Integrand size = 105, antiderivative size = 18 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=4+x+\frac {\log (x)}{2+5 \left (-1+e^x\right ) x} \] Output:
ln(x)/(5*(-1+exp(x))*x+2)+4+x
Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x+\frac {\log (x)}{2-5 x+5 e^x x} \] Input:
Integrate[(2 - x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(5*x + 20*x^2 - 50*x^3) + (5*x + E^x*(-5*x - 5*x^2))*Log[x])/(4*x - 20*x^2 + 25*x^3 + 25*E ^(2*x)*x^3 + E^x*(20*x^2 - 50*x^3)),x]
Output:
x + Log[x]/(2 - 5*x + 5*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.
\(\displaystyle \int \frac {25 e^{2 x} x^3+25 x^3-20 x^2+\left (e^x \left (-5 x^2-5 x\right )+5 x\right ) \log (x)+e^x \left (-50 x^3+20 x^2+5 x\right )-x+2}{25 e^{2 x} x^3+25 x^3-20 x^2+e^x \left (20 x^2-50 x^3\right )+4 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {25 e^{2 x} x^3+25 x^3-20 x^2+\left (e^x \left (-5 x^2-5 x\right )+5 x\right ) \log (x)+e^x \left (-50 x^3+20 x^2+5 x\right )-x+2}{x \left (5 e^x x-5 x+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {\left (5 x^2-2 x-2\right ) \log (x)}{x \left (5 e^x x-5 x+2\right )^2}-\frac {x \log (x)+\log (x)-1}{x \left (5 e^x x-5 x+2\right )}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x \left (5 e^x x-5 x+2\right )}dx-2 \int \frac {\int \frac {1}{\left (5 \left (-1+e^x\right ) x+2\right )^2}dx}{x}dx-2 \int \frac {\int \frac {1}{x \left (5 \left (-1+e^x\right ) x+2\right )^2}dx}{x}dx+5 \int \frac {\int \frac {x}{\left (5 \left (-1+e^x\right ) x+2\right )^2}dx}{x}dx+\int \frac {\int \frac {1}{5 \left (-1+e^x\right ) x+2}dx}{x}dx+\int \frac {\int \frac {1}{x \left (5 \left (-1+e^x\right ) x+2\right )}dx}{x}dx+2 \log (x) \int \frac {1}{\left (5 e^x x-5 x+2\right )^2}dx+2 \log (x) \int \frac {1}{x \left (5 e^x x-5 x+2\right )^2}dx-5 \log (x) \int \frac {x}{\left (5 e^x x-5 x+2\right )^2}dx-\log (x) \int \frac {1}{5 e^x x-5 x+2}dx-\log (x) \int \frac {1}{x \left (5 e^x x-5 x+2\right )}dx+x\) |
Input:
Int[(2 - x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(5*x + 20*x^2 - 50*x^3 ) + (5*x + E^x*(-5*x - 5*x^2))*Log[x])/(4*x - 20*x^2 + 25*x^3 + 25*E^(2*x) *x^3 + E^x*(20*x^2 - 50*x^3)),x]
Output:
$Aborted
Time = 1.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (x \right )}{5 \,{\mathrm e}^{x} x -5 x +2}+x\) | \(18\) |
parallelrisch | \(\frac {25 \,{\mathrm e}^{x} x^{2}-25 x^{2}+10 x +5 \ln \left (x \right )}{25 \,{\mathrm e}^{x} x -25 x +10}\) | \(35\) |
Input:
int((((-5*x^2-5*x)*exp(x)+5*x)*ln(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x)* exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3- 20*x^2+4*x),x,method=_RETURNVERBOSE)
Output:
1/(5*exp(x)*x-5*x+2)*ln(x)+x
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \] Input:
integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^ 2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+ 25*x^3-20*x^2+4*x),x, algorithm="fricas")
Output:
(5*x^2*e^x - 5*x^2 + 2*x + log(x))/(5*x*e^x - 5*x + 2)
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x + \frac {\log {\left (x \right )}}{5 x e^{x} - 5 x + 2} \] Input:
integrate((((-5*x**2-5*x)*exp(x)+5*x)*ln(x)+25*exp(x)**2*x**3+(-50*x**3+20 *x**2+5*x)*exp(x)+25*x**3-20*x**2-x+2)/(25*exp(x)**2*x**3+(-50*x**3+20*x** 2)*exp(x)+25*x**3-20*x**2+4*x),x)
Output:
x + log(x)/(5*x*exp(x) - 5*x + 2)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \] Input:
integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^ 2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+ 25*x^3-20*x^2+4*x),x, algorithm="maxima")
Output:
(5*x^2*e^x - 5*x^2 + 2*x + log(x))/(5*x*e^x - 5*x + 2)
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \] Input:
integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^ 2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+ 25*x^3-20*x^2+4*x),x, algorithm="giac")
Output:
(5*x^2*e^x - 5*x^2 + 2*x + log(x))/(5*x*e^x - 5*x + 2)
Time = 1.76 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x+\frac {\ln \left (x\right )}{5\,x\,{\mathrm {e}}^x-5\,x+2} \] Input:
int((log(x)*(5*x - exp(x)*(5*x + 5*x^2)) - x + 25*x^3*exp(2*x) - 20*x^2 + 25*x^3 + exp(x)*(5*x + 20*x^2 - 50*x^3) + 2)/(4*x + exp(x)*(20*x^2 - 50*x^ 3) + 25*x^3*exp(2*x) - 20*x^2 + 25*x^3),x)
Output:
x + log(x)/(5*x*exp(x) - 5*x + 2)
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 e^{x} x^{2}+\mathrm {log}\left (x \right )-5 x^{2}+2 x}{5 e^{x} x -5 x +2} \] Input:
int((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x) *exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3 -20*x^2+4*x),x)
Output:
(5*e**x*x**2 + log(x) - 5*x**2 + 2*x)/(5*e**x*x - 5*x + 2)