 - E)])/E^x^(-1)

Rule 2206

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

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 2210

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 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 2214

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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-1/x} \left (i \pi -x^3-2 x^4+\log (-2 (1-e))-x (2+i \pi +\log (-2 (1-e)))\right )}{x^3} \, dx\\ &=\int \left (-e^{-1/x}-2 e^{-1/x} x+\frac {i e^{-1/x} (\pi -i \log (-2 (1-e)))}{x^3}-\frac {i e^{-1/x} (\pi -i (2+\log (-2 (1-e))))}{x^2}\right ) \, dx\\ &=-\left (2 \int e^{-1/x} x \, dx\right )+(i \pi +\log (-2 (1-e))) \int \frac {e^{-1/x}}{x^3} \, dx-(2+i \pi +\log (-2 (1-e))) \int \frac {e^{-1/x}}{x^2} \, dx-\int e^{-1/x} \, dx\\ &=-e^{-1/x} x-e^{-1/x} x^2+\frac {e^{-1/x} (i \pi +\log (-2 (1-e)))}{x}-e^{-1/x} (2+i \pi +\log (-2 (1-e)))+(i \pi +\log (-2 (1-e))) \int \frac {e^{-1/x}}{x^2} \, dx+\int e^{-1/x} \, dx+\int \frac {e^{-1/x}}{x} \, dx\\ &=-e^{-1/x} x^2-\text {Ei}\left (-\frac {1}{x}\right )+e^{-1/x} (i \pi +\log (-2 (1-e)))+\frac {e^{-1/x} (i \pi +\log (-2 (1-e)))}{x}-e^{-1/x} (2+i \pi +\log (-2 (1-e)))-\int \frac {e^{-1/x}}{x} \, dx\\ &=-e^{-1/x} x^2+e^{-1/x} (i \pi +\log (-2 (1-e)))+\frac {e^{-1/x} (i \pi +\log (-2 (1-e)))}{x}-e^{-1/x} (2+i \pi +\log (-2 (1-e)))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 31, normalized size = 0.94 \begin {gather*} \frac {e^{-1/x} \left (i \pi -2 x-x^3+\log (2 (-1+e))\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - x^3 - 2*x^4 + (1 - x)*(I*Pi + Log[-2 + 2*E]))/(E^x^(-1)*x^3),x]

[Out]

(I*Pi - 2*x - x^3 + Log[2*(-1 + E)])/(E^x^(-1)*x)

________________________________________________________________________________________

fricas [A]  time = 1.94, size = 35, normalized size = 1.06 \begin {gather*} -\frac {{\left (x^{3} + 2 \, x\right )} e^{\left (-\frac {1}{x}\right )} - e^{\left (-\frac {1}{x}\right )} \log \left (-2 \, e + 2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(-2*exp(1)+2)-2*x^4-x^3-2*x)/x^3/exp(1/x),x, algorithm="fricas")

[Out]

-((x^3 + 2*x)*e^(-1/x) - e^(-1/x)*log(-2*e + 2))/x

________________________________________________________________________________________

giac [A]  time = 0.14, size = 41, normalized size = 1.24 \begin {gather*} -x^{2} {\left (\frac {2 \, e^{\left (-\frac {1}{x}\right )}}{x^{2}} - \frac {e^{\left (-\frac {1}{x}\right )} \log \left (-2 \, e + 2\right )}{x^{3}} + e^{\left (-\frac {1}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(-2*exp(1)+2)-2*x^4-x^3-2*x)/x^3/exp(1/x),x, algorithm="giac")

[Out]

-x^2*(2*e^(-1/x)/x^2 - e^(-1/x)*log(-2*e + 2)/x^3 + e^(-1/x))

