\(\int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx\) [7021]

Optimal result

Integrand size = 49, antiderivative size = 27 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=e^{e^{-x+\frac {-1-e-x+x^2+\log (5)}{x}}}+x \]

[Out]

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

Rubi [F]

\[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=\int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx \]

[In]

Int[(x^2 + E^(E^((-1 - E - x + Log[5])/x) + (-1 - E - x + Log[5])/x)*(1 + E - Log[5]))/x^2,x]

[Out]

x - (1 + E - Log[5])*Defer[Subst][Defer[Int][5^x*E^(-1 + 5^x*E^(-1 - x - E*x) + (-1 - E)*x), x], x, x^(-1)]

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

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=e^{5^{\frac {1}{x}} e^{-\frac {1+e+x}{x}}}+x \]

[In]

Integrate[(x^2 + E^(E^((-1 - E - x + Log[5])/x) + (-1 - E - x + Log[5])/x)*(1 + E - Log[5]))/x^2,x]

[Out]

E^(5^x^(-1)/E^((1 + E + x)/x)) + x

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(x +{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}\)	\(20\)
risch	\(x +{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}+x +1}{x}}}\)	\(21\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}}{x}\)	\(28\)
parts	\(x -\frac {\left (-\ln \left (5\right )+1+{\mathrm e}\right ) {\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}\)	\(40\)
default	\(x -\frac {{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}-\frac {{\mathrm e} \,{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}+\frac {\ln \left (5\right ) {\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}\)	\(114\)

[In]

int(((-ln(5)+1+exp(1))*exp((ln(5)-exp(1)-x-1)/x)*exp(exp((ln(5)-exp(1)-x-1)/x))+x^2)/x^2,x,method=_RETURNVERBO
SE)

[Out]

x+exp(exp((ln(5)-exp(1)-x-1)/x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx={\left (x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} + e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )}\right )} e^{\left (\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x + e^{e^{\frac {- x - e - 1 + \log {\left (5 \right )}}{x}}} \]

[In]

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

[Out]

x + exp(exp((-x - E - 1 + log(5))/x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (26) = 52\).

Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.93 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x - \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )}\right )} \log \left (5\right )}{e - \log \left (5\right ) + 1} + \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )} + 1\right )}}{e - \log \left (5\right ) + 1} + \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )}\right )}}{e - \log \left (5\right ) + 1} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (26) = 52\).

Time = 0.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 8.11 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \log \left (5\right ) - x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x} + 1\right )} - x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} + e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )} \log \left (5\right ) - e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )} - e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - e + \log \left (5\right ) - 1}{x}\right )}}{e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \log \left (5\right ) - e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x} + 1\right )} - e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x+{\mathrm {e}}^{5^{1/x}\,{\mathrm {e}}^{-\frac {\mathrm {e}}{x}}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {1}{x}}} \]

[In]

int((x^2 + exp(exp(-(x + exp(1) - log(5) + 1)/x))*exp(-(x + exp(1) - log(5) + 1)/x)*(exp(1) - log(5) + 1))/x^2
,x)

[Out]

x + exp(5^(1/x)*exp(-exp(1)/x)*exp(-1)*exp(-1/x))