Integrand size = 92, antiderivative size = 24 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x+\frac {e^{-2 e^x} \left (-5-x+\log ^2(x)\right )^2}{x} \] Output:
x+1/x/exp(exp(x))^2*(ln(x)^2-5-x)^2
Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x+\frac {e^{-2 e^x} \left (5+x-\log ^2(x)\right )^2}{x} \] Input:
Integrate[(-25 + x^2 + E^(2*E^x)*x^2 + E^x*(-50*x - 20*x^2 - 2*x^3) + (-20 - 4*x)*Log[x] + (10 + E^x*(20*x + 4*x^2))*Log[x]^2 + 4*Log[x]^3 + (-1 - 2 *E^x*x)*Log[x]^4)/(E^(2*E^x)*x^2),x]
Output:
x + (5 + x - Log[x]^2)^2/(E^(2*E^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.
\(\displaystyle \int \frac {e^{-2 e^x} \left (e^{2 e^x} x^2+x^2+\left (e^x \left (4 x^2+20 x\right )+10\right ) \log ^2(x)+e^x \left (-2 x^3-20 x^2-50 x\right )+\left (-2 e^x x-1\right ) \log ^4(x)+4 \log ^3(x)+(-4 x-20) \log (x)-25\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-2 e^x} \left (e^{2 e^x} x^2+x^2-\log ^4(x)+4 \log ^3(x)+10 \log ^2(x)-4 x \log (x)-20 \log (x)-25\right )}{x^2}-\frac {2 e^{x-2 e^x} \left (x-\log ^2(x)+5\right )^2}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -25 \int \frac {e^{-2 e^x}}{x^2}dx+20 \int \frac {\int \frac {e^{-2 e^x}}{x^2}dx}{x}dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2}dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2}dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2}dx-20 \log (x) \int \frac {e^{-2 e^x}}{x^2}dx-50 \int \frac {e^{x-2 e^x}}{x}dx-2 \int e^{x-2 e^x} xdx+4 \int \frac {\int \frac {e^{-2 e^x}}{x}dx}{x}dx-2 \int \frac {e^{x-2 e^x} \log ^4(x)}{x}dx+4 \int e^{x-2 e^x} \log ^2(x)dx+20 \int \frac {e^{x-2 e^x} \log ^2(x)}{x}dx-4 \log (x) \int \frac {e^{-2 e^x}}{x}dx+\operatorname {ExpIntegralEi}\left (-2 e^x\right )+x+10 e^{-2 e^x}\) |
Input:
Int[(-25 + x^2 + E^(2*E^x)*x^2 + E^x*(-50*x - 20*x^2 - 2*x^3) + (-20 - 4*x )*Log[x] + (10 + E^x*(20*x + 4*x^2))*Log[x]^2 + 4*Log[x]^3 + (-1 - 2*E^x*x )*Log[x]^4)/(E^(2*E^x)*x^2),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
risch | \(x +\frac {\left (\ln \left (x \right )^{4}-2 x \ln \left (x \right )^{2}+x^{2}-10 \ln \left (x \right )^{2}+10 x +25\right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}}}{x}\) | \(37\) |
parallelrisch | \(-\frac {\left (-25-x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x}}-\ln \left (x \right )^{4}+2 x \ln \left (x \right )^{2}-x^{2}+10 \ln \left (x \right )^{2}-10 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}}}{x}\) | \(50\) |
Input:
int((x^2*exp(exp(x))^2+(-2*exp(x)*x-1)*ln(x)^4+4*ln(x)^3+((4*x^2+20*x)*exp (x)+10)*ln(x)^2+(-4*x-20)*ln(x)+(-2*x^3-20*x^2-50*x)*exp(x)+x^2-25)/x^2/ex p(exp(x))^2,x,method=_RETURNVERBOSE)
Output:
x+1/x*(ln(x)^4-2*x*ln(x)^2+x^2-10*ln(x)^2+10*x+25)*exp(-2*exp(x))
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\frac {{\left (\log \left (x\right )^{4} + x^{2} e^{\left (2 \, e^{x}\right )} - 2 \, {\left (x + 5\right )} \log \left (x\right )^{2} + x^{2} + 10 \, x + 25\right )} e^{\left (-2 \, e^{x}\right )}}{x} \] Input:
integrate((x^2*exp(exp(x))^2+(-2*exp(x)*x-1)*log(x)^4+4*log(x)^3+((4*x^2+2 0*x)*exp(x)+10)*log(x)^2+(-4*x-20)*log(x)+(-2*x^3-20*x^2-50*x)*exp(x)+x^2- 25)/x^2/exp(exp(x))^2,x, algorithm="fricas")
Output:
(log(x)^4 + x^2*e^(2*e^x) - 2*(x + 5)*log(x)^2 + x^2 + 10*x + 25)*e^(-2*e^ x)/x
