Integrand size = 144, antiderivative size = 19 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 x}{5 \left (4+x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )\right )} \] Output:
2/5*x/(ln(ln(9*exp(x)+18))+4+x)
Time = 0.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 x}{5 \left (4+x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )\right )} \] Input:
Integrate[(-2*E^x*x + (16 + 8*E^x)*Log[18 + 9*E^x] + (4 + 2*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]])/((160 + 80*x + 10*x^2 + E^x*(80 + 40*x + 5*x^ 2))*Log[18 + 9*E^x] + (80 + 20*x + E^x*(40 + 10*x))*Log[18 + 9*E^x]*Log[Lo g[18 + 9*E^x]] + (10 + 5*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]]^2),x]
Output:
(2*x)/(5*(4 + x + Log[Log[9*(2 + E^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.
| \(\displaystyle \int \frac {-2 e^x x+\left (8 e^x+16\right ) \log \left (9 e^x+18\right )+\left (2 e^x+4\right ) \log \left (9 e^x+18\right ) \log \left (\log \left (9 e^x+18\right )\right )}{\left (10 x^2+e^x \left (5 x^2+40 x+80\right )+80 x+160\right ) \log \left (9 e^x+18\right )+\left (5 e^x+10\right ) \log \left (9 e^x+18\right ) \log ^2\left (\log \left (9 e^x+18\right )\right )+\left (20 x+e^x (10 x+40)+80\right ) \log \left (9 e^x+18\right ) \log \left (\log \left (9 e^x+18\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (e^x+2\right ) \log \left (9 \left (e^x+2\right )\right ) \left (\log \left (\log \left (9 \left (e^x+2\right )\right )\right )+4\right )-2 e^x x}{5 \left (e^x+2\right ) \log \left (9 \left (e^x+2\right )\right ) \left (x+\log \left (\log \left (9 \left (e^x+2\right )\right )\right )+4\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int -\frac {2 \left (e^x x-\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )\right )}{\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \int \frac {e^x x-\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )}{\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {2}{5} \int \left (\frac {x-4 \log \left (9 \left (2+e^x\right )\right )-\log \left (9 \left (2+e^x\right )\right ) \log \left (\log \left (9 \left (2+e^x\right )\right )\right )}{\log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}-\frac {2 x}{\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2}{5} \left (\int \frac {1}{-x-\log \left (\log \left (9 \left (2+e^x\right )\right )\right )-4}dx+\int \frac {x}{\left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}dx+\int \frac {x}{\log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}dx-2 \int \frac {x}{\left (2+e^x\right ) \log \left (9 \left (2+e^x\right )\right ) \left (x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )+4\right )^2}dx\right )\) |
Input:
Int[(-2*E^x*x + (16 + 8*E^x)*Log[18 + 9*E^x] + (4 + 2*E^x)*Log[18 + 9*E^x] *Log[Log[18 + 9*E^x]])/((160 + 80*x + 10*x^2 + E^x*(80 + 40*x + 5*x^2))*Lo g[18 + 9*E^x] + (80 + 20*x + E^x*(40 + 10*x))*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]] + (10 + 5*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]]^2),x]
Output:
$Aborted
Time = 1.54 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
| method | result | size |
| risch | \(\frac {2 x}{5 \left (\ln \left (\ln \left (9 \,{\mathrm e}^{x}+18\right )\right )+4+x \right )}\) | \(17\) |
| parallelrisch | \(\frac {-32-8 \ln \left (\ln \left (9 \,{\mathrm e}^{x}+18\right )\right )}{20 \ln \left (\ln \left (9 \,{\mathrm e}^{x}+18\right )\right )+80+20 x}\) | \(28\) |
Input:
int(((2*exp(x)+4)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+(8*exp(x)+16)*ln(9*e xp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))^2 +((10*x+40)*exp(x)+20*x+80)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+((5*x^2+40 *x+80)*exp(x)+10*x^2+80*x+160)*ln(9*exp(x)+18)),x,method=_RETURNVERBOSE)
Output:
2/5*x/(ln(ln(9*exp(x)+18))+4+x)
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 \, x}{5 \, {\left (x + \log \left (\log \left (9 \, e^{x} + 18\right )\right ) + 4\right )}} \] Input:
integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+1 6)*log(9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*log(9*exp(x)+18)*log(log(9* exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(x) +18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algori thm="fricas")
Output:
2/5*x/(x + log(log(9*e^x + 18)) + 4)
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 x}{5 x + 5 \log {\left (\log {\left (9 e^{x} + 18 \right )} \right )} + 20} \] Input:
integrate(((2*exp(x)+4)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+(8*exp(x)+16)* ln(9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+ 18))**2+((10*x+40)*exp(x)+20*x+80)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+((5 *x**2+40*x+80)*exp(x)+10*x**2+80*x+160)*ln(9*exp(x)+18)),x)
Output:
2*x/(5*x + 5*log(log(9*exp(x) + 18)) + 20)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 \, x}{5 \, {\left (x + \log \left (2 \, \log \left (3\right ) + \log \left (e^{x} + 2\right )\right ) + 4\right )}} \] Input:
integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+1 6)*log(9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*log(9*exp(x)+18)*log(log(9* exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(x) +18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algori thm="maxima")
Output:
2/5*x/(x + log(2*log(3) + log(e^x + 2)) + 4)
Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2 \, x}{5 \, {\left (x + \log \left (\log \left (9 \, e^{x} + 18\right )\right ) + 4\right )}} \] Input:
integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+1 6)*log(9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*log(9*exp(x)+18)*log(log(9* exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(x) +18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algori thm="giac")
Output:
2/5*x/(x + log(log(9*e^x + 18)) + 4)
Time = 3.89 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {2\,x}{5\,\left (x+\ln \left (\ln \left (9\,{\mathrm {e}}^x+18\right )\right )+4\right )} \] Input:
int((log(9*exp(x) + 18)*(8*exp(x) + 16) - 2*x*exp(x) + log(9*exp(x) + 18)* log(log(9*exp(x) + 18))*(2*exp(x) + 4))/(log(9*exp(x) + 18)*(80*x + exp(x) *(40*x + 5*x^2 + 80) + 10*x^2 + 160) + log(9*exp(x) + 18)*log(log(9*exp(x) + 18))*(20*x + exp(x)*(10*x + 40) + 80) + log(9*exp(x) + 18)*log(log(9*ex p(x) + 18))^2*(5*exp(x) + 10)),x)
Output:
(2*x)/(5*(x + log(log(9*exp(x) + 18)) + 4))
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx=\frac {-2 \,\mathrm {log}\left (\mathrm {log}\left (9 e^{x}+18\right )\right )-8}{5 \,\mathrm {log}\left (\mathrm {log}\left (9 e^{x}+18\right )\right )+5 x +20} \] Input:
int(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+16)*log (9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)*log(9*exp(x)+18)*log(log(9*exp(x) +18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+ ((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x)
Output:
(2*( - log(log(9*e**x + 18)) - 4))/(5*(log(log(9*e**x + 18)) + x + 4))