Integrand size = 97, antiderivative size = 21 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=3-3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \] Output:
3+ln(ln(2*exp(exp(1/x)*exp(x))+2*x))-3*ln(x)
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \] Input:
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]
Output:
-3*Log[x] + Log[Log[2*(E^E^(x^(-1) + x) + x)]]
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.
\(\displaystyle \int \frac {x^2+e^{x+e^{x+\frac {1}{x}}+\frac {1}{x}} \left (x^2-1\right )+\left (-3 x^2-3 e^{e^{x+\frac {1}{x}}} x\right ) \log \left (2 x+2 e^{e^{x+\frac {1}{x}}}\right )}{\left (x^3+e^{e^{x+\frac {1}{x}}} x^2\right ) \log \left (2 x+2 e^{e^{x+\frac {1}{x}}}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^2+e^{x+e^{x+\frac {1}{x}}+\frac {1}{x}} \left (x^2-1\right )+\left (-3 x^2-3 e^{e^{x+\frac {1}{x}}} x\right ) \log \left (2 x+2 e^{e^{x+\frac {1}{x}}}\right )}{\left (x^3+e^{e^{x+\frac {1}{x}}} x^2\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+e^{x+\frac {1}{x}}+\frac {1}{x}} (x-1) (x+1)}{x^2 \left (x+e^{e^{x+\frac {1}{x}}}\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}-\frac {-x+3 x \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )+3 e^{e^{x+\frac {1}{x}}} \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}{x \left (x+e^{e^{x+\frac {1}{x}}}\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{x+e^{x+\frac {1}{x}}+\frac {1}{x}}}{x^2 \left (x+e^{e^{x+\frac {1}{x}}}\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}dx+\int \frac {1}{\left (x+e^{e^{x+\frac {1}{x}}}\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}dx+\int \frac {e^{x+e^{x+\frac {1}{x}}+\frac {1}{x}}}{\left (x+e^{e^{x+\frac {1}{x}}}\right ) \log \left (2 \left (x+e^{e^{x+\frac {1}{x}}}\right )\right )}dx-3 \log (x)\) |
Input:
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]
Output:
$Aborted
Time = 6.92 (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 | \(-3 \ln \left (x \right )+\ln \left (\ln \left (2 \,{\mathrm e}^{{\mathrm e}^{\frac {x^{2}+1}{x}}}+2 x \right )\right )\) | \(25\) |
Input:
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)
Output:
-3*ln(x)+ln(ln(2*exp(exp(1/x)*exp(x))+2*x))
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \] Input:
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")
Output:
-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)))
Time = 3.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=- 3 \log {\left (x \right )} + \log {\left (\log {\left (2 x + 2 e^{e^{\frac {1}{x}} e^{x}} \right )} \right )} \] Input:
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)
Output:
-3*log(x) + log(log(2*x + 2*exp(exp(1/x)*exp(x))))
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2\right ) + \log \left (x + e^{\left (e^{\left (x + \frac {1}{x}\right )}\right )}\right )\right ) \] Input:
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")
Output:
-3*log(x) + log(log(2) + log(x + e^(e^(x + 1/x))))
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).
Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=-3 \, \log \left (x\right ) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \] Input:
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")
Output:
-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)))
Time = 3.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=\ln \left (\ln \left (2\,x+2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}\,{\mathrm {e}}^x}\right )\right )-3\,\ln \left (x\right ) \] Input:
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(e xp(1/x)*exp(x)))*(x^2*exp(exp(1/x)*exp(x)) + x^3)),x)
Output:
log(log(2*x + 2*exp(exp(1/x)*exp(x)))) - 3*log(x)
Time = 39.99 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {x^2+e^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (2 e^{e^{\frac {x^{2}+1}{x}}}+2 x \right )\right )-3 \,\mathrm {log}\left (x \right ) \] Input:
int(((-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)
Output:
log(log(2*e**(e**((x**2 + 1)/x)) + 2*x)) - 3*log(x)