Integrand size = 78, antiderivative size = 29 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2}{5} \left (\log \left (-e^x+\frac {1}{x}\right )-\frac {1}{60} \log (2-x-\log (x))\right ) \] Output:
2/5*ln(1/x-exp(x))-1/150*ln(-ln(x)+2-x)
Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {1}{150} \left (-60 \log (x)+60 \log \left (1-e^x x\right )-\log (2-x-\log (x))\right ) \] Input:
Integrate[(-119 + 61*x + E^x*(-x - 121*x^2 + 60*x^3) + (60 + 60*E^x*x^2)*L og[x])/(300*x - 150*x^2 + E^x*(-300*x^2 + 150*x^3) + (-150*x + 150*E^x*x^2 )*Log[x]),x]
Output:
(-60*Log[x] + 60*Log[1 - E^x*x] - Log[2 - x - Log[x]])/150
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.
\(\displaystyle \int \frac {\left (60 e^x x^2+60\right ) \log (x)+e^x \left (60 x^3-121 x^2-x\right )+61 x-119}{-150 x^2+\left (150 e^x x^2-150 x\right ) \log (x)+e^x \left (150 x^3-300 x^2\right )+300 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (60 e^x x^2+60\right ) \log (x)+e^x \left (60 x^3-121 x^2-x\right )+61 x-119}{150 x \left (1-e^x x\right ) (-x-\log (x)+2)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{150} \int -\frac {-61 x+e^x \left (-60 x^3+121 x^2+x\right )-60 \left (e^x x^2+1\right ) \log (x)+119}{x \left (1-e^x x\right ) (-x-\log (x)+2)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{150} \int \frac {-61 x+e^x \left (-60 x^3+121 x^2+x\right )-60 \left (e^x x^2+1\right ) \log (x)+119}{x \left (1-e^x x\right ) (-x-\log (x)+2)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{150} \int \left (\frac {-60 x^2-60 \log (x) x+121 x+1}{x (x+\log (x)-2)}-\frac {60 (x+1)}{x \left (e^x x-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{150} \left (60 \int \frac {1}{e^x x-1}dx+60 \int \frac {1}{x \left (e^x x-1\right )}dx+60 x-\log (-x-\log (x)+2)\right )\) |
Input:
Int[(-119 + 61*x + E^x*(-x - 121*x^2 + 60*x^3) + (60 + 60*E^x*x^2)*Log[x]) /(300*x - 150*x^2 + E^x*(-300*x^2 + 150*x^3) + (-150*x + 150*E^x*x^2)*Log[ x]),x]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {2 \ln \left ({\mathrm e}^{x}-\frac {1}{x}\right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}\) | \(21\) |
norman | \(-\frac {2 \ln \left (x \right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\) | \(23\) |
parallelrisch | \(-\frac {2 \ln \left (x \right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\) | \(23\) |
Input:
int(((60*exp(x)*x^2+60)*ln(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*ex p(x)*x^2-150*x)*ln(x)+(150*x^3-300*x^2)*exp(x)-150*x^2+300*x),x,method=_RE TURNVERBOSE)
Output:
2/5*ln(exp(x)-1/x)-1/150*ln(x+ln(x)-2)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=-\frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \] Input:
integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/( (150*exp(x)*x^2-150*x)*log(x)+(150*x^3-300*x^2)*exp(x)-150*x^2+300*x),x, a lgorithm="fricas")
Output:
-1/150*log(x + log(x) - 2) + 2/5*log((x*e^x - 1)/x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2 \log {\left (e^{x} - \frac {1}{x} \right )}}{5} - \frac {\log {\left (x + \log {\left (x \right )} - 2 \right )}}{150} \] Input:
integrate(((60*exp(x)*x**2+60)*ln(x)+(60*x**3-121*x**2-x)*exp(x)+61*x-119) /((150*exp(x)*x**2-150*x)*ln(x)+(150*x**3-300*x**2)*exp(x)-150*x**2+300*x) ,x)
Output:
2*log(exp(x) - 1/x)/5 - log(x + log(x) - 2)/150
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=-\frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \] Input:
integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/( (150*exp(x)*x^2-150*x)*log(x)+(150*x^3-300*x^2)*exp(x)-150*x^2+300*x),x, a lgorithm="maxima")
Output:
-1/150*log(x + log(x) - 2) + 2/5*log((x*e^x - 1)/x)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2}{5} \, \log \left (x e^{x} - 1\right ) - \frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) - \frac {2}{5} \, \log \left (x\right ) \] Input:
integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/( (150*exp(x)*x^2-150*x)*log(x)+(150*x^3-300*x^2)*exp(x)-150*x^2+300*x),x, a lgorithm="giac")
Output:
2/5*log(x*e^x - 1) - 1/150*log(x + log(x) - 2) - 2/5*log(x)
Time = 2.73 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2\,\ln \left (\frac {x\,{\mathrm {e}}^x-1}{x}\right )}{5}-\frac {\ln \left (x+\ln \left (x\right )-2\right )}{150} \] Input:
int(-(61*x - exp(x)*(x + 121*x^2 - 60*x^3) + log(x)*(60*x^2*exp(x) + 60) - 119)/(exp(x)*(300*x^2 - 150*x^3) - 300*x + log(x)*(150*x - 150*x^2*exp(x) ) + 150*x^2),x)
Output:
(2*log((x*exp(x) - 1)/x))/5 - log(x + log(x) - 2)/150
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=-\frac {\mathrm {log}\left (\mathrm {log}\left (x \right )+x -2\right )}{150}+\frac {2 \,\mathrm {log}\left (e^{x} x -1\right )}{5}-\frac {2 \,\mathrm {log}\left (x \right )}{5} \] Input:
int(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*e xp(x)*x^2-150*x)*log(x)+(150*x^3-300*x^2)*exp(x)-150*x^2+300*x),x)
Output:
( - log(log(x) + x - 2) + 60*log(e**x*x - 1) - 60*log(x))/150