________________________________________________________________________________________

maple [A]  time = 0.16, size = 27, normalized size = 0.82

method	result	size
gosper	\(\frac {\left (-x^{3}+\ln \left (-2 \,{\mathrm e}+2\right )-2 x \right ) {\mathrm e}^{-\frac {1}{x}}}{x}\)	\(27\)
risch	\(\frac {\left (-x^{3}+\ln \relax (2)+\ln \left (1-{\mathrm e}\right )-2 x \right ) {\mathrm e}^{-\frac {1}{x}}}{x}\)	\(29\)
norman	\(\frac {\left (\left (\ln \relax (2)+\ln \left (1-{\mathrm e}\right )\right ) x -2 x^{2}-x^{4}\right ) {\mathrm e}^{-\frac {1}{x}}}{x^{2}}\)	\(34\)
derivativedivides	\(\frac {{\mathrm e}^{-\frac {1}{x}} \ln \relax (2)}{x}+\frac {{\mathrm e}^{-\frac {1}{x}} \ln \left (1-{\mathrm e}\right )}{x}-2 \,{\mathrm e}^{-\frac {1}{x}}-x^{2} {\mathrm e}^{-\frac {1}{x}}\)	\(50\)
default	\(\frac {{\mathrm e}^{-\frac {1}{x}} \ln \relax (2)}{x}+\frac {{\mathrm e}^{-\frac {1}{x}} \ln \left (1-{\mathrm e}\right )}{x}-2 \,{\mathrm e}^{-\frac {1}{x}}-x^{2} {\mathrm e}^{-\frac {1}{x}}\)	\(50\)
meijerg	\(-\left (-\ln \left (-2 \,{\mathrm e}+2\right )-2\right ) \left (1-{\mathrm e}^{-\frac {1}{x}}\right )+\frac {x^{2} \left (\frac {9}{x^{2}}-\frac {12}{x}+6\right )}{6}-\frac {x^{2} \left (3-\frac {3}{x}\right ) {\mathrm e}^{-\frac {1}{x}}}{3}-\frac {1}{2}+x -x^{2}+\frac {x \left (2-\frac {2}{x}\right )}{2}-x \,{\mathrm e}^{-\frac {1}{x}}-\ln \left (-2 \,{\mathrm e}+2\right ) \left (1-\frac {\left (\frac {2}{x}+2\right ) {\mathrm e}^{-\frac {1}{x}}}{2}\right )\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*ln(-2*exp(1)+2)-2*x^4-x^3-2*x)/x^3/exp(1/x),x,method=_RETURNVERBOSE)

[Out]

(-x^3+ln(-2*exp(1)+2)-2*x)/exp(1/x)/x

________________________________________________________________________________________

maxima [C]  time = 0.42, size = 51, normalized size = 1.55 \begin {gather*} -e^{\left (-\frac {1}{x}\right )} \log \left (-2 \, e + 2\right ) + \Gamma \left (2, \frac {1}{x}\right ) \log \left (-2 \, e + 2\right ) - 2 \, e^{\left (-\frac {1}{x}\right )} - \Gamma \left (-1, \frac {1}{x}\right ) - 2 \, \Gamma \left (-2, \frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(-2*exp(1)+2)-2*x^4-x^3-2*x)/x^3/exp(1/x),x, algorithm="maxima")

[Out]

-e^(-1/x)*log(-2*e + 2) + gamma(2, 1/x)*log(-2*e + 2) - 2*e^(-1/x) - gamma(-1, 1/x) - 2*gamma(-2, 1/x)

________________________________________________________________________________________

mupad [B]  time = 3.41, size = 27, normalized size = 0.82 \begin {gather*} -\frac {{\mathrm {e}}^{-\frac {1}{x}}\,\left (x^3+2\,x-\ln \left (2-2\,\mathrm {e}\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1/x)*(2*x + log(2 - 2*exp(1))*(x - 1) + x^3 + 2*x^4))/x^3,x)

[Out]

-(exp(-1/x)*(2*x - log(2 - 2*exp(1)) + x^3))/x

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 27, normalized size = 0.82 \begin {gather*} - \frac {\left (x^{3} + 2 x - \log {\relax (2 )} - \log {\left (-1 + e \right )} - i \pi \right ) e^{- \frac {1}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*ln(-2*exp(1)+2)-2*x**4-x**3-2*x)/x**3/exp(1/x),x)

[Out]

-(x**3 + 2*x - log(2) - log(-1 + E) - I*pi)*exp(-1/x)/x

________________________________________________________________________________________