Integrand size = 78, antiderivative size = 21 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{4+x^{e^{46+\frac {1}{5} \left (e^5-x\right )}}} \] Output:
exp(exp(exp(1/5*exp(5)-1/5*x+46)*ln(x))+4)
Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} \] Input:
Integrate[(E^(4 + x^E^((230 + E^5 - x)/5))*x^(-1 + E^((230 + E^5 - x)/5))* (5*E^((230 + E^5 - x)/5) - E^((230 + E^5 - x)/5)*x*Log[x]))/5,x]
Output:
E^(4 + x^E^((230 + E^5 - x)/5))
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.
\(\displaystyle \int \frac {1}{5} e^{x^{e^{\frac {1}{5} \left (-x+e^5+230\right )}}+4} x^{e^{\frac {1}{5} \left (-x+e^5+230\right )}-1} \left (5 e^{\frac {1}{5} \left (-x+e^5+230\right )}-e^{\frac {1}{5} \left (-x+e^5+230\right )} x \log (x)\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int e^{x^{e^{\frac {1}{5} \left (-x+e^5+230\right )}}+4} x^{-1+e^{\frac {1}{5} \left (-x+e^5+230\right )}} \left (5 e^{\frac {1}{5} \left (-x+e^5+230\right )}-e^{\frac {1}{5} \left (-x+e^5+230\right )} x \log (x)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{5} \int e^{x^{e^{\frac {1}{5} \left (-x+e^5+230\right )}}-\frac {x}{5}+\frac {e^5}{5}+50} x^{-1+e^{\frac {1}{5} \left (-x+e^5+230\right )}} (5-x \log (x))dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{5} \int \exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{-1+e^{\frac {1}{5} \left (-x+e^5+230\right )}} (5-x \log (x))dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (5 \exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{-1+e^{\frac {1}{5} \left (-x+e^5+230\right )}}-\exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{e^{\frac {1}{5} \left (-x+e^5+230\right )}} \log (x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\int \frac {\int \exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{e^{\frac {1}{5} \left (230+e^5\right )-\frac {x}{5}}}dx}{x}dx-\log (x) \int \exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{e^{\frac {1}{5} \left (230+e^5\right )-\frac {x}{5}}}dx+5 \int \exp \left (\frac {1}{5} \left (5 x^{e^{-\frac {x}{5}+\frac {e^5}{5}+46}}-x+250 \left (1+\frac {e^5}{250}\right )\right )\right ) x^{-1+e^{\frac {1}{5} \left (-x+e^5+230\right )}}dx\right )\) |
Input:
Int[(E^(4 + x^E^((230 + E^5 - x)/5))*x^(-1 + E^((230 + E^5 - x)/5))*(5*E^( (230 + E^5 - x)/5) - E^((230 + E^5 - x)/5)*x*Log[x]))/5,x]
Output:
$Aborted
Time = 0.48 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76
method | result | size |
risch | \({\mathrm e}^{x^{{\mathrm e}^{\frac {{\mathrm e}^{5}}{5}-\frac {x}{5}+46}}+4}\) | \(16\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{5}}{5}-\frac {x}{5}+46} \ln \left (x \right )}+4}\) | \(18\) |
Input:
int(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*ln(x)+5*exp(1/5*exp(5)-1/5*x+46))*exp (exp(1/5*exp(5)-1/5*x+46)*ln(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)*ln(x))+4 )/x,x,method=_RETURNVERBOSE)
Output:
exp(x^exp(1/5*exp(5)-1/5*x+46)+4)
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{\left (x^{e^{\left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{5} + 46\right )}} + 4\right )} \] Input:
integrate(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*log(x)+5*exp(1/5*exp(5)-1/5*x+4 6))*exp(exp(1/5*exp(5)-1/5*x+46)*log(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)* log(x))+4)/x,x, algorithm="fricas")
Output:
e^(x^e^(-1/5*x + 1/5*e^5 + 46) + 4)
Time = 0.91 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{e^{e^{- \frac {x}{5} + \frac {e^{5}}{5} + 46} \log {\left (x \right )}} + 4} \] Input:
integrate(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*ln(x)+5*exp(1/5*exp(5)-1/5*x+46 ))*exp(exp(1/5*exp(5)-1/5*x+46)*ln(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)*ln (x))+4)/x,x)
Output:
exp(exp(exp(-x/5 + exp(5)/5 + 46)*log(x)) + 4)
Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{\left (x^{e^{\left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{5} + 46\right )}} + 4\right )} \] Input:
integrate(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*log(x)+5*exp(1/5*exp(5)-1/5*x+4 6))*exp(exp(1/5*exp(5)-1/5*x+46)*log(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)* log(x))+4)/x,x, algorithm="maxima")
Output:
e^(x^e^(-1/5*x + 1/5*e^5 + 46) + 4)
Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{\left (x^{e^{\left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{5} + 46\right )}} + 4\right )} \] Input:
integrate(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*log(x)+5*exp(1/5*exp(5)-1/5*x+4 6))*exp(exp(1/5*exp(5)-1/5*x+46)*log(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)* log(x))+4)/x,x, algorithm="giac")
Output:
e^(x^e^(-1/5*x + 1/5*e^5 + 46) + 4)
Time = 4.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx={\mathrm {e}}^4\,{\mathrm {e}}^{x^{{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{5}}\,{\mathrm {e}}^{-\frac {x}{5}}\,{\mathrm {e}}^{46}}} \] Input:
int((exp(exp(exp(5)/5 - x/5 + 46)*log(x))*exp(exp(exp(exp(5)/5 - x/5 + 46) *log(x)) + 4)*(5*exp(exp(5)/5 - x/5 + 46) - x*exp(exp(5)/5 - x/5 + 46)*log (x)))/(5*x),x)
Output:
exp(4)*exp(x^(exp(exp(5)/5)*exp(-x/5)*exp(46)))
Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {1}{5} e^{4+x^{e^{\frac {1}{5} \left (230+e^5-x\right )}}} x^{-1+e^{\frac {1}{5} \left (230+e^5-x\right )}} \left (5 e^{\frac {1}{5} \left (230+e^5-x\right )}-e^{\frac {1}{5} \left (230+e^5-x\right )} x \log (x)\right ) \, dx=e^{e^{\frac {e^{\frac {e^{5}}{5}} \mathrm {log}\left (x \right ) e^{46}}{e^{\frac {x}{5}}}}} e^{4} \] Input:
int(1/5*(-x*exp(1/5*exp(5)-1/5*x+46)*log(x)+5*exp(1/5*exp(5)-1/5*x+46))*ex p(exp(1/5*exp(5)-1/5*x+46)*log(x))*exp(exp(exp(1/5*exp(5)-1/5*x+46)*log(x) )+4)/x,x)
Output:
e**(e**((e**(e**5/5)*log(x)*e**46)/e**(x/5)))*e**4