Integrand size = 62, antiderivative size = 29 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\log \left (\frac {5}{2}\right ) \left (-2+\frac {1}{2} \left (-3+x+\frac {x}{e^5}+x^2+\frac {x}{\log (\log (x))}\right )\right ) \] Output:
(1/2*x+1/2*x^2+1/2*x/ln(ln(x))-7/2+1/2*x/exp(5))*ln(5/2)
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\frac {x \log \left (\frac {5}{2}\right ) \left (1+e^5 (1+x)+\frac {e^5}{\log (\log (x))}\right )}{2 e^5} \] Input:
Integrate[(-(E^5*Log[5/2]) + E^5*Log[5/2]*Log[x]*Log[Log[x]] + (1 + E^5*(1 + 2*x))*Log[5/2]*Log[x]*Log[Log[x]]^2)/(2*E^5*Log[x]*Log[Log[x]]^2),x]
Output:
(x*Log[5/2]*(1 + E^5*(1 + x) + E^5/Log[Log[x]]))/(2*E^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.
| \(\displaystyle \int \frac {\left (e^5 (2 x+1)+1\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))-e^5 \log \left (\frac {5}{2}\right )}{2 e^5 \log (x) \log ^2(\log (x))} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {-\left (\left (e^5 (2 x+1)+1\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))\right )-e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+e^5 \log \left (\frac {5}{2}\right )}{\log (x) \log ^2(\log (x))}dx}{2 e^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {-\left (\left (e^5 (2 x+1)+1\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))\right )-e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+e^5 \log \left (\frac {5}{2}\right )}{\log (x) \log ^2(\log (x))}dx}{2 e^5}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\int \left (-\log \left (\frac {5}{2}\right ) \left (2 e^5 x+e^5+1\right )-\frac {e^5 \log \left (\frac {5}{2}\right )}{\log (\log (x))}+\frac {e^5 \log \left (\frac {5}{2}\right )}{\log (x) \log ^2(\log (x))}\right )dx}{2 e^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e^5 \log \left (\frac {5}{2}\right ) \int \frac {1}{\log (x) \log ^2(\log (x))}dx-e^5 \log \left (\frac {5}{2}\right ) \int \frac {1}{\log (\log (x))}dx-\frac {\left (2 e^5 x+e^5+1\right )^2 \log \left (\frac {5}{2}\right )}{4 e^5}}{2 e^5}\) |
Input:
Int[(-(E^5*Log[5/2]) + E^5*Log[5/2]*Log[x]*Log[Log[x]] + (1 + E^5*(1 + 2*x ))*Log[5/2]*Log[x]*Log[Log[x]]^2)/(2*E^5*Log[x]*Log[Log[x]]^2),x]
Output:
$Aborted
Time = 0.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(\frac {x \left (\ln \left (5\right )-\ln \left (2\right )\right ) \left (x \ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )+{\mathrm e}^{-5} \ln \left (\ln \left (x \right )\right )+1\right )}{2 \ln \left (\ln \left (x \right )\right )}\) | \(32\) |
| parallelrisch | \(\frac {{\mathrm e}^{-5} \left ({\mathrm e}^{5} \ln \left (\frac {5}{2}\right ) \ln \left (\ln \left (x \right )\right ) x^{2}+{\mathrm e}^{5} \ln \left (\frac {5}{2}\right ) x \ln \left (\ln \left (x \right )\right )+{\mathrm e}^{5} \ln \left (\frac {5}{2}\right ) x +\ln \left (\frac {5}{2}\right ) x \ln \left (\ln \left (x \right )\right )\right )}{2 \ln \left (\ln \left (x \right )\right )}\) | \(46\) |
| norman | \(\frac {\left (-\frac {\ln \left (2\right )}{2}+\frac {\ln \left (5\right )}{2}\right ) x +\left (-\frac {\ln \left (2\right )}{2}+\frac {\ln \left (5\right )}{2}\right ) x^{2} \ln \left (\ln \left (x \right )\right )+\frac {\left ({\mathrm e}^{5} \ln \left (5\right )+\ln \left (5\right )-{\mathrm e}^{5} \ln \left (2\right )-\ln \left (2\right )\right ) {\mathrm e}^{-5} x \ln \left (\ln \left (x \right )\right )}{2}}{\ln \left (\ln \left (x \right )\right )}\) | \(63\) |
| default | \(\frac {{\mathrm e}^{-5} \left (\frac {\left (-{\mathrm e}^{5} \ln \left (2\right )-\ln \left (2\right )\right ) x \ln \left (\ln \left (x \right )\right )-\ln \left (2\right ) {\mathrm e}^{5} x -{\mathrm e}^{5} \ln \left (2\right ) x^{2} \ln \left (\ln \left (x \right )\right )}{\ln \left (\ln \left (x \right )\right )}+\frac {\left ({\mathrm e}^{5} \ln \left (5\right )+\ln \left (5\right )\right ) x \ln \left (\ln \left (x \right )\right )+x \,{\mathrm e}^{5} \ln \left (5\right )+{\mathrm e}^{5} \ln \left (5\right ) x^{2} \ln \left (\ln \left (x \right )\right )}{\ln \left (\ln \left (x \right )\right )}\right )}{2}\) | \(87\) |
Input:
int(1/2*(((1+2*x)*exp(5)+1)*ln(5/2)*ln(x)*ln(ln(x))^2+exp(5)*ln(5/2)*ln(x) *ln(ln(x))-exp(5)*ln(5/2))/exp(5)/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOS E)
