Integrand size = 82, antiderivative size = 23 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=-2-16 \left (5+\frac {e^{e^x}}{\log \left (\frac {\log (\log (\log (x)))}{x}\right )}\right ) \]
Time = 0.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=-\frac {16 e^{e^x}}{\log \left (\frac {\log (\log (\log (x)))}{x}\right )} \]
Integrate[(16*E^E^x - 16*E^E^x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]] - 16*E^ (E^x + x)*x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]*Log[Log[Log[Log[x]]]/x])/( x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]*Log[Log[Log[Log[x]]]/x]^2),x]
Time = 3.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {7292, 27, 2726}
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.
| \(\displaystyle \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{x+e^x} x \log (x) \log (\log (x)) \log \left (\frac {\log (\log (\log (x)))}{x}\right ) \log (\log (\log (x)))}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {16 e^{e^x} \left (-\log (x) \log (\log (x)) \log (\log (\log (x)))-e^x x \log (x) \log (\log (x)) \log \left (\frac {\log (\log (\log (x)))}{x}\right ) \log (\log (\log (x)))+1\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 16 \int \frac {e^{e^x} \left (-\log (x) \log (\log (x)) \log (\log (\log (x)))-e^x x \log (x) \log (\log (x)) \log \left (\frac {\log (\log (\log (x)))}{x}\right ) \log (\log (\log (x)))+1\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle -\frac {16 e^{e^x}}{\log \left (\frac {\log (\log (\log (x)))}{x}\right )}\) |
Int[(16*E^E^x - 16*E^E^x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]] - 16*E^(E^x + x)*x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]*Log[Log[Log[Log[x]]]/x])/(x*Log[ x]*Log[Log[x]]*Log[Log[Log[x]]]*Log[Log[Log[Log[x]]]/x]^2),x]
3.4.27.3.1 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]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 4.91
\[-\frac {32 i {\mathrm e}^{{\mathrm e}^{x}}}{\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (\ln \left (\ln \left (x \right )\right )\right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\ln \left (x \right )\right )\right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\ln \left (x \right )\right )\right )}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (\ln \left (\ln \left (x \right )\right )\right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\ln \left (x \right )\right )\right )}{x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\ln \left (x \right )\right )\right )}{x}\right )^{3}-2 i \ln \left (x \right )+2 i \ln \left (\ln \left (\ln \left (\ln \left (x \right )\right )\right )\right )}\]
int((-16*x*exp(x)*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))*ln(ln(ln(ln(x) ))/x)-16*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))+16*exp(exp(x)))/x/ln(x) /ln(ln(x))/ln(ln(ln(x)))/ln(ln(ln(ln(x)))/x)^2,x)
-32*I*exp(exp(x))/(Pi*csgn(I/x)*csgn(I*ln(ln(ln(x))))*csgn(I/x*ln(ln(ln(x) )))-Pi*csgn(I/x)*csgn(I/x*ln(ln(ln(x))))^2-Pi*csgn(I*ln(ln(ln(x))))*csgn(I /x*ln(ln(ln(x))))^2+Pi*csgn(I/x*ln(ln(ln(x))))^3-2*I*ln(x)+2*I*ln(ln(ln(ln (x)))))
Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=-\frac {16 \, e^{\left (e^{x}\right )}}{\log \left (\frac {\log \left (\log \left (\log \left (x\right )\right )\right )}{x}\right )} \]
integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*lo g(log(log(log(x)))/x)-16*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))+1 6*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x)))/ x)^2,x, algorithm=\
Time = 95.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=- \frac {16 e^{e^{x}}}{\log {\left (\frac {\log {\left (\log {\left (\log {\left (x \right )} \right )} \right )}}{x} \right )}} \]
integrate((-16*x*exp(x)*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))*ln(ln(ln (ln(x)))/x)-16*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))+16*exp(exp(x)))/x /ln(x)/ln(ln(x))/ln(ln(ln(x)))/ln(ln(ln(ln(x)))/x)**2,x)
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=\frac {16 \, e^{\left (e^{x}\right )}}{\log \left (x\right ) - \log \left (\log \left (\log \left (\log \left (x\right )\right )\right )\right )} \]
integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*lo g(log(log(log(x)))/x)-16*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))+1 6*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x)))/ x)^2,x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=\frac {16 \, e^{\left (e^{x}\right )}}{\log \left (x\right ) - \log \left (\log \left (\log \left (\log \left (x\right )\right )\right )\right )} \]
integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*lo g(log(log(log(x)))/x)-16*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))+1 6*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x)))/ x)^2,x, algorithm=\
Timed out. \[ \int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx=\int -\frac {16\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \left (\ln \left (\ln \left (x\right )\right )\right )\,\ln \left (x\right )-16\,{\mathrm {e}}^{{\mathrm {e}}^x}+16\,x\,\ln \left (\ln \left (x\right )\right )\,\ln \left (\frac {\ln \left (\ln \left (\ln \left (x\right )\right )\right )}{x}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\ln \left (\ln \left (\ln \left (x\right )\right )\right )\,\ln \left (x\right )}{x\,\ln \left (\ln \left (x\right )\right )\,{\ln \left (\frac {\ln \left (\ln \left (\ln \left (x\right )\right )\right )}{x}\right )}^2\,\ln \left (\ln \left (\ln \left (x\right )\right )\right )\,\ln \left (x\right )} \,d x \]
int(-(16*log(log(x))*exp(exp(x))*log(log(log(x)))*log(x) - 16*exp(exp(x)) + 16*x*log(log(x))*log(log(log(log(x)))/x)*exp(exp(x))*exp(x)*log(log(log( x)))*log(x))/(x*log(log(x))*log(log(log(log(x)))/x)^2*log(log(log(x)))*log (x)),x)