Integrand size = 120, antiderivative size = 34 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=1+\log \left (\frac {\left (\frac {e^{e^{5-x}}}{x}+x\right ) \left (e^{2-x}-\log (x)\right )}{x^2}\right ) \] Output:
1+ln((exp(2-x)-ln(x))*(x+exp(exp(5-x))/x)/x^2)
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=-x-3 \log (x)+\log \left (e^{e^{5-x}}+x^2\right )+\log \left (e^2-e^x \log (x)\right ) \] Input:
Integrate[(x^2 + E^(2 - x)*(x^2 + x^3) - x^2*Log[x] + E^E^(5 - x)*(1 + E^( 7 - 2*x)*x + E^(2 - x)*(3 + x) + (-3 - E^(5 - x)*x)*Log[x]))/(-(E^(2 - x)* x^3) + x^3*Log[x] + E^E^(5 - x)*(-(E^(2 - x)*x) + x*Log[x])),x]
Output:
-x - 3*Log[x] + Log[E^E^(5 - x) + x^2] + Log[E^2 - E^x*Log[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.
\(\displaystyle \int \frac {x^2+x^2 (-\log (x))+e^{2-x} \left (x^3+x^2\right )+e^{e^{5-x}} \left (e^{7-2 x} x+e^{2-x} (x+3)+\left (-e^{5-x} x-3\right ) \log (x)+1\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (x \log (x)-e^{2-x} x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-x^2+x^2 \log (x)-e^{2-x} \left (x^3+x^2\right )-e^{e^{5-x}} \left (e^{7-2 x} x+e^{2-x} (x+3)+\left (-e^{5-x} x-3\right ) \log (x)+1\right )\right )}{x \left (x^2+e^{e^{5-x}}\right ) \left (e^2-e^x \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^{-x+e^{5-x}+5}}{x^2+e^{e^{5-x}}}-\frac {x^3+x^2+e^{e^{5-x}} x+3 e^{e^{5-x}}}{x \left (x^2+e^{e^{5-x}}\right )}-\frac {e^x (x \log (x)+1)}{x \left (e^2-e^x \log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{-x+e^{5-x}+5}}{x^2+e^{e^{5-x}}}dx+2 \int \frac {x}{x^2+e^{e^{5-x}}}dx-x-3 \log (x)+\log \left (e^2-e^x \log (x)\right )\) |
Input:
Int[(x^2 + E^(2 - x)*(x^2 + x^3) - x^2*Log[x] + E^E^(5 - x)*(1 + E^(7 - 2* x)*x + E^(2 - x)*(3 + x) + (-3 - E^(5 - x)*x)*Log[x]))/(-(E^(2 - x)*x^3) + x^3*Log[x] + E^E^(5 - x)*(-(E^(2 - x)*x) + x*Log[x])),x]
Output:
$Aborted
Time = 2.55 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-3 \ln \left (x \right )+\ln \left (-{\mathrm e}^{2-x}+\ln \left (x \right )\right )+\ln \left (x^{2}+{\mathrm e}^{{\mathrm e}^{5-x}}\right )\) | \(30\) |
parallelrisch | \(-3 \ln \left (x \right )+\ln \left (-{\mathrm e}^{2-x}+\ln \left (x \right )\right )+\ln \left (x^{2}+{\mathrm e}^{{\mathrm e}^{5-x}}\right )\) | \(30\) |
Input:
int((((-x*exp(5-x)-3)*ln(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*exp(exp( 5-x))-x^2*ln(x)+(x^3+x^2)*exp(2-x)+x^2)/((x*ln(x)-x*exp(2-x))*exp(exp(5-x) )+x^3*ln(x)-x^3*exp(2-x)),x,method=_RETURNVERBOSE)
Output:
-3*ln(x)+ln(-exp(2-x)+ln(x))+ln(x^2+exp(exp(5-x)))
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=\log \left (x^{2} + e^{\left (e^{\left (-x + 5\right )}\right )}\right ) + \log \left (e^{3} \log \left (x\right ) - e^{\left (-x + 5\right )}\right ) - 3 \, \log \left (x\right ) \] Input:
integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*e xp(exp(5-x))-x^2*log(x)+(x^3+x^2)*exp(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp (exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="fricas")
Output:
log(x^2 + e^(e^(-x + 5))) + log(e^3*log(x) - e^(-x + 5)) - 3*log(x)
