Integrand size = 81, antiderivative size = 26 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=4-e^x+3 (-1+x)+\frac {5}{\left (-3+e^x\right ) \log (\log (x))} \] Output:
3*x+1-exp(x)+5/ln(ln(x))/(exp(x)-3)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=-e^x+3 x+\frac {5}{\left (-3+e^x\right ) \log (\log (x))} \] Input:
Integrate[(15 - 5*E^x - 5*E^x*x*Log[x]*Log[Log[x]] + (27*x - 27*E^x*x + 9* E^(2*x)*x - E^(3*x)*x)*Log[x]*Log[Log[x]]^2)/((9*x - 6*E^x*x + E^(2*x)*x)* Log[x]*Log[Log[x]]^2),x]
Output:
-E^x + 3*x + 5/((-3 + E^x)*Log[Log[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.
\(\displaystyle \int \frac {-5 e^x+\left (-27 e^x x+9 e^{2 x} x-e^{3 x} x+27 x\right ) \log (x) \log ^2(\log (x))-5 e^x x \log (x) \log (\log (x))+15}{\left (-6 e^x x+e^{2 x} x+9 x\right ) \log (x) \log ^2(\log (x))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-5 e^x+\left (-27 e^x x+9 e^{2 x} x-e^{3 x} x+27 x\right ) \log (x) \log ^2(\log (x))-5 e^x x \log (x) \log (\log (x))+15}{\left (3-e^x\right )^2 x \log (x) \log ^2(\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^x-\frac {5 (x \log (x) \log (\log (x))+1)}{\left (e^x-3\right ) x \log (x) \log ^2(\log (x))}-\frac {15}{\left (e^x-3\right )^2 \log (\log (x))}+3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {1}{\left (-3+e^x\right ) x \log (x) \log ^2(\log (x))}dx-15 \int \frac {1}{\left (-3+e^x\right )^2 \log (\log (x))}dx-5 \int \frac {1}{\left (-3+e^x\right ) \log (\log (x))}dx+3 x-e^x\) |
Input:
Int[(15 - 5*E^x - 5*E^x*x*Log[x]*Log[Log[x]] + (27*x - 27*E^x*x + 9*E^(2*x )*x - E^(3*x)*x)*Log[x]*Log[Log[x]]^2)/((9*x - 6*E^x*x + E^(2*x)*x)*Log[x] *Log[Log[x]]^2),x]
Output:
$Aborted
Time = 4.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \(3 x -{\mathrm e}^{x}+\frac {5}{\ln \left (\ln \left (x \right )\right ) \left ({\mathrm e}^{x}-3\right )}\) | \(22\) |
parallelrisch | \(-\frac {-9 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right ) x +3 \ln \left (\ln \left (x \right )\right ) {\mathrm e}^{2 x}-15+27 x \ln \left (\ln \left (x \right )\right )-9 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{3 \ln \left (\ln \left (x \right )\right ) \left ({\mathrm e}^{x}-3\right )}\) | \(46\) |
Input:
int(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*ln(x)*ln(ln(x))^2-5*x*exp (x)*ln(x)*ln(ln(x))-5*exp(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/ln(x)/ln(ln(x ))^2,x,method=_RETURNVERBOSE)
Output:
3*x-exp(x)+5/ln(ln(x))/(exp(x)-3)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {{\left (3 \, {\left (x + 1\right )} e^{x} - 9 \, x - e^{\left (2 \, x\right )}\right )} \log \left (\log \left (x\right )\right ) + 5}{{\left (e^{x} - 3\right )} \log \left (\log \left (x\right )\right )} \] Input:
integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^ 2-5*x*exp(x)*log(x)*log(log(x))-5*exp(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/l og(x)/log(log(x))^2,x, algorithm="fricas")
Output:
((3*(x + 1)*e^x - 9*x - e^(2*x))*log(log(x)) + 5)/((e^x - 3)*log(log(x)))
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=3 x - e^{x} + \frac {5}{e^{x} \log {\left (\log {\left (x \right )} \right )} - 3 \log {\left (\log {\left (x \right )} \right )}} \] Input:
integrate(((-x*exp(x)**3+9*x*exp(x)**2-27*exp(x)*x+27*x)*ln(x)*ln(ln(x))** 2-5*x*exp(x)*ln(x)*ln(ln(x))-5*exp(x)+15)/(x*exp(x)**2-6*exp(x)*x+9*x)/ln( x)/ln(ln(x))**2,x)
Output:
3*x - exp(x) + 5/(exp(x)*log(log(x)) - 3*log(log(x)))
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {{\left (3 \, {\left (x + 1\right )} e^{x} - 9 \, x - e^{\left (2 \, x\right )}\right )} \log \left (\log \left (x\right )\right ) + 5}{{\left (e^{x} - 3\right )} \log \left (\log \left (x\right )\right )} \] Input:
integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^ 2-5*x*exp(x)*log(x)*log(log(x))-5*exp(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/l og(x)/log(log(x))^2,x, algorithm="maxima")
Output:
((3*(x + 1)*e^x - 9*x - e^(2*x))*log(log(x)) + 5)/((e^x - 3)*log(log(x)))
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {3 \, x e^{x} \log \left (\log \left (x\right )\right ) - 9 \, x \log \left (\log \left (x\right )\right ) - e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right ) + 3 \, e^{x} \log \left (\log \left (x\right )\right ) + 5}{e^{x} \log \left (\log \left (x\right )\right ) - 3 \, \log \left (\log \left (x\right )\right )} \] Input:
integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^ 2-5*x*exp(x)*log(x)*log(log(x))-5*exp(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/l og(x)/log(log(x))^2,x, algorithm="giac")
Output:
(3*x*e^x*log(log(x)) - 9*x*log(log(x)) - e^(2*x)*log(log(x)) + 3*e^x*log(l og(x)) + 5)/(e^x*log(log(x)) - 3*log(log(x)))
Time = 4.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=3\,x-{\mathrm {e}}^x-\frac {5}{3\,\ln \left (\ln \left (x\right )\right )-\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^x} \] Input:
int(-(5*exp(x) - log(log(x))^2*log(x)*(27*x + 9*x*exp(2*x) - x*exp(3*x) - 27*x*exp(x)) + 5*x*log(log(x))*exp(x)*log(x) - 15)/(log(log(x))^2*log(x)*( 9*x + x*exp(2*x) - 6*x*exp(x))),x)
Output:
3*x - exp(x) - 5/(3*log(log(x)) - log(log(x))*exp(x))
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {-e^{2 x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+3 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +3 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )-9 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +5}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \left (e^{x}-3\right )} \] Input:
int(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^2-5*x* exp(x)*log(x)*log(log(x))-5*exp(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/log(x)/ log(log(x))^2,x)
Output:
( - e**(2*x)*log(log(x)) + 3*e**x*log(log(x))*x + 3*e**x*log(log(x)) - 9*l og(log(x))*x + 5)/(log(log(x))*(e**x - 3))