Integrand size = 68, antiderivative size = 26 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=-1+\frac {x \log (\log (\log (5)))}{5-e^{(7-x) \log ^2(x)}} \] Output:
x*ln(ln(ln(5)))/(5-exp((-x+7)*ln(x)^2))-1
Time = 1.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=\frac {1}{5} \left (1+\frac {1}{-1+5 e^{(-7+x) \log ^2(x)}}\right ) x \log (\log (\log (5))) \] Input:
Integrate[((5 + E^((7 - x)*Log[x]^2)*(-1 + (14 - 2*x)*Log[x] - x*Log[x]^2) )*Log[Log[Log[5]]])/(25 - 10*E^((7 - x)*Log[x]^2) + E^(2*(7 - x)*Log[x]^2) ),x]
Output:
((1 + (-1 + 5*E^((-7 + x)*Log[x]^2))^(-1))*x*Log[Log[Log[5]]])/5
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.
\(\displaystyle \int \frac {\log (\log (\log (5))) \left (e^{(7-x) \log ^2(x)} \left (-x \log ^2(x)+(14-2 x) \log (x)-1\right )+5\right )}{-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}+25} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (\log (\log (5))) \int \frac {5-e^{(7-x) \log ^2(x)} \left (x \log ^2(x)-2 (7-x) \log (x)+1\right )}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \log (\log (\log (5))) \int \frac {e^{2 (x-7) \log ^2(x)} \left (5-e^{(7-x) \log ^2(x)} \left (x \log ^2(x)-2 (7-x) \log (x)+1\right )\right )}{\left (1-5 e^{(x-7) \log ^2(x)}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \log (\log (\log (5))) \int \left (-\frac {5 e^{2 (x-7) \log ^2(x)} \log (x) (\log (x) x+2 x-14)}{\left (-1+5 e^{(x-7) \log ^2(x)}\right )^2}-e^{(x-7) \log ^2(x)} \left (x \log ^2(x)+2 x \log (x)-14 \log (x)+1\right )+\frac {5 e^{2 (x-7) \log ^2(x)} \left (x \log ^2(x)+2 x \log (x)-14 \log (x)+1\right )}{-1+5 e^{(x-7) \log ^2(x)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (\log (\log (5))) \left (\int e^{(x-7) \log ^2(x)}dx+\frac {1}{5} \int \frac {1}{-1+5 e^{(x-7) \log ^2(x)}}dx+70 \int \frac {e^{2 (x-7) \log ^2(x)} \log (x)}{\left (-1+5 e^{(x-7) \log ^2(x)}\right )^2}dx-70 \int \frac {e^{2 (x-7) \log ^2(x)} \log (x)}{-1+5 e^{(x-7) \log ^2(x)}}dx-10 \int \frac {e^{2 (x-7) \log ^2(x)} x \log (x)}{\left (-1+5 e^{(x-7) \log ^2(x)}\right )^2}dx+10 \int \frac {e^{2 (x-7) \log ^2(x)} x \log (x)}{-1+5 e^{(x-7) \log ^2(x)}}dx-5 \int \frac {e^{2 (x-7) \log ^2(x)} x \log ^2(x)}{\left (-1+5 e^{(x-7) \log ^2(x)}\right )^2}dx+5 \int \frac {e^{2 (x-7) \log ^2(x)} x \log ^2(x)}{-1+5 e^{(x-7) \log ^2(x)}}dx+\frac {x}{5}-\frac {e^{-\left ((7-x) \log ^2(x)\right )} \left (-x \log ^2(x)-2 x \log (x)+14 \log (x)\right )}{\frac {2 (7-x) \log (x)}{x}-\log ^2(x)}\right )\) |
Input:
Int[((5 + E^((7 - x)*Log[x]^2)*(-1 + (14 - 2*x)*Log[x] - x*Log[x]^2))*Log[ Log[Log[5]]])/(25 - 10*E^((7 - x)*Log[x]^2) + E^(2*(7 - x)*Log[x]^2)),x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {\ln \left (\ln \left (\ln \left (5\right )\right )\right ) x}{{\mathrm e}^{-\left (-7+x \right ) \ln \left (x \right )^{2}}-5}\) | \(22\) |
norman | \(-\frac {\ln \left (\ln \left (\ln \left (5\right )\right )\right ) x}{{\mathrm e}^{\left (-x +7\right ) \ln \left (x \right )^{2}}-5}\) | \(23\) |
parallelrisch | \(-\frac {\ln \left (\ln \left (\ln \left (5\right )\right )\right ) x}{{\mathrm e}^{\left (-x +7\right ) \ln \left (x \right )^{2}}-5}\) | \(23\) |
