\(\int \frac {e^{-x-e^{-x} x} (-4 x^3+4 x^4+e^{e^x+e^{-x} x} (-2 e^x+e^{2 x} x)+e^{x+e^{-x} x} (-x-2 \log (5)))}{4 x^3} \, dx\) [5269]

Optimal result

Integrand size = 79, antiderivative size = 27 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=e^{-e^{-x} x}+\frac {e^{e^x}+x+\log (5)}{4 x^2} \]

[Out]

1/4*(x+ln(5)+exp(exp(x)))/x^2+1/exp(x/exp(x))

Rubi [F]

\[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx \]

[In]

Int[(E^(-x - x/E^x)*(-4*x^3 + 4*x^4 + E^(E^x + x/E^x)*(-2*E^x + E^(2*x)*x) + E^(x + x/E^x)*(-x - 2*Log[5])))/(
4*x^3),x]

[Out]

(x + Log[25])^2/(8*x^2*Log[25]) - Defer[Int][E^(-(((1 + E^x)*x)/E^x)), x] - Defer[Int][E^E^x/x^3, x]/2 + Defer
[Int][E^(E^x + x)/x^2, x]/4 + Defer[Int][x/E^(((1 + E^x)*x)/E^x), x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{x^3} \, dx \\ & = \frac {1}{4} \int \frac {-2 e^{e^x}-x+e^{e^x+x} x+4 e^{\left (-1-e^{-x}\right ) x} (-1+x) x^3-2 \log (5)}{x^3} \, dx \\ & = \frac {1}{4} \int \left (4 e^{-e^{-x} \left (1+e^x\right ) x} (-1+x)+\frac {-2 e^{e^x}-x+e^{e^x+x} x-\log (25)}{x^3}\right ) \, dx \\ & = \frac {1}{4} \int \frac {-2 e^{e^x}-x+e^{e^x+x} x-\log (25)}{x^3} \, dx+\int e^{-e^{-x} \left (1+e^x\right ) x} (-1+x) \, dx \\ & = \frac {1}{4} \int \left (\frac {e^{e^x+x}}{x^2}-\frac {2 e^{e^x}+x+\log (25)}{x^3}\right ) \, dx+\int \left (-e^{-e^{-x} \left (1+e^x\right ) x}+e^{-e^{-x} \left (1+e^x\right ) x} x\right ) \, dx \\ & = \frac {1}{4} \int \frac {e^{e^x+x}}{x^2} \, dx-\frac {1}{4} \int \frac {2 e^{e^x}+x+\log (25)}{x^3} \, dx-\int e^{-e^{-x} \left (1+e^x\right ) x} \, dx+\int e^{-e^{-x} \left (1+e^x\right ) x} x \, dx \\ & = \frac {1}{4} \int \frac {e^{e^x+x}}{x^2} \, dx-\frac {1}{4} \int \left (\frac {2 e^{e^x}}{x^3}+\frac {x+\log (25)}{x^3}\right ) \, dx-\int e^{-e^{-x} \left (1+e^x\right ) x} \, dx+\int e^{-e^{-x} \left (1+e^x\right ) x} x \, dx \\ & = \frac {1}{4} \int \frac {e^{e^x+x}}{x^2} \, dx-\frac {1}{4} \int \frac {x+\log (25)}{x^3} \, dx-\frac {1}{2} \int \frac {e^{e^x}}{x^3} \, dx-\int e^{-e^{-x} \left (1+e^x\right ) x} \, dx+\int e^{-e^{-x} \left (1+e^x\right ) x} x \, dx \\ & = \frac {(x+\log (25))^2}{8 x^2 \log (25)}+\frac {1}{4} \int \frac {e^{e^x+x}}{x^2} \, dx-\frac {1}{2} \int \frac {e^{e^x}}{x^3} \, dx-\int e^{-e^{-x} \left (1+e^x\right ) x} \, dx+\int e^{-e^{-x} \left (1+e^x\right ) x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {1}{4} \left (4 e^{-e^{-x} x}+\frac {e^{e^x}}{x^2}+\frac {1}{x}+\frac {\log (25)}{2 x^2}\right ) \]