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x + \frac {\left (x^{2} - 2 x \log {\left (x \right )}^{2} + 10 x + \log {\left (x \right )}^{4} - 10 \log {\left (x \right )}^{2} + 25\right ) e^{- 2 e^{x}}}{x} \] Input:
integrate((x**2*exp(exp(x))**2+(-2*exp(x)*x-1)*ln(x)**4+4*ln(x)**3+((4*x** 2+20*x)*exp(x)+10)*ln(x)**2+(-4*x-20)*ln(x)+(-2*x**3-20*x**2-50*x)*exp(x)+ x**2-25)/x**2/exp(exp(x))**2,x)
Output:
x + (x**2 - 2*x*log(x)**2 + 10*x + log(x)**4 - 10*log(x)**2 + 25)*exp(-2*e xp(x))/x
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x + \frac {{\left (\log \left (x\right )^{4} - 2 \, {\left (x + 5\right )} \log \left (x\right )^{2} + x^{2} + 25\right )} e^{\left (-2 \, e^{x}\right )}}{x} + 10 \, e^{\left (-2 \, e^{x}\right )} \] Input:
integrate((x^2*exp(exp(x))^2+(-2*exp(x)*x-1)*log(x)^4+4*log(x)^3+((4*x^2+2 0*x)*exp(x)+10)*log(x)^2+(-4*x-20)*log(x)+(-2*x^3-20*x^2-50*x)*exp(x)+x^2- 25)/x^2/exp(exp(x))^2,x, algorithm="maxima")
Output:
x + (log(x)^4 - 2*(x + 5)*log(x)^2 + x^2 + 25)*e^(-2*e^x)/x + 10*e^(-2*e^x )
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.50 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\frac {{\left (e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{4} - 2 \, x e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{2} + x^{2} e^{\left (x - 2 \, e^{x}\right )} + x^{2} e^{x} - 10 \, e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{2} + 10 \, x e^{\left (x - 2 \, e^{x}\right )} + 25 \, e^{\left (x - 2 \, e^{x}\right )}\right )} e^{\left (-x\right )}}{x} \] Input:
integrate((x^2*exp(exp(x))^2+(-2*exp(x)*x-1)*log(x)^4+4*log(x)^3+((4*x^2+2 0*x)*exp(x)+10)*log(x)^2+(-4*x-20)*log(x)+(-2*x^3-20*x^2-50*x)*exp(x)+x^2- 25)/x^2/exp(exp(x))^2,x, algorithm="giac")
Output:
(e^(x - 2*e^x)*log(x)^4 - 2*x*e^(x - 2*e^x)*log(x)^2 + x^2*e^(x - 2*e^x) + x^2*e^x - 10*e^(x - 2*e^x)*log(x)^2 + 10*x*e^(x - 2*e^x) + 25*e^(x - 2*e^ x))*e^(-x)/x
Timed out. \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\int \frac {{\mathrm {e}}^{-2\,{\mathrm {e}}^x}\,\left (4\,{\ln \left (x\right )}^3-\ln \left (x\right )\,\left (4\,x+20\right )+{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^x\,\left (4\,x^2+20\,x\right )+10\right )-{\ln \left (x\right )}^4\,\left (2\,x\,{\mathrm {e}}^x+1\right )+x^2+x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-{\mathrm {e}}^x\,\left (2\,x^3+20\,x^2+50\,x\right )-25\right )}{x^2} \,d x \] Input:
int((exp(-2*exp(x))*(4*log(x)^3 - log(x)*(4*x + 20) + log(x)^2*(exp(x)*(20 *x + 4*x^2) + 10) - log(x)^4*(2*x*exp(x) + 1) + x^2 + x^2*exp(2*exp(x)) - exp(x)*(50*x + 20*x^2 + 2*x^3) - 25))/x^2,x)
Output:
int((exp(-2*exp(x))*(4*log(x)^3 - log(x)*(4*x + 20) + log(x)^2*(exp(x)*(20 *x + 4*x^2) + 10) - log(x)^4*(2*x*exp(x) + 1) + x^2 + x^2*exp(2*exp(x)) - exp(x)*(50*x + 20*x^2 + 2*x^3) - 25))/x^2, x)
Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\frac {e^{2 e^{x}} x^{2}+\mathrm {log}\left (x \right )^{4}-2 \mathrm {log}\left (x \right )^{2} x -10 \mathrm {log}\left (x \right )^{2}+x^{2}+10 x +25}{e^{2 e^{x}} x} \] Input:
int((x^2*exp(exp(x))^2+(-2*exp(x)*x-1)*log(x)^4+4*log(x)^3+((4*x^2+20*x)*e xp(x)+10)*log(x)^2+(-4*x-20)*log(x)+(-2*x^3-20*x^2-50*x)*exp(x)+x^2-25)/x^ 2/exp(exp(x))^2,x)
Output:
(e**(2*e**x)*x**2 + log(x)**4 - 2*log(x)**2*x - 10*log(x)**2 + x**2 + 10*x + 25)/(e**(2*e**x)*x)