Integrand size = 64, antiderivative size = 33 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=-e+\left (-2+\frac {x}{3}\right ) \left (-\left (\left (3 e^4+3 e^{-e^x}\right ) x\right )+\log (x)\right ) \] Output:
(ln(x)-x*(3/exp(exp(x))+3*exp(2)^2))*(1/3*x-2)-exp(1)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=-\frac {1}{3} e^{-e^x} (-6+x) \left (3 \left (1+e^{4+e^x}\right ) x-e^{e^x} \log (x)\right ) \] Input:
Integrate[(18*x - 6*x^2 + E^x*(-18*x^2 + 3*x^3) + E^E^x*(-6 + x + E^4*(18* x - 6*x^2) + x*Log[x]))/(3*E^E^x*x),x]
Output:
-1/3*((-6 + x)*(3*(1 + E^(4 + E^x))*x - E^E^x*Log[x]))/E^E^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.
\(\displaystyle \int \frac {e^{-e^x} \left (-6 x^2+e^{e^x} \left (e^4 \left (18 x-6 x^2\right )+x+x \log (x)-6\right )+e^x \left (3 x^3-18 x^2\right )+18 x\right )}{3 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-e^x} \left (-6 x^2+18 x-3 e^x \left (6 x^2-x^3\right )-e^{e^x} \left (-\log (x) x-x-6 e^4 \left (3 x-x^2\right )+6\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (3 e^{x-e^x} (x-6) x+\frac {e^{-e^x} \left (-6 e^{4+e^x} x^2-6 x^2+e^{e^x} \log (x) x+e^{e^x} \left (1+18 e^4\right ) x+18 x-6 e^{e^x}\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (3 \int e^{x-e^x} x^2dx-6 \int e^{-e^x} xdx-18 \int e^{x-e^x} xdx+18 \operatorname {ExpIntegralEi}\left (-e^x\right )-3 e^4 (3-x)^2+x \log (x)-6 \log (x)\right )\) |
Input:
Int[(18*x - 6*x^2 + E^x*(-18*x^2 + 3*x^3) + E^E^x*(-6 + x + E^4*(18*x - 6* x^2) + x*Log[x]))/(3*E^E^x*x),x]
Output:
$Aborted
Time = 0.46 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {x \ln \left (x \right )}{3}-x^{2} {\mathrm e}^{4}+6 x \,{\mathrm e}^{4}-2 \ln \left (x \right )+\frac {\left (-3 x^{2}+18 x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{3}\) | \(39\) |
parallelrisch | \(-\frac {\left (3 \,{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{x}} x^{2}-18 \,{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{x}} x -\ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x}} x +3 x^{2}+6 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (x \right )-18 x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{3}\) | \(54\) |
Input:
int(1/3*((x*ln(x)+(-6*x^2+18*x)*exp(2)^2+x-6)*exp(exp(x))+(3*x^3-18*x^2)*e xp(x)-6*x^2+18*x)/x/exp(exp(x)),x,method=_RETURNVERBOSE)
Output:
1/3*x*ln(x)-x^2*exp(4)+6*x*exp(4)-2*ln(x)+1/3*(-3*x^2+18*x)*exp(-exp(x))
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=-\frac {1}{3} \, {\left (3 \, x^{2} + {\left (3 \, {\left (x^{2} - 6 \, x\right )} e^{4} - {\left (x - 6\right )} \log \left (x\right )\right )} e^{\left (e^{x}\right )} - 18 \, x\right )} e^{\left (-e^{x}\right )} \] Input:
integrate(1/3*((x*log(x)+(-6*x^2+18*x)*exp(2)^2+x-6)*exp(exp(x))+(3*x^3-18 *x^2)*exp(x)-6*x^2+18*x)/x/exp(exp(x)),x, algorithm="fricas")