[In]

Integrate[(E^(-x - x/E^x)*(-4*x^3 + 4*x^4 + E^(E^x + x/E^x)*(-2*E^x + E^(2*x)*x) + E^(x + x/E^x)*(-x - 2*Log[5
])))/(4*x^3),x]

[Out]

(4/E^(x/E^x) + E^E^x/x^2 + x^(-1) + Log[25]/(2*x^2))/4

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\ln \left (5\right )+x}{4 x^{2}}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4 x^{2}}+{\mathrm e}^{-x \,{\mathrm e}^{-x}}\)	\(27\)
parallelrisch	\(\frac {{\mathrm e}^{-x} \left ({\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}} \ln \left (5\right )+4 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}} x +{\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}} {\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-x \,{\mathrm e}^{-x}}}{4 x^{2}}\)	\(63\)

[In]

int(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*ln(5)-x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/ex
p(x)/exp(x/exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/4*(ln(5)+x)/x^2+1/4*exp(exp(x))/x^2+exp(-x*exp(-x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {{\left (4 \, x^{2} e^{x} + {\left (x + \log \left (5\right )\right )} e^{\left ({\left (x e^{x} + x\right )} e^{\left (-x\right )}\right )} + e^{\left ({\left (x + e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + x\right )}\right )} e^{\left (-{\left (x e^{x} + x\right )} e^{\left (-x\right )}\right )}}{4 \, x^{2}} \]

[In]

integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)-x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)
/x^3/exp(x)/exp(x/exp(x)),x, algorithm="fricas")

[Out]

1/4*(4*x^2*e^x + (x + log(5))*e^((x*e^x + x)*e^(-x)) + e^((x + e^(2*x))*e^(-x) + x))*e^(-(x*e^x + x)*e^(-x))/x
^2

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=e^{- x e^{- x}} - \frac {- x - \log {\left (5 \right )}}{4 x^{2}} + \frac {e^{e^{x}}}{4 x^{2}} \]

[In]

integrate(1/4*((x*exp(x)**2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*ln(5)-x)*exp(x)*exp(x/exp(x))+4*x**4-4*x**
3)/x**3/exp(x)/exp(x/exp(x)),x)

[Out]

exp(-x*exp(-x)) - (-x - log(5))/(4*x**2) + exp(exp(x))/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {{\left (4 \, x^{2} + e^{\left (x e^{\left (-x\right )} + e^{x}\right )}\right )} e^{\left (-x e^{\left (-x\right )}\right )}}{4 \, x^{2}} + \frac {1}{4 \, x} + \frac {\log \left (5\right )}{4 \, x^{2}} \]

[In]

integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)-x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)
/x^3/exp(x)/exp(x/exp(x)),x, algorithm="maxima")

[Out]

1/4*(4*x^2 + e^(x*e^(-x) + e^x))*e^(-x*e^(-x))/x^2 + 1/4/x + 1/4*log(5)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {4 \, x^{2} e^{\left (-x e^{\left (-x\right )}\right )} + x + e^{\left (e^{x}\right )} + \log \left (5\right )}{4 \, x^{2}} \]

[In]

integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)-x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)
/x^3/exp(x)/exp(x/exp(x)),x, algorithm="giac")

[Out]

1/4*(4*x^2*e^(-x*e^(-x)) + x + e^(e^x) + log(5))/x^2

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx={\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}}+\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{4\,x^2}+\frac {x+\ln \left (5\right )}{4\,x^2} \]

[In]

int(-(exp(-x)*exp(-x*exp(-x))*(x^3 - x^4 + (exp(x*exp(-x))*exp(x)*(x + 2*log(5)))/4 + (exp(exp(x))*exp(x*exp(-
x))*(2*exp(x) - x*exp(2*x)))/4))/x^3,x)

[Out]

exp(-x*exp(-x)) + exp(exp(x))/(4*x^2) + (x + log(5))/(4*x^2)