Input:
int(((-x*ln(x)^2+(-2*x+14)*ln(x)-1)*exp((-x+7)*ln(x)^2)+5)*ln(ln(ln(5)))/( exp((-x+7)*ln(x)^2)^2-10*exp((-x+7)*ln(x)^2)+25),x,method=_RETURNVERBOSE)
Output:
-ln(ln(ln(5)))*x/(exp(-(-7+x)*ln(x)^2)-5)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=-\frac {x \log \left (\log \left (\log \left (5\right )\right )\right )}{e^{\left (-{\left (x - 7\right )} \log \left (x\right )^{2}\right )} - 5} \] Input:
integrate(((-x*log(x)^2+(-2*x+14)*log(x)-1)*exp((-x+7)*log(x)^2)+5)*log(lo g(log(5)))/(exp((-x+7)*log(x)^2)^2-10*exp((-x+7)*log(x)^2)+25),x, algorith m="fricas")
Output:
-x*log(log(log(5)))/(e^(-(x - 7)*log(x)^2) - 5)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=- \frac {x \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )}}{e^{\left (7 - x\right ) \log {\left (x \right )}^{2}} - 5} \] Input:
integrate(((-x*ln(x)**2+(-2*x+14)*ln(x)-1)*exp((-x+7)*ln(x)**2)+5)*ln(ln(l n(5)))/(exp((-x+7)*ln(x)**2)**2-10*exp((-x+7)*ln(x)**2)+25),x)
Output:
-x*log(log(log(5)))/(exp((7 - x)*log(x)**2) - 5)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=\frac {x e^{\left (x \log \left (x\right )^{2}\right )} \log \left (\log \left (\log \left (5\right )\right )\right )}{5 \, e^{\left (x \log \left (x\right )^{2}\right )} - e^{\left (7 \, \log \left (x\right )^{2}\right )}} \] Input:
integrate(((-x*log(x)^2+(-2*x+14)*log(x)-1)*exp((-x+7)*log(x)^2)+5)*log(lo g(log(5)))/(exp((-x+7)*log(x)^2)^2-10*exp((-x+7)*log(x)^2)+25),x, algorith m="maxima")
Output:
x*e^(x*log(x)^2)*log(log(log(5)))/(5*e^(x*log(x)^2) - e^(7*log(x)^2))
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=-\frac {x \log \left (\log \left (\log \left (5\right )\right )\right )}{e^{\left (-x \log \left (x\right )^{2} + 7 \, \log \left (x\right )^{2}\right )} - 5} \] Input:
integrate(((-x*log(x)^2+(-2*x+14)*log(x)-1)*exp((-x+7)*log(x)^2)+5)*log(lo g(log(5)))/(exp((-x+7)*log(x)^2)^2-10*exp((-x+7)*log(x)^2)+25),x, algorith m="giac")
Output:
-x*log(log(log(5)))/(e^(-x*log(x)^2 + 7*log(x)^2) - 5)
Time = 3.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=-\frac {x\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )}{{\mathrm {e}}^{7\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-x\,{\ln \left (x\right )}^2}-5} \] Input:
int(-(log(log(log(5)))*(exp(-log(x)^2*(x - 7))*(x*log(x)^2 + log(x)*(2*x - 14) + 1) - 5))/(exp(-2*log(x)^2*(x - 7)) - 10*exp(-log(x)^2*(x - 7)) + 25 ),x)
Output:
-(x*log(log(log(5))))/(exp(7*log(x)^2)*exp(-x*log(x)^2) - 5)
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {\left (5+e^{(7-x) \log ^2(x)} \left (-1+(14-2 x) \log (x)-x \log ^2(x)\right )\right ) \log (\log (\log (5)))}{25-10 e^{(7-x) \log ^2(x)}+e^{2 (7-x) \log ^2(x)}} \, dx=-\frac {e^{\mathrm {log}\left (x \right )^{2} x} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (5\right )\right )\right ) x}{e^{7 \mathrm {log}\left (x \right )^{2}}-5 e^{\mathrm {log}\left (x \right )^{2} x}} \] Input:
int(((-x*log(x)^2+(-2*x+14)*log(x)-1)*exp((-x+7)*log(x)^2)+5)*log(log(log( 5)))/(exp((-x+7)*log(x)^2)^2-10*exp((-x+7)*log(x)^2)+25),x)
Output:
( - e**(log(x)**2*x)*log(log(log(5)))*x)/(e**(7*log(x)**2) - 5*e**(log(x)* *2*x))