Integrand size = 103, antiderivative size = 17 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=5 x+\frac {x}{e^2+\log (1+\log (x))} \]
Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=5 x+\frac {x}{e^2+\log (1+\log (x))} \]
Integrate[(-1 + E^2 + 5*E^4 + (E^2 + 5*E^4)*Log[x] + (1 + 10*E^2 + (1 + 10 *E^2)*Log[x])*Log[1 + Log[x]] + (5 + 5*Log[x])*Log[1 + Log[x]]^2)/(E^4 + E ^4*Log[x] + (2*E^2 + 2*E^2*Log[x])*Log[1 + Log[x]] + (1 + Log[x])*Log[1 + Log[x]]^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.
\(\displaystyle \int \frac {(5 \log (x)+5) \log ^2(\log (x)+1)+\left (\left (1+10 e^2\right ) \log (x)+10 e^2+1\right ) \log (\log (x)+1)+\left (e^2+5 e^4\right ) \log (x)+5 e^4+e^2-1}{(\log (x)+1) \log ^2(\log (x)+1)+\left (2 e^2 \log (x)+2 e^2\right ) \log (\log (x)+1)+e^4 \log (x)+e^4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(5 \log (x)+5) \log ^2(\log (x)+1)+\left (\left (1+10 e^2\right ) \log (x)+10 e^2+1\right ) \log (\log (x)+1)+\left (e^2+5 e^4\right ) \log (x)+e^2 \left (1+5 e^2\right )-1}{(\log (x)+1) \left (\log (\log (x)+1)+e^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{\log (\log (x)+1)+e^2}-\frac {1}{\left (\log (\log (x)+1)+e^2\right )^2 (\log (x)+1)}+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {1}{(\log (x)+1) \left (\log (\log (x)+1)+e^2\right )^2}dx+\int \frac {1}{\log (\log (x)+1)+e^2}dx+5 x\) |
Int[(-1 + E^2 + 5*E^4 + (E^2 + 5*E^4)*Log[x] + (1 + 10*E^2 + (1 + 10*E^2)* Log[x])*Log[1 + Log[x]] + (5 + 5*Log[x])*Log[1 + Log[x]]^2)/(E^4 + E^4*Log [x] + (2*E^2 + 2*E^2*Log[x])*Log[1 + Log[x]] + (1 + Log[x])*Log[1 + Log[x] ]^2),x]
3.20.49.3.1 Defintions of rubi rules used
Time = 0.67 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x}{{\mathrm e}^{2}+\ln \left (\ln \left (x \right )+1\right )}+5 x\) | \(17\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{2} x +5 x \ln \left (\ln \left (x \right )+1\right )+x}{{\mathrm e}^{2}+\ln \left (\ln \left (x \right )+1\right )}\) | \(27\) |
default | \(\frac {\left (5 \,{\mathrm e}^{2}+1\right ) x +5 x \ln \left (\ln \left (x \right )+1\right )}{{\mathrm e}^{2}+\ln \left (\ln \left (x \right )+1\right )}\) | \(29\) |
norman | \(\frac {\left (5 \,{\mathrm e}^{2}+1\right ) x +5 x \ln \left (\ln \left (x \right )+1\right )}{{\mathrm e}^{2}+\ln \left (\ln \left (x \right )+1\right )}\) | \(29\) |
int(((5*ln(x)+5)*ln(ln(x)+1)^2+((10*exp(2)+1)*ln(x)+10*exp(2)+1)*ln(ln(x)+ 1)+(5*exp(2)^2+exp(2))*ln(x)+5*exp(2)^2+exp(2)-1)/((ln(x)+1)*ln(ln(x)+1)^2 +(2*exp(2)*ln(x)+2*exp(2))*ln(ln(x)+1)+exp(2)^2*ln(x)+exp(2)^2),x,method=_ RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=\frac {5 \, x e^{2} + 5 \, x \log \left (\log \left (x\right ) + 1\right ) + x}{e^{2} + \log \left (\log \left (x\right ) + 1\right )} \]
integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1) *log(log(x)+1)+(5*exp(2)^2+exp(2))*log(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1) *log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x)+ exp(2)^2),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=5 x + \frac {x}{\log {\left (\log {\left (x \right )} + 1 \right )} + e^{2}} \]
integrate(((5*ln(x)+5)*ln(ln(x)+1)**2+((10*exp(2)+1)*ln(x)+10*exp(2)+1)*ln (ln(x)+1)+(5*exp(2)**2+exp(2))*ln(x)+5*exp(2)**2+exp(2)-1)/((ln(x)+1)*ln(l n(x)+1)**2+(2*exp(2)*ln(x)+2*exp(2))*ln(ln(x)+1)+exp(2)**2*ln(x)+exp(2)**2 ),x)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=\frac {x {\left (5 \, e^{2} + 1\right )} + 5 \, x \log \left (\log \left (x\right ) + 1\right )}{e^{2} + \log \left (\log \left (x\right ) + 1\right )} \]
integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1) *log(log(x)+1)+(5*exp(2)^2+exp(2))*log(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1) *log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x)+ exp(2)^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (16) = 32\).
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.71 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=\frac {5 \, x e^{2}}{e^{2} + \log \left (\log \left (x\right ) + 1\right )} + \frac {5 \, x \log \left (\log \left (x\right ) + 1\right )}{e^{2} + \log \left (\log \left (x\right ) + 1\right )} + \frac {x}{e^{2} + \log \left (\log \left (x\right ) + 1\right )} \]
integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1) *log(log(x)+1)+(5*exp(2)^2+exp(2))*log(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1) *log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x)+ exp(2)^2),x, algorithm=\
5*x*e^2/(e^2 + log(log(x) + 1)) + 5*x*log(log(x) + 1)/(e^2 + log(log(x) + 1)) + x/(e^2 + log(log(x) + 1))
Time = 9.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx=\frac {x\,\left (5\,{\mathrm {e}}^2+5\,\ln \left (\ln \left (x\right )+1\right )+1\right )}{{\mathrm {e}}^2+\ln \left (\ln \left (x\right )+1\right )} \]