Integrand size = 88, antiderivative size = 21 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 x}{5 \left (-2+e^x+x (3+x)-\log (x)\right )} \] Output:
18/5*x/(exp(x)+(3+x)*x-ln(x)-2)
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=-\frac {18 x}{5 \left (2-e^x-3 x-x^2+\log (x)\right )} \] Input:
Integrate[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 6 0*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*(-20 + 30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]
Output:
(-18*x)/(5*(2 - E^x - 3*x - x^2 + Log[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.
\(\displaystyle \int \frac {-18 x^2+e^x (18-18 x)-18 \log (x)-18}{5 x^4+30 x^3+25 x^2+e^x \left (10 x^2+30 x-20\right )+\left (-10 x^2-30 x-10 e^x+20\right ) \log (x)-60 x+5 e^{2 x}+5 \log ^2(x)+20} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {18 \left (-x^2-e^x (x-1)-\log (x)-1\right )}{5 \left (-x^2-3 x-e^x+\log (x)+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {18}{5} \int -\frac {x^2-e^x (1-x)+\log (x)+1}{\left (-x^2-3 x-e^x+\log (x)+2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {18}{5} \int \frac {x^2-e^x (1-x)+\log (x)+1}{\left (-x^2-3 x-e^x+\log (x)+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {18}{5} \int \left (\frac {x-1}{x^2+3 x+e^x-\log (x)-2}-\frac {x^3+x^2-\log (x) x-5 x+1}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {18}{5} \left (-\int \frac {1}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}dx+5 \int \frac {x}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}dx-\int \frac {x^2}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}dx-\int \frac {1}{x^2+3 x+e^x-\log (x)-2}dx+\int \frac {x}{x^2+3 x+e^x-\log (x)-2}dx+\int \frac {x \log (x)}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}dx-\int \frac {x^3}{\left (x^2+3 x+e^x-\log (x)-2\right )^2}dx\right )\) |
Input:
Int[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 60*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*(-20 + 30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]
Output:
$Aborted
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {18 x}{5 \left (x^{2}+3 x +{\mathrm e}^{x}-\ln \left (x \right )-2\right )}\) | \(20\) |
parallelrisch | \(\frac {18 x}{5 \left (x^{2}+3 x +{\mathrm e}^{x}-\ln \left (x \right )-2\right )}\) | \(20\) |
Input:
int((-18*ln(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*ln(x)^2+(-10*exp(x)-10*x^2- 30*x+20)*ln(x)+5*exp(x)^2+(10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x +20),x,method=_RETURNVERBOSE)
Output:
18/5*x/(x^2+3*x+exp(x)-ln(x)-2)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \] Input:
integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x) -10*x^2-30*x+20)*log(x)+5*exp(x)^2+(10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25 *x^2-60*x+20),x, algorithm="fricas")
Output:
18/5*x/(x^2 + 3*x + e^x - log(x) - 2)
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 x}{5 x^{2} + 15 x + 5 e^{x} - 5 \log {\left (x \right )} - 10} \] Input:
integrate((-18*ln(x)+(-18*x+18)*exp(x)-18*x**2-18)/(5*ln(x)**2+(-10*exp(x) -10*x**2-30*x+20)*ln(x)+5*exp(x)**2+(10*x**2+30*x-20)*exp(x)+5*x**4+30*x** 3+25*x**2-60*x+20),x)
Output:
18*x/(5*x**2 + 15*x + 5*exp(x) - 5*log(x) - 10)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \] Input:
integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x) -10*x^2-30*x+20)*log(x)+5*exp(x)^2+(10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25 *x^2-60*x+20),x, algorithm="maxima")
Output:
18/5*x/(x^2 + 3*x + e^x - log(x) - 2)
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \] Input:
integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x) -10*x^2-30*x+20)*log(x)+5*exp(x)^2+(10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25 *x^2-60*x+20),x, algorithm="giac")
Output:
18/5*x/(x^2 + 3*x + e^x - log(x) - 2)
Time = 0.85 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18\,x}{5\,\left (3\,x+{\mathrm {e}}^x-\ln \left (x\right )+x^2-2\right )} \] Input:
int(-(18*log(x) + exp(x)*(18*x - 18) + 18*x^2 + 18)/(5*exp(2*x) - 60*x + 5 *log(x)^2 + exp(x)*(30*x + 10*x^2 - 20) + 25*x^2 + 30*x^3 + 5*x^4 - log(x) *(30*x + 10*exp(x) + 10*x^2 - 20) + 20),x)
Output:
(18*x)/(5*(3*x + exp(x) - log(x) + x^2 - 2))
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 x}{5 e^{x}-5 \,\mathrm {log}\left (x \right )+5 x^{2}+15 x -10} \] Input:
int((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^ 2-30*x+20)*log(x)+5*exp(x)^2+(10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-6 0*x+20),x)
Output:
(18*x)/(5*(e**x - log(x) + x**2 + 3*x - 2))