Integrand size = 74, antiderivative size = 20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (1+e^x+x-\frac {8 \log (2) (-5+\log (x))}{\log (x)}\right )\right ) \]
Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )\right ) \]
Integrate[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + E^x*x + x^2 - 8*x*Log[2])*Log[x]^2)*Log[(40*Log[2] + (1 + E^x + x - 8*Log[ 2])*Log[x])/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.
\(\displaystyle \int \frac {\left (e^x x+x\right ) \log ^2(x)-40 \log (2)}{\left (\left (x^2+e^x x+x-8 x \log (2)\right ) \log ^2(x)+40 x \log (2) \log (x)\right ) \log \left (\frac {\left (x+e^x+1-8 \log (2)\right ) \log (x)+40 \log (2)}{\log (x)}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (e^x x+x\right ) \log ^2(x)-40 \log (2)}{x \log (x) \left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (\frac {\left (x+e^x+1-8 \log (2)\right ) \log (x)+40 \log (2)}{\log (x)}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-x^2 \log ^2(x)+8 x \log (2) \log ^2(x)-40 x \log (2) \log (x)-40 \log (2)}{x \log (x) \left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}+1-8 \log (2)\right )}+\frac {1}{\log \left (x+e^x+\frac {40 \log (2)}{\log (x)}+1-8 \log (2)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+40 \log (2) \int \frac {1}{\left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+40 \log (2) \int \frac {1}{x \log (x) \left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+\int \frac {x \log (x)}{\left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+8 \log (2) \int \frac {\log (x)}{\left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx\) |
Int[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + E^x*x + x^2 - 8*x*Log[2])*Log[x]^2)*Log[(40*Log[2] + (1 + E^x + x - 8*Log[2])*Lo g[x])/Log[x]]),x]
3.1.61.3.1 Defintions of rubi rules used
Time = 33.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25
method | result | size |
parallelrisch | \(\ln \left (\ln \left (\frac {\left ({\mathrm e}^{x}-8 \ln \left (2\right )+x +1\right ) \ln \left (x \right )+40 \ln \left (2\right )}{\ln \left (x \right )}\right )\right )\) | \(25\) |
risch | \(\ln \left (\ln \left (\left (\ln \left (x \right )-5\right ) \ln \left (2\right )-\frac {x \ln \left (x \right )}{8}-\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}-\frac {\ln \left (x \right )}{8}\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}+2 \pi {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{3}+6 i \ln \left (2\right )-2 i \ln \left (\ln \left (x \right )\right )-2 \pi \right )}{2}\right )\) | \(297\) |
int(((exp(x)*x+x)*ln(x)^2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x^2+x)*ln(x)^2+40 *x*ln(2)*ln(x))/ln(((exp(x)-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)),x,method=_ RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (\frac {{\left (x + e^{x} - 8 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) + 40 \, \log \left (2\right )}{\log \left (x\right )}\right )\right ) \]
integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l og(x)),x, algorithm=\
Time = 2.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log {\left (\log {\left (\frac {\left (x + e^{x} - 8 \log {\left (2 \right )} + 1\right ) \log {\left (x \right )} + 40 \log {\left (2 \right )}}{\log {\left (x \right )}} \right )} \right )} \]
integrate(((exp(x)*x+x)*ln(x)**2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x**2+x)*ln (x)**2+40*x*ln(2)*ln(x))/ln(((exp(x)-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)),x )
Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left ({\left (x + e^{x} - 8 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) + 40 \, \log \left (2\right )\right ) - \log \left (\log \left (x\right )\right )\right ) \]
integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l og(x)),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (-\log \left (x \log \left (x\right ) + e^{x} \log \left (x\right ) - 8 \, \log \left (2\right ) \log \left (x\right ) + 40 \, \log \left (2\right ) + \log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )\right ) \]
integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l og(x)),x, algorithm=\
Time = 12.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\ln \left (\ln \left (\frac {40\,\ln \left (2\right )+\ln \left (x\right )\,\left (x-8\,\ln \left (2\right )+{\mathrm {e}}^x+1\right )}{\ln \left (x\right )}\right )\right ) \]