Integrand size = 85, antiderivative size = 23 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\frac {1}{-5 e^{2-x}+\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )} \]
Time = 1.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\frac {e^x}{-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )} \]
Integrate[(-1 - 5*E^(2 - x) - Log[x*Log[17/7]])/(25*E^(4 - 2*x) - 10*E^(2 - x)*Log[4] + Log[4]^2 + (-10*E^(2 - x)*x + 2*x*Log[4])*Log[x*Log[17/7]] + x^2*Log[x*Log[17/7]]^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.
\(\displaystyle \int \frac {-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )-1}{x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )+25 e^{4-2 x}+\left (2 x \log (4)-10 e^{2-x} x\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )-10 e^{2-x} \log (4)+\log ^2(4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} \left (-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )-1\right )}{\left (-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )-e^x \log (4)+5 e^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^{x-2}}{5}-\frac {e^{2 x-2} \left (x \log \left (x \log \left (\frac {17}{7}\right )\right )+\log (4)\right )}{5 \left (-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )-e^x \log (4)+5 e^2\right )}-\frac {e^{2 x} \left (x \log \left (x \log \left (\frac {17}{7}\right )\right )+\log \left (x \log \left (\frac {17}{7}\right )\right )+1+\log (4)\right )}{\left (e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )+e^x \log (4)-5 e^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left ((1+\log (4)) \int \frac {e^{2 x}}{\left (-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )-e^x \log (4)+5 e^2\right )^2}dx\right )-\int \frac {e^{2 x} \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )+e^x \log (4)-5 e^2\right )^2}dx-\int \frac {e^{2 x} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )+e^x \log (4)-5 e^2\right )^2}dx+\frac {1}{5} \log (4) \int \frac {e^{2 x-2}}{e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )+e^x \log (4)-5 e^2}dx+\frac {1}{5} \int \frac {e^{2 x-2} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )+e^x \log (4)-5 e^2}dx-\frac {e^{x-2}}{5}\) |
Int[(-1 - 5*E^(2 - x) - Log[x*Log[17/7]])/(25*E^(4 - 2*x) - 10*E^(2 - x)*L og[4] + Log[4]^2 + (-10*E^(2 - x)*x + 2*x*Log[4])*Log[x*Log[17/7]] + x^2*L og[x*Log[17/7]]^2),x]
3.1.13.3.1 Defintions of rubi rules used
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {1}{\ln \left (x \ln \left (\frac {17}{7}\right )\right ) x -5 \,{\mathrm e}^{2-x}+2 \ln \left (2\right )}\) | \(23\) |
risch | \(\frac {1}{x \ln \left (x \left (\ln \left (17\right )-\ln \left (7\right )\right )\right )+2 \ln \left (2\right )-5 \,{\mathrm e}^{2-x}}\) | \(28\) |
int((-ln(x*ln(17/7))-5*exp(2-x)-1)/(x^2*ln(x*ln(17/7))^2+(-10*x*exp(2-x)+4 *x*ln(2))*ln(x*ln(17/7))+25*exp(2-x)^2-20*exp(2-x)*ln(2)+4*ln(2)^2),x,meth od=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\frac {1}{x \log \left (x \log \left (\frac {17}{7}\right )\right ) - 5 \, e^{\left (-x + 2\right )} + 2 \, \log \left (2\right )} \]
integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x* exp(2-x)+4*x*log(2))*log(x*log(17/7))+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*l og(2)^2),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=- \frac {1}{- x \log {\left (x \log {\left (\frac {17}{7} \right )} \right )} + 5 e^{2 - x} - 2 \log {\left (2 \right )}} \]
integrate((-ln(x*ln(17/7))-5*exp(2-x)-1)/(x**2*ln(x*ln(17/7))**2+(-10*x*ex p(2-x)+4*x*ln(2))*ln(x*ln(17/7))+25*exp(2-x)**2-20*exp(2-x)*ln(2)+4*ln(2)* *2),x)
Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\frac {e^{x}}{{\left (x \log \left (x\right ) + x \log \left (\log \left (17\right ) - \log \left (7\right )\right ) + 2 \, \log \left (2\right )\right )} e^{x} - 5 \, e^{2}} \]
integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x* exp(2-x)+4*x*log(2))*log(x*log(17/7))+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*l og(2)^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\frac {1}{{\left (x - 2\right )} \log \left ({\left (x - 2\right )} \log \left (\frac {17}{7}\right ) + 2 \, \log \left (\frac {17}{7}\right )\right ) - 5 \, e^{\left (-x + 2\right )} + 2 \, \log \left (2\right ) + 2 \, \log \left ({\left (x - 2\right )} \log \left (\frac {17}{7}\right ) + 2 \, \log \left (\frac {17}{7}\right )\right )} \]
integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x* exp(2-x)+4*x*log(2))*log(x*log(17/7))+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*l og(2)^2),x, algorithm=\
1/((x - 2)*log((x - 2)*log(17/7) + 2*log(17/7)) - 5*e^(-x + 2) + 2*log(2) + 2*log((x - 2)*log(17/7) + 2*log(17/7)))
Timed out. \[ \int \frac {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx=\int -\frac {5\,{\mathrm {e}}^{2-x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )+1}{25\,{\mathrm {e}}^{4-2\,x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )\,\left (4\,x\,\ln \left (2\right )-10\,x\,{\mathrm {e}}^{2-x}\right )+x^2\,{\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )}^2-20\,{\mathrm {e}}^{2-x}\,\ln \left (2\right )+4\,{\ln \left (2\right )}^2} \,d x \]
int(-(5*exp(2 - x) + log(x*log(17/7)) + 1)/(25*exp(4 - 2*x) + log(x*log(17 /7))*(4*x*log(2) - 10*x*exp(2 - x)) + x^2*log(x*log(17/7))^2 - 20*exp(2 - x)*log(2) + 4*log(2)^2),x)