Integrand size = 86, antiderivative size = 27 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=3+\frac {4+e^{-e^5} x^2 \log (4)}{4+\log ^2(\log (x))} \]
Time = 0.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {e^{-e^5} \left (4 e^{e^5}+x^2 \log (4)\right )}{4+\log ^2(\log (x))} \]
Integrate[(8*x^2*Log[4]*Log[x] + (-8*E^E^5 - 2*x^2*Log[4])*Log[Log[x]] + 2 *x^2*Log[4]*Log[x]*Log[Log[x]]^2)/(16*E^E^5*x*Log[x] + 8*E^E^5*x*Log[x]*Lo g[Log[x]]^2 + E^E^5*x*Log[x]*Log[Log[x]]^4),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.
\(\displaystyle \int \frac {2 x^2 \log (4) \log (x) \log ^2(\log (x))+8 x^2 \log (4) \log (x)+\left (-2 x^2 \log (4)-8 e^{e^5}\right ) \log (\log (x))}{e^{e^5} x \log (x) \log ^4(\log (x))+8 e^{e^5} x \log (x) \log ^2(\log (x))+16 e^{e^5} x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 e^{-e^5} \left (x^2 \log (4) \log (x) \log ^2(\log (x))+4 x^2 \log (4) \log (x)-x^2 \log (4) \log (\log (x))-4 e^{e^5} \log (\log (x))\right )}{x \log (x) \left (\log ^2(\log (x))+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 e^{-e^5} \int \frac {\log (4) \log (x) \log ^2(\log (x)) x^2+4 \log (4) \log (x) x^2-\log (4) \log (\log (x)) x^2-4 e^{e^5} \log (\log (x))}{x \log (x) \left (\log ^2(\log (x))+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 e^{-e^5} \int \left (\frac {x \log (4)}{\log ^2(\log (x))+4}-\frac {\left (\log (4) x^2+4 e^{e^5}\right ) \log (\log (x))}{x \log (x) \left (\log ^2(\log (x))+4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^{-e^5} \left (-\log (4) \int \frac {x \log (\log (x))}{\log (x) \left (\log ^2(\log (x))+4\right )^2}dx+\log (4) \int \frac {x}{\log ^2(\log (x))+4}dx+\frac {2 e^{e^5}}{\log ^2(\log (x))+4}\right )\) |
Int[(8*x^2*Log[4]*Log[x] + (-8*E^E^5 - 2*x^2*Log[4])*Log[Log[x]] + 2*x^2*L og[4]*Log[x]*Log[Log[x]]^2)/(16*E^E^5*x*Log[x] + 8*E^E^5*x*Log[x]*Log[Log[ x]]^2 + E^E^5*x*Log[x]*Log[Log[x]]^4),x]
3.10.23.3.1 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 = 3.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {2 \left (x^{2} \ln \left (2\right )+2 \,{\mathrm e}^{{\mathrm e}^{5}}\right ) {\mathrm e}^{-{\mathrm e}^{5}}}{4+\ln \left (\ln \left (x \right )\right )^{2}}\) | \(29\) |
parallelrisch | \(\frac {\left (2 x^{2} \ln \left (2\right )+4 \,{\mathrm e}^{{\mathrm e}^{5}}\right ) {\mathrm e}^{-{\mathrm e}^{5}}}{4+\ln \left (\ln \left (x \right )\right )^{2}}\) | \(29\) |
int((4*x^2*ln(2)*ln(x)*ln(ln(x))^2+(-8*exp(exp(5))-4*x^2*ln(2))*ln(ln(x))+ 16*x^2*ln(2)*ln(x))/(x*exp(exp(5))*ln(x)*ln(ln(x))^4+8*x*exp(exp(5))*ln(x) *ln(ln(x))^2+16*x*exp(exp(5))*ln(x)),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {2 \, {\left (x^{2} \log \left (2\right ) + 2 \, e^{\left (e^{5}\right )}\right )}}{e^{\left (e^{5}\right )} \log \left (\log \left (x\right )\right )^{2} + 4 \, e^{\left (e^{5}\right )}} \]
integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2)) *log(log(x))+16*x^2*log(2)*log(x))/(x*exp(exp(5))*log(x)*log(log(x))^4+8*x *exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {2 x^{2} \log {\left (2 \right )} + 4 e^{e^{5}}}{e^{e^{5}} \log {\left (\log {\left (x \right )} \right )}^{2} + 4 e^{e^{5}}} \]
integrate((4*x**2*ln(2)*ln(x)*ln(ln(x))**2+(-8*exp(exp(5))-4*x**2*ln(2))*l n(ln(x))+16*x**2*ln(2)*ln(x))/(x*exp(exp(5))*ln(x)*ln(ln(x))**4+8*x*exp(ex p(5))*ln(x)*ln(ln(x))**2+16*x*exp(exp(5))*ln(x)),x)
Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {2 \, {\left (x^{2} \log \left (2\right ) + 2 \, e^{\left (e^{5}\right )}\right )}}{e^{\left (e^{5}\right )} \log \left (\log \left (x\right )\right )^{2} + 4 \, e^{\left (e^{5}\right )}} \]
integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2)) *log(log(x))+16*x^2*log(2)*log(x))/(x*exp(exp(5))*log(x)*log(log(x))^4+8*x *exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {2 \, {\left (x^{2} e^{\left (-e^{5}\right )} \log \left (2\right ) + 2\right )}}{\log \left (\log \left (x\right )\right )^{2} + 4} \]
integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2)) *log(log(x))+16*x^2*log(2)*log(x))/(x*exp(exp(5))*log(x)*log(log(x))^4+8*x *exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=\
Time = 12.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx=\frac {2\,{\mathrm {e}}^{-{\mathrm {e}}^5}\,\left (\ln \left (2\right )\,x^2+2\,{\mathrm {e}}^{{\mathrm {e}}^5}\right )}{{\ln \left (\ln \left (x\right )\right )}^2+4} \]