Integrand size = 104, antiderivative size = 25 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=\log \left (\frac {e^{-e^x+x} \log (x)}{-4+\log \left (\frac {5+x}{x}\right )}\right ) \] Output:
ln(exp(x-exp(x))*ln(x)/(ln(1/x*(5+x))-4))
Time = 1.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=-e^x+x+\log (\log (x))-\log \left (4-\log \left (\frac {5+x}{x}\right )\right ) \] Input:
Integrate[(-20 - 4*x + (5 + x)*Log[(5 + x)/x] + Log[x]*(5 - 20*x - 4*x^2 + E^x*(20*x + 4*x^2) + (5*x + x^2 + E^x*(-5*x - x^2))*Log[(5 + x)/x]))/(Log [x]*(-20*x - 4*x^2 + (5*x + x^2)*Log[(5 + x)/x])),x]
Output:
-E^x + x + Log[Log[x]] - Log[4 - Log[(5 + 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.
\(\displaystyle \int \frac {\log (x) \left (-4 x^2+e^x \left (4 x^2+20 x\right )+\left (x^2+e^x \left (-x^2-5 x\right )+5 x\right ) \log \left (\frac {x+5}{x}\right )-20 x+5\right )-4 x+(x+5) \log \left (\frac {x+5}{x}\right )-20}{\log (x) \left (-4 x^2+\left (x^2+5 x\right ) \log \left (\frac {x+5}{x}\right )-20 x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\log (x) \left (-4 x^2+e^x \left (4 x^2+20 x\right )+\left (x^2+e^x \left (-x^2-5 x\right )+5 x\right ) \log \left (\frac {x+5}{x}\right )-20 x+5\right )+4 x-(x+5) \log \left (\frac {x+5}{x}\right )+20}{x (x+5) \log (x) \left (4-\log \left (\frac {x+5}{x}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^x+\frac {x \log \left (\frac {5}{x}+1\right )}{(-x-5) \left (4-\log \left (\frac {x+5}{x}\right )\right )}-\frac {4 x}{(x+5) \left (\log \left (\frac {x+5}{x}\right )-4\right )}+\frac {5 \log \left (\frac {5}{x}+1\right )}{(-x-5) \left (4-\log \left (\frac {x+5}{x}\right )\right )}-\frac {20}{(x+5) \left (\log \left (\frac {x+5}{x}\right )-4\right )}-\frac {4}{(x+5) \log (x) \left (\log \left (\frac {x+5}{x}\right )-4\right )}+\frac {5}{(x+5) x \left (\log \left (\frac {x+5}{x}\right )-4\right )}+\frac {\log \left (\frac {5}{x}+1\right )}{x \log (x) \left (\log \left (\frac {x+5}{x}\right )-4\right )}-\frac {20}{(x+5) x \log (x) \left (\log \left (\frac {x+5}{x}\right )-4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {1}{\log \left (1+\frac {5}{x}\right )-4}dx+\int \frac {1}{x \left (\log \left (1+\frac {5}{x}\right )-4\right )}dx-\int \frac {1}{(x+5) \left (\log \left (1+\frac {5}{x}\right )-4\right )}dx+5 \int \frac {\log \left (1+\frac {5}{x}\right )}{(-x-5) \left (4-\log \left (\frac {x+5}{x}\right )\right )}dx+5 \int \frac {\log \left (1+\frac {5}{x}\right )}{(x+5) \left (4-\log \left (\frac {x+5}{x}\right )\right )}dx+\int \frac {\log \left (1+\frac {5}{x}\right )}{\log \left (\frac {x+5}{x}\right )-4}dx-4 \int \frac {1}{x \log (x) \left (\log \left (\frac {x+5}{x}\right )-4\right )}dx+\int \frac {\log \left (1+\frac {5}{x}\right )}{x \log (x) \left (\log \left (\frac {x+5}{x}\right )-4\right )}dx-e^x\) |
Input:
