Integrand size = 68, antiderivative size = 34 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=4+5 \log \left (\frac {x^2 \left (4-\left (-e^{e^x}-\frac {x}{2}\right ) x\right )}{4 \log (x)}\right ) \] Output:
4+5*ln(1/4*(4-x*(-1/2*x-exp(exp(x))))*x^2/ln(x))
Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=5 \left (2 \log (x)+\log \left (8+2 e^{e^x} x+x^2\right )-\log (\log (x))\right ) \] Input:
Integrate[(-40 - 5*x^2 + (80 + 20*x^2)*Log[x] + E^E^x*(-10*x + (30*x + 10* E^x*x^2)*Log[x]))/(2*E^E^x*x^2*Log[x] + (8*x + x^3)*Log[x]),x]
Output:
5*(2*Log[x] + Log[8 + 2*E^E^x*x + x^2] - Log[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.
\(\displaystyle \int \frac {-5 x^2+\left (20 x^2+80\right ) \log (x)+e^{e^x} \left (\left (10 e^x x^2+30 x\right ) \log (x)-10 x\right )-40}{\left (x^3+8 x\right ) \log (x)+2 e^{e^x} x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-5 x^2+\left (20 x^2+80\right ) \log (x)+e^{e^x} \left (\left (10 e^x x^2+30 x\right ) \log (x)-10 x\right )-40}{x \left (x^2+2 e^{e^x} x+8\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {10 e^{x+e^x} x}{x^2+2 e^{e^x} x+8}+\frac {5 \left (-x^2+4 x^2 \log (x)-2 e^{e^x} x+6 e^{e^x} x \log (x)+16 \log (x)-8\right )}{\left (x^2+2 e^{e^x} x+8\right ) x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -40 \int \frac {1}{x \left (x^2+2 e^{e^x} x+8\right )}dx+5 \int \frac {x}{x^2+2 e^{e^x} x+8}dx+10 \int \frac {e^{x+e^x} x}{x^2+2 e^{e^x} x+8}dx+15 \log (x)-5 \log (\log (x))\) |
Input:
Int[(-40 - 5*x^2 + (80 + 20*x^2)*Log[x] + E^E^x*(-10*x + (30*x + 10*E^x*x^ 2)*Log[x]))/(2*E^E^x*x^2*Log[x] + (8*x + x^3)*Log[x]),x]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(10 \ln \left (x \right )-5 \ln \left (\ln \left (x \right )\right )+5 \ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{x}}+x^{2}+8\right )\) | \(25\) |
risch | \(15 \ln \left (x \right )-5 \ln \left (\ln \left (x \right )\right )+5 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {x^{2}+8}{2 x}\right )\) | \(28\) |
Input:
int((((10*exp(x)*x^2+30*x)*ln(x)-10*x)*exp(exp(x))+(20*x^2+80)*ln(x)-5*x^2 -40)/(2*x^2*ln(x)*exp(exp(x))+(x^3+8*x)*ln(x)),x,method=_RETURNVERBOSE)
Output:
10*ln(x)-5*ln(ln(x))+5*ln(2*x*exp(exp(x))+x^2+8)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=15 \, \log \left (x\right ) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{x}\right ) - 5 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log( x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x))+(x^3+8*x)*log(x)),x, algorithm="fri cas")
Output:
15*log(x) + 5*log((x^2 + 2*x*e^(e^x) + 8)/x) - 5*log(log(x))
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=15 \log {\left (x \right )} + 5 \log {\left (e^{e^{x}} + \frac {x^{2} + 8}{2 x} \right )} - 5 \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate((((10*exp(x)*x**2+30*x)*ln(x)-10*x)*exp(exp(x))+(20*x**2+80)*ln( x)-5*x**2-40)/(2*x**2*ln(x)*exp(exp(x))+(x**3+8*x)*ln(x)),x)
Output:
15*log(x) + 5*log(exp(exp(x)) + (x**2 + 8)/(2*x)) - 5*log(log(x))
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=15 \, \log \left (x\right ) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{2 \, x}\right ) - 5 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log( x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x))+(x^3+8*x)*log(x)),x, algorithm="max ima")
Output:
15*log(x) + 5*log(1/2*(x^2 + 2*x*e^(e^x) + 8)/x) - 5*log(log(x))
Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=-5 \, x + 5 \, \log \left (x^{2} e^{x} + 2 \, x e^{\left (x + e^{x}\right )} + 8 \, e^{x}\right ) + 10 \, \log \left (x\right ) - 5 \, \log \left (\log \left (x\right )\right ) \] Input:
integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log( x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x))+(x^3+8*x)*log(x)),x, algorithm="gia c")
Output:
-5*x + 5*log(x^2*e^x + 2*x*e^(x + e^x) + 8*e^x) + 10*log(x) - 5*log(log(x) )
Time = 2.41 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=15\,\ln \left (x\right )-5\,\ln \left (\ln \left (x\right )\right )+5\,\ln \left (\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+x^2+8}{x}\right ) \] Input:
int(-(exp(exp(x))*(10*x - log(x)*(30*x + 10*x^2*exp(x))) + 5*x^2 - log(x)* (20*x^2 + 80) + 40)/(log(x)*(8*x + x^3) + 2*x^2*exp(exp(x))*log(x)),x)
Output:
15*log(x) - 5*log(log(x)) + 5*log((2*x*exp(exp(x)) + x^2 + 8)/x)
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx=-5 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+5 \,\mathrm {log}\left (2 e^{e^{x}} x +x^{2}+8\right )+10 \,\mathrm {log}\left (x \right ) \] Input:
int((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log(x)-5*x ^2-40)/(2*x^2*log(x)*exp(exp(x))+(x^3+8*x)*log(x)),x)
Output:
5*( - log(log(x)) + log(2*e**(e**x)*x + x**2 + 8) + 2*log(x))