Output:
1/2*x*(ln(5)-ln(2))*(x*ln(ln(x))+ln(ln(x))+exp(-5)*ln(ln(x))+1)/ln(ln(x))
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\frac {{\left (x e^{5} \log \left (\frac {5}{2}\right ) + {\left ({\left (x^{2} + x\right )} e^{5} + x\right )} \log \left (\frac {5}{2}\right ) \log \left (\log \left (x\right )\right )\right )} e^{\left (-5\right )}}{2 \, \log \left (\log \left (x\right )\right )} \] Input:
integrate(1/2*(((1+2*x)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log (5/2)*log(x)*log(log(x))-exp(5)*log(5/2))/exp(5)/log(x)/log(log(x))^2,x, a lgorithm="fricas")
Output:
1/2*(x*e^5*log(5/2) + ((x^2 + x)*e^5 + x)*log(5/2)*log(log(x)))*e^(-5)/log (log(x))
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=x^{2} \left (- \frac {\log {\left (2 \right )}}{2} + \frac {\log {\left (5 \right )}}{2}\right ) + \frac {x \left (- e^{5} \log {\left (2 \right )} - \log {\left (2 \right )} + \log {\left (5 \right )} + e^{5} \log {\left (5 \right )}\right )}{2 e^{5}} + \frac {- x \log {\left (2 \right )} + x \log {\left (5 \right )}}{2 \log {\left (\log {\left (x \right )} \right )}} \] Input:
integrate(1/2*(((1+2*x)*exp(5)+1)*ln(5/2)*ln(x)*ln(ln(x))**2+exp(5)*ln(5/2 )*ln(x)*ln(ln(x))-exp(5)*ln(5/2))/exp(5)/ln(x)/ln(ln(x))**2,x)
Output:
x**2*(-log(2)/2 + log(5)/2) + x*(-exp(5)*log(2) - log(2) + log(5) + exp(5) *log(5))*exp(-5)/2 + (-x*log(2) + x*log(5))/(2*log(log(x)))
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2} \, {\left (x^{2} e^{5} \log \left (\frac {5}{2}\right ) + x e^{5} \log \left (\frac {5}{2}\right ) + x \log \left (\frac {5}{2}\right ) + \frac {x {\left (\log \left (5\right ) - \log \left (2\right )\right )} e^{5}}{\log \left (\log \left (x\right )\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(1/2*(((1+2*x)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log (5/2)*log(x)*log(log(x))-exp(5)*log(5/2))/exp(5)/log(x)/log(log(x))^2,x, a lgorithm="maxima")
Output:
1/2*(x^2*e^5*log(5/2) + x*e^5*log(5/2) + x*log(5/2) + x*(log(5) - log(2))* e^5/log(log(x)))*e^(-5)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (21) = 42\).
Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2} \, {\left (x^{2} e^{5} \log \left (5\right ) - x^{2} e^{5} \log \left (2\right ) + x e^{5} \log \left (5\right ) - x e^{5} \log \left (2\right ) + x \log \left (5\right ) - x \log \left (2\right ) + \frac {x e^{5} \log \left (5\right )}{\log \left (\log \left (x\right )\right )} - \frac {x e^{5} \log \left (2\right )}{\log \left (\log \left (x\right )\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(1/2*(((1+2*x)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log (5/2)*log(x)*log(log(x))-exp(5)*log(5/2))/exp(5)/log(x)/log(log(x))^2,x, a lgorithm="giac")
Output:
1/2*(x^2*e^5*log(5) - x^2*e^5*log(2) + x*e^5*log(5) - x*e^5*log(2) + x*log (5) - x*log(2) + x*e^5*log(5)/log(log(x)) - x*e^5*log(2)/log(log(x)))*e^(- 5)
Time = 1.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=x\,\left (\frac {\ln \left (\frac {5}{2}\right )}{2}+\frac {{\mathrm {e}}^{-5}\,\ln \left (\frac {5}{2}\right )}{2}\right )+\frac {x^2\,\ln \left (\frac {5}{2}\right )}{2}+\frac {x\,\ln \left (\frac {5}{2}\right )}{2\,\ln \left (\ln \left (x\right )\right )} \] Input:
int((exp(-5)*((log(log(x))*exp(5)*log(5/2)*log(x))/2 - (exp(5)*log(5/2))/2 + (log(log(x))^2*log(5/2)*log(x)*(exp(5)*(2*x + 1) + 1))/2))/(log(log(x)) ^2*log(x)),x)
Output:
x*(log(5/2)/2 + (exp(-5)*log(5/2))/2) + (x^2*log(5/2))/2 + (x*log(5/2))/(2 *log(log(x)))
Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx=\frac {\mathrm {log}\left (\frac {5}{2}\right ) x \left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) e^{5} x +\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) e^{5}+\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+e^{5}\right )}{2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) e^{5}} \] Input:
int(1/2*(((1+2*x)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log(5/2)* log(x)*log(log(x))-exp(5)*log(5/2))/exp(5)/log(x)/log(log(x))^2,x)
Output:
(log(5/2)*x*(log(log(x))*e**5*x + log(log(x))*e**5 + log(log(x)) + e**5))/ (2*log(log(x))*e**5)