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=- 3 \log {\left (x \right )} + \log {\left (x^{2} + e^{e^{3} e^{2 - x}} \right )} + \log {\left (e^{2 - x} - \log {\left (x \right )} \right )} \] Input:
integrate((((-x*exp(5-x)-3)*ln(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*ex p(exp(5-x))-x**2*ln(x)+(x**3+x**2)*exp(2-x)+x**2)/((x*ln(x)-x*exp(2-x))*ex p(exp(5-x))+x**3*ln(x)-x**3*exp(2-x)),x)
Output:
-3*log(x) + log(x**2 + exp(exp(3)*exp(2 - x))) + log(exp(2 - x) - log(x))
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=-x + \log \left (x^{2} + e^{\left (e^{\left (-x + 5\right )}\right )}\right ) - 3 \, \log \left (x\right ) + \log \left (\frac {e^{x} \log \left (x\right ) - e^{2}}{\log \left (x\right )}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*e xp(exp(5-x))-x^2*log(x)+(x^3+x^2)*exp(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp (exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="maxima")
Output:
-x + log(x^2 + e^(e^(-x + 5))) - 3*log(x) + log((e^x*log(x) - e^2)/log(x)) + log(log(x))
\[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=\int { \frac {x^{2} \log \left (x\right ) - x^{2} - {\left (x^{3} + x^{2}\right )} e^{\left (-x + 2\right )} - {\left ({\left (x + 3\right )} e^{\left (-x + 2\right )} + x e^{\left (-2 \, x + 7\right )} - {\left (x e^{\left (-x + 5\right )} + 3\right )} \log \left (x\right ) + 1\right )} e^{\left (e^{\left (-x + 5\right )}\right )}}{x^{3} e^{\left (-x + 2\right )} - x^{3} \log \left (x\right ) + {\left (x e^{\left (-x + 2\right )} - x \log \left (x\right )\right )} e^{\left (e^{\left (-x + 5\right )}\right )}} \,d x } \] Input:
integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*e xp(exp(5-x))-x^2*log(x)+(x^3+x^2)*exp(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp (exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="giac")
Output:
integrate((x^2*log(x) - x^2 - (x^3 + x^2)*e^(-x + 2) - ((x + 3)*e^(-x + 2) + x*e^(-2*x + 7) - (x*e^(-x + 5) + 3)*log(x) + 1)*e^(e^(-x + 5)))/(x^3*e^ (-x + 2) - x^3*log(x) + (x*e^(-x + 2) - x*log(x))*e^(e^(-x + 5))), x)
Time = 4.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=\ln \left (\left (\ln \left (x\right )-{\mathrm {e}}^{-x}\,{\mathrm {e}}^2\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}+x^2\right )\right )-3\,\ln \left (x\right ) \] Input:
int(-(exp(2 - x)*(x^2 + x^3) - x^2*log(x) + exp(exp(5 - x))*(exp(2 - x)*(x + 3) - log(x)*(x*exp(5 - x) + 3) + x*exp(2 - x)*exp(5 - x) + 1) + x^2)/(x ^3*exp(2 - x) - x^3*log(x) + exp(exp(5 - x))*(x*exp(2 - x) - x*log(x))),x)
Output:
log((log(x) - exp(-x)*exp(2))*(exp(exp(-x)*exp(5)) + x^2)) - 3*log(x)
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx=\mathrm {log}\left (e^{\frac {e^{5}}{e^{x}}}+x^{2}\right )+\mathrm {log}\left (e^{x} \mathrm {log}\left (x \right )-e^{2}\right )-3 \,\mathrm {log}\left (x \right )-x \] Input:
int((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*exp(exp (5-x))-x^2*log(x)+(x^3+x^2)*exp(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp(exp(5 -x))+x^3*log(x)-x^3*exp(2-x)),x)
Output:
log(e**(e**5/e**x) + x**2) + log(e**x*log(x) - e**2) - 3*log(x) - x