Int[(-20 - 4*x + (5 + x)*Log[(5 + x)/x] + Log[x]*(5 - 20*x - 4*x^2 + E^x*( 20*x + 4*x^2) + (5*x + x^2 + E^x*(-5*x - x^2))*Log[(5 + x)/x]))/(Log[x]*(- 20*x - 4*x^2 + (5*x + x^2)*Log[(5 + x)/x])),x]
Output:
$Aborted
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
\[x -\ln \left (\ln \left (1+\frac {5}{x}\right )-4\right )+\ln \left (\ln \left (x \right )\right )-{\mathrm e}^{x}\]
Input:
int(((((-x^2-5*x)*exp(x)+x^2+5*x)*ln(1/x*(5+x))+(4*x^2+20*x)*exp(x)-4*x^2- 20*x+5)*ln(x)+(5+x)*ln(1/x*(5+x))-4*x-20)/((x^2+5*x)*ln(1/x*(5+x))-4*x^2-2 0*x)/ln(x),x)
Output:
x-ln(ln(1+5/x)-4)+ln(ln(x))-exp(x)
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (\frac {x + 5}{x}\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate(((((-x^2-5*x)*exp(x)+x^2+5*x)*log(1/x*(5+x))+(4*x^2+20*x)*exp(x) -4*x^2-20*x+5)*log(x)+(5+x)*log(1/x*(5+x))-4*x-20)/((x^2+5*x)*log(1/x*(5+x ))-4*x^2-20*x)/log(x),x, algorithm="fricas")
Output:
x - e^x - log(log((x + 5)/x) - 4) + log(log(x))
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log {\left (\log {\left (\frac {x + 5}{x} \right )} - 4 \right )} + \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate(((((-x**2-5*x)*exp(x)+x**2+5*x)*ln(1/x*(5+x))+(4*x**2+20*x)*exp( x)-4*x**2-20*x+5)*ln(x)+(5+x)*ln(1/x*(5+x))-4*x-20)/((x**2+5*x)*ln(1/x*(5+ x))-4*x**2-20*x)/ln(x),x)
Output:
x - exp(x) - log(log((x + 5)/x) - 4) + log(log(x))
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \left (x\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate(((((-x^2-5*x)*exp(x)+x^2+5*x)*log(1/x*(5+x))+(4*x^2+20*x)*exp(x) -4*x^2-20*x+5)*log(x)+(5+x)*log(1/x*(5+x))-4*x-20)/((x^2+5*x)*log(1/x*(5+x ))-4*x^2-20*x)/log(x),x, algorithm="maxima")
Output:
x - e^x - log(log(x + 5) - log(x) - 4) + log(log(x))
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \left (x\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate(((((-x^2-5*x)*exp(x)+x^2+5*x)*log(1/x*(5+x))+(4*x^2+20*x)*exp(x) -4*x^2-20*x+5)*log(x)+(5+x)*log(1/x*(5+x))-4*x-20)/((x^2+5*x)*log(1/x*(5+x ))-4*x^2-20*x)/log(x),x, algorithm="giac")
Output:
x - e^x - log(log(x + 5) - log(x) - 4) + log(log(x))
Time = 3.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x-\ln \left (\ln \left (\frac {x+5}{x}\right )-4\right )+\ln \left (\ln \left (x\right )\right )-{\mathrm {e}}^x \] Input:
int((4*x - log(x)*(log((x + 5)/x)*(5*x - exp(x)*(5*x + x^2) + x^2) - 20*x + exp(x)*(20*x + 4*x^2) - 4*x^2 + 5) - log((x + 5)/x)*(x + 5) + 20)/(log(x )*(20*x + 4*x^2 - log((x + 5)/x)*(5*x + x^2))),x)
Output:
x - log(log((x + 5)/x) - 4) + log(log(x)) - exp(x)
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=-e^{x}+\mathrm {log}\left (\mathrm {log}\left (x \right )\right )-\mathrm {log}\left (\mathrm {log}\left (\frac {x +5}{x}\right )-4\right )+x \] Input:
int(((((-x^2-5*x)*exp(x)+x^2+5*x)*log(1/x*(5+x))+(4*x^2+20*x)*exp(x)-4*x^2 -20*x+5)*log(x)+(5+x)*log(1/x*(5+x))-4*x-20)/((x^2+5*x)*log(1/x*(5+x))-4*x ^2-20*x)/log(x),x)
Output:
- e**x + log(log(x)) - log(log((x + 5)/x) - 4) + x