Integrand size = 60, antiderivative size = 26 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=\frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \]
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=\frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )} \]
Integrate[(-2*x + 2*x*Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4 *E^4))])/(Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4*E^4))]^3),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.
\(\displaystyle \int \frac {2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (4 e^4-8\right )}\right )-2 x}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (4 e^4-8\right )}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x \left (\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )-1\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {x \left (1-\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {x \left (1-\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}-\frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\int \frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}dx-\int \frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}dx\right )\) |
Int[(-2*x + 2*x*Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4*E^4)) ])/(Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4*E^4))]^3),x]
3.4.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x^{2}}{\ln \left (-\frac {5 \ln \left (\ln \left (x \right )\right ) {\mathrm e}^{-5}}{4 \,{\mathrm e}^{4}-8}\right )^{2}}\) | \(23\) |
parallelrisch | \(\frac {x^{2}}{\ln \left (-\frac {5 \ln \left (\ln \left (x \right )\right ) {\mathrm e}^{-5}}{4 \,{\mathrm e}^{4}-8}\right )^{2}}\) | \(25\) |
default | \(\frac {2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) x^{2}}{-\ln \left (5\right )+2 \ln \left (2\right )+\ln \left ({\mathrm e}^{4}-2\right )+5-\ln \left (-\ln \left (\ln \left (x \right )\right )\right )}-\frac {x^{2} \left (2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left ({\mathrm e}^{4}-2\right )-2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left (5\right )+4 \ln \left (2\right ) \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left (-\ln \left (\ln \left (x \right )\right )\right )+10 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-1\right )}{{\left (-\ln \left (5\right )+2 \ln \left (2\right )+\ln \left ({\mathrm e}^{4}-2\right )+5-\ln \left (-\ln \left (\ln \left (x \right )\right )\right )\right )}^{2}}\) | \(119\) |
parts | \(\frac {2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) x^{2}}{-\ln \left (5\right )+2 \ln \left (2\right )+\ln \left ({\mathrm e}^{4}-2\right )+5-\ln \left (-\ln \left (\ln \left (x \right )\right )\right )}-\frac {x^{2} \left (2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left ({\mathrm e}^{4}-2\right )-2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left (5\right )+4 \ln \left (2\right ) \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right ) \ln \left (-\ln \left (\ln \left (x \right )\right )\right )+10 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-1\right )}{{\left (-\ln \left (5\right )+2 \ln \left (2\right )+\ln \left ({\mathrm e}^{4}-2\right )+5-\ln \left (-\ln \left (\ln \left (x \right )\right )\right )\right )}^{2}}\) | \(119\) |
int((2*x*ln(x)*ln(ln(x))*ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))-2*x)/ln(x)/l n(ln(x))/ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))^3,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=\frac {x^{2}}{\log \left (-\frac {5 \, \log \left (\log \left (x\right )\right )}{4 \, {\left (e^{9} - 2 \, e^{5}\right )}}\right )^{2}} \]
integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))- 2*x)/log(x)/log(log(x))/log(-5*log(log(x))/(4*exp(4)-8)/exp(5))^3,x, algor ithm=\
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=\frac {x^{2}}{\log {\left (- \frac {5 \log {\left (\log {\left (x \right )} \right )}}{\left (-8 + 4 e^{4}\right ) e^{5}} \right )}^{2}} \]
integrate((2*x*ln(x)*ln(ln(x))*ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))-2*x)/l n(x)/ln(ln(x))/ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (20) = 40\).
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.92 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=-\frac {x^{2}}{2 \, {\left (\log \left (-e^{4} + 2\right ) + 5\right )} \log \left (5\right ) - \log \left (5\right )^{2} + 4 \, {\left (\log \left (5\right ) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - \log \left (-e^{4} + 2\right )^{2} - 2 \, {\left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \left (\log \left (\log \left (x\right )\right )\right ) - \log \left (\log \left (\log \left (x\right )\right )\right )^{2} - 10 \, \log \left (-e^{4} + 2\right ) - 25} \]
integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))- 2*x)/log(x)/log(log(x))/log(-5*log(log(x))/(4*exp(4)-8)/exp(5))^3,x, algor ithm=\
-x^2/(2*(log(-e^4 + 2) + 5)*log(5) - log(5)^2 + 4*(log(5) - log(-e^4 + 2) - 5)*log(2) - 4*log(2)^2 - log(-e^4 + 2)^2 - 2*(log(5) - 2*log(2) - log(-e ^4 + 2) - 5)*log(log(log(x))) - log(log(log(x)))^2 - 10*log(-e^4 + 2) - 25 )
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (20) = 40\).
Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.08 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx=\frac {x^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (-5 \, \log \left (\log \left (x\right )\right )\right ) + \log \left (-5 \, \log \left (\log \left (x\right )\right )\right )^{2} + 4 \, \log \left (2\right ) \log \left (e^{4} - 2\right ) - 2 \, \log \left (-5 \, \log \left (\log \left (x\right )\right )\right ) \log \left (e^{4} - 2\right ) + \log \left (e^{4} - 2\right )^{2} + 20 \, \log \left (2\right ) - 10 \, \log \left (-5 \, \log \left (\log \left (x\right )\right )\right ) + 10 \, \log \left (e^{4} - 2\right ) + 25} \]
integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))- 2*x)/log(x)/log(log(x))/log(-5*log(log(x))/(4*exp(4)-8)/exp(5))^3,x, algor ithm=\
x^2/(4*log(2)^2 - 4*log(2)*log(-5*log(log(x))) + log(-5*log(log(x)))^2 + 4 *log(2)*log(e^4 - 2) - 2*log(-5*log(log(x)))*log(e^4 - 2) + log(e^4 - 2)^2 + 20*log(2) - 10*log(-5*log(log(x))) + 10*log(e^4 - 2) + 25)
Time = 14.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 5.73 \[ \int \frac {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx={\ln \left (\ln \left (x\right )\right )}^2\,\left (2\,x^2\,{\ln \left (x\right )}^2+x^2\,\ln \left (x\right )\right )+\frac {x^2-x^2\,\ln \left (-\frac {5\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )}{{\ln \left (-\frac {5\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}^2}+\frac {x^2\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )-x\,\ln \left (-\frac {5\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (x+x\,\ln \left (\ln \left (x\right )\right )+2\,x\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\right )}{\ln \left (-\frac {5\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}+x^2\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right ) \]
int(-(2*x - 2*x*log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8))*log(log(x))*l og(x))/(log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8))^3*log(log(x))*log(x)) ,x)
log(log(x))^2*(x^2*log(x) + 2*x^2*log(x)^2) + (x^2 - x^2*log(-(5*log(log(x ))*exp(-5))/(4*exp(4) - 8))*log(log(x))*log(x))/log(-(5*log(log(x))*exp(-5 ))/(4*exp(4) - 8))^2 + (x^2*log(log(x))*log(x) - x*log(-(5*log(log(x))*exp (-5))/(4*exp(4) - 8))*log(log(x))*log(x)*(x + x*log(log(x)) + 2*x*log(log( x))*log(x)))/log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8)) + x^2*log(log(x) )*log(x)