Output:
-1/3*(3*x^2 + (3*(x^2 - 6*x)*e^4 - (x - 6)*log(x))*e^(e^x) - 18*x)*e^(-e^x )
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=- x^{2} e^{4} + \frac {x \log {\left (x \right )}}{3} + 6 x e^{4} + \left (- x^{2} + 6 x\right ) e^{- e^{x}} - 2 \log {\left (x \right )} \] Input:
integrate(1/3*((x*ln(x)+(-6*x**2+18*x)*exp(2)**2+x-6)*exp(exp(x))+(3*x**3- 18*x**2)*exp(x)-6*x**2+18*x)/x/exp(exp(x)),x)
Output:
-x**2*exp(4) + x*log(x)/3 + 6*x*exp(4) + (-x**2 + 6*x)*exp(-exp(x)) - 2*lo g(x)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=-x^{2} e^{4} + 6 \, x e^{4} - {\left (x^{2} - 6 \, x\right )} e^{\left (-e^{x}\right )} + \frac {1}{3} \, x \log \left (x\right ) - 2 \, \log \left (x\right ) \] Input:
integrate(1/3*((x*log(x)+(-6*x^2+18*x)*exp(2)^2+x-6)*exp(exp(x))+(3*x^3-18 *x^2)*exp(x)-6*x^2+18*x)/x/exp(exp(x)),x, algorithm="maxima")
Output:
-x^2*e^4 + 6*x*e^4 - (x^2 - 6*x)*e^(-e^x) + 1/3*x*log(x) - 2*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=-\frac {1}{3} \, {\left (3 \, x^{2} e^{\left (x - e^{x}\right )} + 3 \, x^{2} e^{\left (x + 4\right )} - x e^{x} \log \left (x\right ) - 18 \, x e^{\left (x - e^{x}\right )} - 18 \, x e^{\left (x + 4\right )} + 6 \, e^{x} \log \left (x\right )\right )} e^{\left (-x\right )} \] Input:
integrate(1/3*((x*log(x)+(-6*x^2+18*x)*exp(2)^2+x-6)*exp(exp(x))+(3*x^3-18 *x^2)*exp(x)-6*x^2+18*x)/x/exp(exp(x)),x, algorithm="giac")
Output:
-1/3*(3*x^2*e^(x - e^x) + 3*x^2*e^(x + 4) - x*e^x*log(x) - 18*x*e^(x - e^x ) - 18*x*e^(x + 4) + 6*e^x*log(x))*e^(-x)
Timed out. \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=\int \frac {{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (6\,x-\frac {{\mathrm {e}}^x\,\left (18\,x^2-3\,x^3\right )}{3}+\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x+{\mathrm {e}}^4\,\left (18\,x-6\,x^2\right )+x\,\ln \left (x\right )-6\right )}{3}-2\,x^2\right )}{x} \,d x \] Input:
int((exp(-exp(x))*(6*x - (exp(x)*(18*x^2 - 3*x^3))/3 + (exp(exp(x))*(x + e xp(4)*(18*x - 6*x^2) + x*log(x) - 6))/3 - 2*x^2))/x,x)
Output:
int((exp(-exp(x))*(6*x - (exp(x)*(18*x^2 - 3*x^3))/3 + (exp(exp(x))*(x + e xp(4)*(18*x - 6*x^2) + x*log(x) - 6))/3 - 2*x^2))/x, x)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {e^{-e^x} \left (18 x-6 x^2+e^x \left (-18 x^2+3 x^3\right )+e^{e^x} \left (-6+x+e^4 \left (18 x-6 x^2\right )+x \log (x)\right )\right )}{3 x} \, dx=\frac {e^{e^{x}} \mathrm {log}\left (x \right ) x -6 e^{e^{x}} \mathrm {log}\left (x \right )-3 e^{e^{x}} e^{4} x^{2}+18 e^{e^{x}} e^{4} x -3 x^{2}+18 x}{3 e^{e^{x}}} \] Input:
int(1/3*((x*log(x)+(-6*x^2+18*x)*exp(2)^2+x-6)*exp(exp(x))+(3*x^3-18*x^2)* exp(x)-6*x^2+18*x)/x/exp(exp(x)),x)
Output:
(e**(e**x)*log(x)*x - 6*e**(e**x)*log(x) - 3*e**(e**x)*e**4*x**2 + 18*e**( e**x)*e**4*x - 3*x**2 + 18*x)/(3*e**(e**x))