Integrand size = 89, antiderivative size = 28 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log \left (2 e^2-2 x-\log (5)+\frac {e^2-\log (x)}{x}\right ) \end {dmath*}
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log \left (-\frac {-e^2 (1+2 x)+x (2 x+\log (5))+\log (x)}{x}\right ) \end {dmath*}
Integrate[(1 + E^2 + 2*x^2 - Log[x] + (E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x])*Log[(-2*x^2 + E^2*(1 + 2*x) - x*Log[5] - Log[x])/x])/(E^2*(-1 - 2 *x) + 2*x^2 + x*Log[5] + Log[x]),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.
\(\displaystyle \int \frac {2 x^2+\left (2 x^2+e^2 (-2 x-1)+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (2 x+1)-x \log (5)-\log (x)}{x}\right )-\log (x)+e^2+1}{2 x^2+e^2 (-2 x-1)+x \log (5)+\log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x^2+\log (x)-e^2-1}{-2 x^2+2 e^2 x \left (1-\frac {\log (5)}{2 e^2}\right )-\log (x)+e^2}+\log \left (-2 x+\frac {e^2 (2 x+1)}{x}-\frac {\log (x)}{x}-\log (5)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (2 e^2-\log (5)\right ) \int \frac {x}{-2 x^2+2 e^2 \left (1-\frac {\log (5)}{2 e^2}\right ) x-\log (x)+e^2}dx+\int \frac {1}{2 x^2-2 e^2 \left (1-\frac {\log (5)}{2 e^2}\right ) x+\log (x)-e^2}dx+4 \int \frac {x^2}{2 x^2-2 e^2 \left (1-\frac {\log (5)}{2 e^2}\right ) x+\log (x)-e^2}dx+\int \log \left (-2 x-\log (5)+\frac {e^2 (2 x+1)}{x}-\frac {\log (x)}{x}\right )dx-x\) |
Int[(1 + E^2 + 2*x^2 - Log[x] + (E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x ])*Log[(-2*x^2 + E^2*(1 + 2*x) - x*Log[5] - Log[x])/x])/(E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x]),x]
3.6.98.3.1 Defintions of rubi rules used
Time = 2.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
default | \(\ln \left (\frac {2 \,{\mathrm e}^{2} x -x \ln \left (5\right )-2 x^{2}+{\mathrm e}^{2}-\ln \left (x \right )}{x}\right ) x\) | \(30\) |
norman | \(x \ln \left (\frac {-\ln \left (x \right )-x \ln \left (5\right )+\left (1+2 x \right ) {\mathrm e}^{2}-2 x^{2}}{x}\right )\) | \(31\) |
parallelrisch | \(x \ln \left (\frac {-\ln \left (x \right )-x \ln \left (5\right )+\left (1+2 x \right ) {\mathrm e}^{2}-2 x^{2}}{x}\right )\) | \(31\) |
risch | \(\frac {i \pi x \,\operatorname {csgn}\left (i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )}{x}\right )}^{2}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+x \ln \left (2\right )-x \ln \left (x \right )+\ln \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right ) x -\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \left (5\right )}{2}-x^{2}-\frac {\ln \left (x \right )}{2}\right )}{x}\right )}^{3}}{2}\) | \(237\) |
int(((ln(x)+x*ln(5)+(-1-2*x)*exp(2)+2*x^2)*ln((-ln(x)-x*ln(5)+(1+2*x)*exp( 2)-2*x^2)/x)-ln(x)+exp(2)+2*x^2+1)/(ln(x)+x*ln(5)+(-1-2*x)*exp(2)+2*x^2),x ,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log \left (-\frac {2 \, x^{2} - {\left (2 \, x + 1\right )} e^{2} + x \log \left (5\right ) + \log \left (x\right )}{x}\right ) \end {dmath*}
integrate(((log(x)+x*log(5)+(-1-2*x)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+( 1+2*x)*exp(2)-2*x^2)/x)-log(x)+exp(2)+2*x^2+1)/(log(x)+x*log(5)+(-1-2*x)*e xp(2)+2*x^2),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log {\left (\frac {- 2 x^{2} - x \log {\left (5 \right )} + \left (2 x + 1\right ) e^{2} - \log {\left (x \right )}}{x} \right )} \end {dmath*}
integrate(((ln(x)+x*ln(5)+(-1-2*x)*exp(2)+2*x**2)*ln((-ln(x)-x*ln(5)+(1+2* x)*exp(2)-2*x**2)/x)-ln(x)+exp(2)+2*x**2+1)/(ln(x)+x*ln(5)+(-1-2*x)*exp(2) +2*x**2),x)
Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log \left (-2 \, x^{2} + x {\left (2 \, e^{2} - \log \left (5\right )\right )} + e^{2} - \log \left (x\right )\right ) - x \log \left (x\right ) \end {dmath*}
integrate(((log(x)+x*log(5)+(-1-2*x)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+( 1+2*x)*exp(2)-2*x^2)/x)-log(x)+exp(2)+2*x^2+1)/(log(x)+x*log(5)+(-1-2*x)*e xp(2)+2*x^2),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x \log \left (-2 \, x^{2} + 2 \, x e^{2} - x \log \left (5\right ) + e^{2} - \log \left (x\right )\right ) - x \log \left (x\right ) \end {dmath*}
integrate(((log(x)+x*log(5)+(-1-2*x)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+( 1+2*x)*exp(2)-2*x^2)/x)-log(x)+exp(2)+2*x^2+1)/(log(x)+x*log(5)+(-1-2*x)*e xp(2)+2*x^2),x, algorithm=\
Time = 15.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {1+e^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx=x\,\ln \left (-\frac {\ln \left (x\right )+x\,\ln \left (5\right )+2\,x^2-{\mathrm {e}}^2\,\left (2\,x+1\right )}{x}\right ) \end {dmath*}