\(\int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} (-1+x^2)+(-3 e^{e^{\frac {1}{x}+x}} x-3 x^2) \log (2 e^{e^{\frac {1}{x}+x}}+2 x)}{(e^{e^{\frac {1}{x}+x}} x^2+x^3) \log (2 e^{e^{\frac {1}{x}+x}}+2 x)} \, dx\) [7749]

Optimal result

Integrand size = 97, antiderivative size = 21 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=3-3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=\int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
parallelrisch	\(-3 \ln \left (x \right )+\ln \left (\ln \left (2 \,{\mathrm e}^{{\mathrm e}^{\frac {1}{x}} {\mathrm e}^{x}}+2 x \right )\right )\)	\(22\)
risch	\(-3 \ln \left (x \right )+\ln \left (\ln \left (2 \,{\mathrm e}^{{\mathrm e}^{\frac {x^{2}+1}{x}}}+2 x \right )\right )\)	\(25\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).

Time = 0.38 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=- 3 \log {\left (x \right )} + \log {\left (\log {\left (2 x + 2 e^{e^{\frac {1}{x}} e^{x}} \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2\right ) + \log \left (x + e^{\left (e^{\left (x + \frac {1}{x}\right )}\right )}\right )\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).

Time = 0.39 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=\ln \left (\ln \left (2\,x+2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}\,{\mathrm {e}}^x}\right )\right )-3\,\ln \left (x\right ) \]

[In]

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

[Out]

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