Integrand size = 105, antiderivative size = 29 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=\frac {2 e^{e^8} x}{\left (5-e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \]
Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=-\frac {2 e^{e^8} x}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \]
Integrate[(E^E^8*(20 - 4*E^x) + E^E^8*(-20 + 4*E^x)*Log[x] + E^E^8*(10 + E ^x*(-2 + 2*x))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2])/((25 - 10*E^x + E^(2*x))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2),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.
\(\displaystyle \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (e^x (2 x-2)+10\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )+e^{e^8} \left (4 e^x-20\right ) \log (x)}{\left (-10 e^x+e^{2 x}+25\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (e^x (2 x-2)+10\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )+e^{e^8} \left (4 e^x-20\right ) \log (x)}{\left (5-e^x\right )^2 \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {10 e^{e^8} x}{\left (e^x-5\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}+\frac {2 e^{e^8} \left (x \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right ) \log (x)-\log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right ) \log (x)+2 \log (x)-2\right )}{\left (e^x-5\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 e^{e^8} \int \frac {1}{\left (-5+e^x\right ) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx-4 e^{e^8} \int \frac {1}{\left (-5+e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx-2 e^{e^8} \int \frac {1}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx+10 e^{e^8} \int \frac {x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx+2 e^{e^8} \int \frac {x}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}dx\) |
Int[(E^E^8*(20 - 4*E^x) + E^E^8*(-20 + 4*E^x)*Log[x] + E^E^8*(10 + E^x*(-2 + 2*x))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2])/((25 - 10*E^x + E^(2*x ))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2),x]
3.14.71.3.1 Defintions of rubi rules used
Time = 157.73 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{{\mathrm e}^{8}} x}{\ln \left (\ln \left (\frac {x}{\ln \left (x \right )}\right )^{2}\right ) \left ({\mathrm e}^{x}-5\right )}\) | \(25\) |
risch | \(\text {Expression too large to display}\) | \(712\) |
int((((-2+2*x)*exp(x)+10)*exp(exp(8))*ln(x)*ln(x/ln(x))*ln(ln(x/ln(x))^2)+ (4*exp(x)-20)*exp(exp(8))*ln(x)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x)^2-10*e xp(x)+25)/ln(x)/ln(x/ln(x))/ln(ln(x/ln(x))^2)^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=-\frac {2 \, x e^{\left (e^{8}\right )}}{{\left (e^{x} - 5\right )} \log \left (\log \left (\frac {x}{\log \left (x\right )}\right )^{2}\right )} \]
integrate((((-2+2*x)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x /log(x))^2)+(4*exp(x)-20)*exp(exp(8))*log(x)+(-4*exp(x)+20)*exp(exp(8)))/( exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, algo rithm=\
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=- \frac {2 x e^{e^{8}}}{e^{x} \log {\left (\log {\left (\frac {x}{\log {\left (x \right )}} \right )}^{2} \right )} - 5 \log {\left (\log {\left (\frac {x}{\log {\left (x \right )}} \right )}^{2} \right )}} \]
integrate((((-2+2*x)*exp(x)+10)*exp(exp(8))*ln(x)*ln(x/ln(x))*ln(ln(x/ln(x ))**2)+(4*exp(x)-20)*exp(exp(8))*ln(x)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x) **2-10*exp(x)+25)/ln(x)/ln(x/ln(x))/ln(ln(x/ln(x))**2)**2,x)
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=-\frac {x e^{\left (e^{8}\right )}}{{\left (e^{x} - 5\right )} \log \left (\log \left (x\right ) - \log \left (\log \left (x\right )\right )\right )} \]
integrate((((-2+2*x)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x /log(x))^2)+(4*exp(x)-20)*exp(exp(8))*log(x)+(-4*exp(x)+20)*exp(exp(8)))/( exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, algo rithm=\
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 49.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=-\frac {2 \, x e^{\left (e^{8}\right )}}{e^{x} \log \left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right ) - 5 \, \log \left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right )} \]
integrate((((-2+2*x)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x /log(x))^2)+(4*exp(x)-20)*exp(exp(8))*log(x)+(-4*exp(x)+20)*exp(exp(8)))/( exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, algo rithm=\
-2*x*e^(e^8)/(e^x*log(log(x)^2 - 2*log(x)*log(log(x)) + log(log(x))^2) - 5 *log(log(x)^2 - 2*log(x)*log(log(x)) + log(log(x))^2))
Time = 14.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx=-\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^8}}{\ln \left ({\ln \left (\frac {x}{\ln \left (x\right )}\right )}^2\right )\,\left ({\mathrm {e}}^x-5\right )} \]