Integrand size = 60, antiderivative size = 25 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log \left (\frac {\left (x^2+\frac {2 x \left (-4+e^x+\log (x)\right )}{\log (x)}\right )^2}{x}\right ) \] Output:
ln((x^2+2*(exp(x)+ln(x)-4)*x/ln(x))^2/x)
Time = 0.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log (x)-2 \log (\log (x))+2 \log \left (8-2 e^x-2 \log (x)-x \log (x)\right ) \] Input:
Integrate[(16 - 4*E^x + (-8 + E^x*(2 + 4*x))*Log[x] + (2 + 3*x)*Log[x]^2)/ ((-8*x + 2*E^x*x)*Log[x] + (2*x + x^2)*Log[x]^2),x]
Output:
Log[x] - 2*Log[Log[x]] + 2*Log[8 - 2*E^x - 2*Log[x] - x*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.
\(\displaystyle \int \frac {-4 e^x+(3 x+2) \log ^2(x)+\left (e^x (4 x+2)-8\right ) \log (x)+16}{\left (x^2+2 x\right ) \log ^2(x)+\left (2 e^x x-8 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 e^x-\left ((3 x+2) \log ^2(x)\right )-\left (e^x (4 x+2)-8\right ) \log (x)-16}{x \log (x) \left (-2 e^x-x \log (x)-2 \log (x)+8\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x \log (x)+\log (x)-2}{x \log (x)}-\frac {2 \left (x^2 \log (x)-9 x+x \log (x)-2\right )}{x \left (2 e^x+x \log (x)+2 \log (x)-8\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 18 \int \frac {1}{x \log (x)+2 \log (x)+2 e^x-8}dx+4 \int \frac {1}{x \left (x \log (x)+2 \log (x)+2 e^x-8\right )}dx-2 \int \frac {\log (x)}{x \log (x)+2 \log (x)+2 e^x-8}dx-2 \int \frac {x \log (x)}{x \log (x)+2 \log (x)+2 e^x-8}dx+2 x+\log (x)-2 \log (\log (x))\) |
Input:
Int[(16 - 4*E^x + (-8 + E^x*(2 + 4*x))*Log[x] + (2 + 3*x)*Log[x]^2)/((-8*x + 2*E^x*x)*Log[x] + (2*x + x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.70 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\ln \left (x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \ln \left (x \right )+2 \ln \left (x \right )+2 \,{\mathrm e}^{x}-8\right )\) | \(26\) |
parallelrisch | \(\ln \left (x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \ln \left (x \right )+2 \ln \left (x \right )+2 \,{\mathrm e}^{x}-8\right )\) | \(26\) |
risch | \(2 \ln \left (2+x \right )+\ln \left (x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (\ln \left (x \right )+\frac {2 \,{\mathrm e}^{x}-8}{2+x}\right )\) | \(32\) |
Input:
int(((2+3*x)*ln(x)^2+((4*x+2)*exp(x)-8)*ln(x)-4*exp(x)+16)/((x^2+2*x)*ln(x )^2+(2*exp(x)*x-8*x)*ln(x)),x,method=_RETURNVERBOSE)
Output:
ln(x)-2*ln(ln(x))+2*ln(x*ln(x)+2*ln(x)+2*exp(x)-8)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (x + 2\right ) + \log \left (x\right ) + 2 \, \log \left (\frac {{\left (x + 2\right )} \log \left (x\right ) + 2 \, e^{x} - 8}{x + 2}\right ) - 2 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate(((2+3*x)*log(x)^2+((2+4*x)*exp(x)-8)*log(x)-4*exp(x)+16)/((x^2+2 *x)*log(x)^2+(2*exp(x)*x-8*x)*log(x)),x, algorithm="fricas")
Output:
2*log(x + 2) + log(x) + 2*log(((x + 2)*log(x) + 2*e^x - 8)/(x + 2)) - 2*lo g(log(x))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log {\left (x \right )} + 2 \log {\left (\frac {x \log {\left (x \right )}}{2} + e^{x} + \log {\left (x \right )} - 4 \right )} - 2 \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate(((2+3*x)*ln(x)**2+((2+4*x)*exp(x)-8)*ln(x)-4*exp(x)+16)/((x**2+2 *x)*ln(x)**2+(2*exp(x)*x-8*x)*ln(x)),x)
Output:
log(x) + 2*log(x*log(x)/2 + exp(x) + log(x) - 4) - 2*log(log(x))
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (\frac {1}{2} \, {\left (x + 2\right )} \log \left (x\right ) + e^{x} - 4\right ) + \log \left (x\right ) - 2 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate(((2+3*x)*log(x)^2+((2+4*x)*exp(x)-8)*log(x)-4*exp(x)+16)/((x^2+2 *x)*log(x)^2+(2*exp(x)*x-8*x)*log(x)),x, algorithm="maxima")
Output:
2*log(1/2*(x + 2)*log(x) + e^x - 4) + log(x) - 2*log(log(x))
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (x \log \left (x\right ) + 2 \, e^{x} + 2 \, \log \left (x\right ) - 8\right ) + \log \left (x\right ) - 2 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate(((2+3*x)*log(x)^2+((2+4*x)*exp(x)-8)*log(x)-4*exp(x)+16)/((x^2+2 *x)*log(x)^2+(2*exp(x)*x-8*x)*log(x)),x, algorithm="giac")
Output:
2*log(x*log(x) + 2*e^x + 2*log(x) - 8) + log(x) - 2*log(log(x))
Time = 0.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2\,\ln \left (2\,{\mathrm {e}}^x+2\,\ln \left (x\right )+x\,\ln \left (x\right )-8\right )-2\,\ln \left (\ln \left (x\right )\right )+\ln \left (x\right ) \] Input:
int(-(log(x)*(exp(x)*(4*x + 2) - 8) - 4*exp(x) + log(x)^2*(3*x + 2) + 16)/ (log(x)*(8*x - 2*x*exp(x)) - log(x)^2*(2*x + x^2)),x)
Output:
2*log(2*exp(x) + 2*log(x) + x*log(x) - 8) - 2*log(log(x)) + log(x)
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=-2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+2 \,\mathrm {log}\left (2 e^{x}+\mathrm {log}\left (x \right ) x +2 \,\mathrm {log}\left (x \right )-8\right )+\mathrm {log}\left (x \right ) \] Input:
int(((2+3*x)*log(x)^2+((2+4*x)*exp(x)-8)*log(x)-4*exp(x)+16)/((x^2+2*x)*lo g(x)^2+(2*exp(x)*x-8*x)*log(x)),x)
Output:
- 2*log(log(x)) + 2*log(2*e**x + log(x)*x + 2*log(x) - 8) + log(x)