Integrand size = 72, antiderivative size = 26 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \] Output:
exp(5/exp(ln(50/x^2)+x+exp(2))+ln(2*ln(2)/x))
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \] Input:
Integrate[(E^(-E^2 - x + (E^(-E^2 - x)*x^2*(5 + (50*E^(E^2 + x)*Log[Log[4] /x])/x^2))/50)*(10 - (50*E^(E^2 + x))/x^2 - 5*x)*x)/50,x]
Output:
(E^((E^(-E^2 - x)*x^2)/10)*Log[4])/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.
| \(\displaystyle \int \frac {1}{50} \left (-\frac {50 e^{x+e^2}}{x^2}-5 x+10\right ) x \exp \left (\frac {1}{50} e^{-x-e^2} x^2 \left (\frac {50 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+5\right )-x-e^2\right ) \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{50} \int 5 \exp \left (\frac {1}{10} e^{-x-e^2} \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right ) x^2-x-e^2\right ) \left (-x+2-\frac {10 e^{x+e^2}}{x^2}\right ) xdx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \int \exp \left (\frac {1}{10} e^{-x-e^2} \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right ) x^2-x-e^2\right ) \left (-x+2-\frac {10 e^{x+e^2}}{x^2}\right ) xdx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{10} \int \left (-\exp \left (\frac {1}{10} e^{-x-e^2} \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right ) x^2-x-e^2\right ) (x-2) x-\frac {10 \exp \left (\frac {1}{10} e^{-x-e^2} x^2 \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right )\right )}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{10} \left (2 \int \exp \left (\frac {1}{10} e^{-x-e^2} \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right ) x^2-x-e^2\right ) xdx-\int \exp \left (\frac {1}{10} e^{-x-e^2} \left (\frac {10 e^{x+e^2} \log \left (\frac {\log (4)}{x}\right )}{x^2}+1\right ) x^2-x-e^2\right ) x^2dx-10 \log (4) \int \frac {e^{\frac {1}{10} e^{-x-e^2} x^2}}{x^2}dx\right )\) |
Input:
Int[(E^(-E^2 - x + (E^(-E^2 - x)*x^2*(5 + (50*E^(E^2 + x)*Log[Log[4]/x])/x ^2))/50)*(10 - (50*E^(E^2 + x))/x^2 - 5*x)*x)/50,x]
Output:
$Aborted
Time = 0.74 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46
| method | result | size |
| default | \({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \left (2\right )}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\) | \(38\) |
| parallelrisch | \({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \left (2\right )}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\) | \(38\) |
| risch | \(x^{-\left (-1\right )^{-\frac {\operatorname {csgn}\left (i x^{2}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right )}{2}}} 2^{\left (-1\right )^{-\frac {\operatorname {csgn}\left (i x^{2}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right )}{2}}} \ln \left (2\right )^{\left (-1\right )^{-\frac {\operatorname {csgn}\left (i x^{2}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right )}{2}}} {\mathrm e}^{-\frac {x^{2} \left (-1\right )^{-\frac {\operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-x -{\mathrm e}^{2}}}{10}}\) | \(128\) |
Input:
int((-exp(ln(50/x^2)+x+exp(2))-5*x+10)*exp((ln(2*ln(2)/x)*exp(ln(50/x^2)+x +exp(2))+5)/exp(ln(50/x^2)+x+exp(2)))/x/exp(ln(50/x^2)+x+exp(2)),x,method= _RETURNVERBOSE)
Output:
exp((ln(2*ln(2)/x)*exp(ln(50/x^2)+x+exp(2))+5)/exp(ln(50/x^2)+x+exp(2)))
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (29) = 58\).
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=e^{\left (-\frac {1}{2} \, {\left ({\left (2 \, x + 2 \, e^{2} - \log \left (\frac {2}{25} \, \log \left (2\right )^{2}\right ) + \log \left (\frac {50}{x^{2}}\right )\right )} e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} + x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \] Input:
integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log (50/x^2)+x+exp(2))+5)/exp(log(50/x^2)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2 )),x, algorithm="fricas")
Output:
e^(-1/2*((2*x + 2*e^2 - log(2/25*log(2)^2) + log(50/x^2))*e^(x + e^2 + log (50/x^2)) - 10)*e^(-x - e^2 - log(50/x^2)) + x + e^2 + log(50/x^2))
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=e^{\frac {x^{2} \cdot \left (5 + \frac {50 e^{x + e^{2}} \log {\left (\frac {2 \log {\left (2 \right )}}{x} \right )}}{x^{2}}\right ) e^{- x - e^{2}}}{50}} \] Input:
integrate((-exp(ln(50/x**2)+x+exp(2))-5*x+10)*exp((ln(2*ln(2)/x)*exp(ln(50 /x**2)+x+exp(2))+5)/exp(ln(50/x**2)+x+exp(2)))/x/exp(ln(50/x**2)+x+exp(2)) ,x)
Output:
exp(x**2*(5 + 50*exp(x + exp(2))*log(2*log(2)/x)/x**2)*exp(-x - exp(2))/50 )
\[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\int { -\frac {{\left (5 \, x + e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left ({\left (e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) + 5\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} - x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )}}{x} \,d x } \] Input:
integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log (50/x^2)+x+exp(2))+5)/exp(log(50/x^2)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2 )),x, algorithm="maxima")
Output:
-1/50*integrate(5*(x + 10*e^(x + e^2)/x^2 - 2)*x*e^(1/10*x^2*(10*e^(x + e^ 2)*log(2*log(2)/x)/x^2 + 1)*e^(-x - e^2) - x - e^2), x)
\[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\int { -\frac {{\left (5 \, x + e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left ({\left (e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) + 5\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} - x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )}}{x} \,d x } \] Input:
integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log (50/x^2)+x+exp(2))+5)/exp(log(50/x^2)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2 )),x, algorithm="giac")
Output:
integrate(-(5*x + e^(x + e^2 + log(50/x^2)) - 10)*e^((e^(x + e^2 + log(50/ x^2))*log(2*log(2)/x) + 5)*e^(-x - e^2 - log(50/x^2)) - x - e^2 - log(50/x ^2))/x, x)
Timed out. \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=-\int \frac {x\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \left (2\right )}{5}-\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \left (2\right )}{5}+\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}}\,\ln \left (2\right )}{x^2} \,d x \] Input:
int(-(exp(exp(- x - exp(2) - log(50/x^2))*(log((2*log(2))/x)*exp(x + exp(2 ) + log(50/x^2)) + 5))*exp(- x - exp(2) - log(50/x^2))*(5*x + exp(x + exp( 2) + log(50/x^2)) - 10))/x,x)
Output:
-int((x*exp((x^2*exp(-exp(2))*exp(-x))/10 - exp(2) - x)*log(2))/5 - (2*exp ((x^2*exp(-exp(2))*exp(-x))/10 - exp(2) - x)*log(2))/5 + (2*exp((x^2*exp(- exp(2))*exp(-x))/10)*log(2))/x^2, x)
Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\frac {2 e^{\frac {x^{2}}{10 e^{e^{2}+x}}} \mathrm {log}\left (2\right )}{x} \] Input:
int((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log(50/x^ 2)+x+exp(2))+5)/exp(log(50/x^2)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2)),x)
Output:
(2*e**(x**2/(10*e**(e**2 + x)))*log(2))/x