Integrand size = 63, antiderivative size = 25 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=e^{-1-e^x+x+x^2+x \left (4 x+\log \left (\log ^2(\log (x))\right )\right )} \] Output:
exp(x-exp(x)+(ln(ln(ln(x))^2)+4*x)*x+x^2-1)
Time = 0.52 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=e^{-1-e^x+x+5 x^2} \log ^2(\log (x))^x \] Input:
Integrate[(E^(-1 - E^x + x + 5*x^2 + x*Log[Log[Log[x]]^2])*(2 + (1 - E^x + 10*x)*Log[x]*Log[Log[x]] + Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2]))/(Log[x ]*Log[Log[x]]),x]
Output:
E^(-1 - E^x + x + 5*x^2)*(Log[Log[x]]^2)^x
Time = 0.76 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{5 x^2+x-e^x+x \log \left (\log ^2(\log (x))\right )-1} \left (\log (x) \log \left (\log ^2(\log (x))\right ) \log (\log (x))+\left (10 x-e^x+1\right ) \log (x) \log (\log (x))+2\right )}{\log (x) \log (\log (x))} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{5 x^2+x-e^x-1} \log ^2(\log (x))^x\) |
Input:
Int[(E^(-1 - E^x + x + 5*x^2 + x*Log[Log[Log[x]]^2])*(2 + (1 - E^x + 10*x) *Log[x]*Log[Log[x]] + Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2]))/(Log[x]*Log[ Log[x]]),x]
Output:
E^(-1 - E^x + x + 5*x^2)*(Log[Log[x]]^2)^x
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 5.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \({\mathrm e}^{x \ln \left (\ln \left (\ln \left (x \right )\right )^{2}\right )-{\mathrm e}^{x}+5 x^{2}+x -1}\) | \(22\) |
risch | \(\ln \left (\ln \left (x \right )\right )^{2 x} {\mathrm e}^{-1-\frac {i x \pi \operatorname {csgn}\left (i \ln \left (\ln \left (x \right )\right )^{2}\right )^{3}}{2}+i x \pi \operatorname {csgn}\left (i \ln \left (\ln \left (x \right )\right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (\ln \left (x \right )\right )\right )-\frac {i x \pi \,\operatorname {csgn}\left (i \ln \left (\ln \left (x \right )\right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (\ln \left (x \right )\right )\right )^{2}}{2}-{\mathrm e}^{x}+5 x^{2}+x}\) | \(84\) |
Input:
int((ln(x)*ln(ln(x))*ln(ln(ln(x))^2)+(-exp(x)+10*x+1)*ln(x)*ln(ln(x))+2)*e xp(x*ln(ln(ln(x))^2)-exp(x)+5*x^2+x-1)/ln(x)/ln(ln(x)),x,method=_RETURNVER BOSE)
Output:
exp(x*ln(ln(ln(x))^2)-exp(x)+5*x^2+x-1)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=e^{\left (5 \, x^{2} + x \log \left (\log \left (\log \left (x\right )\right )^{2}\right ) + x - e^{x} - 1\right )} \] Input:
integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*l og(log(x))+2)*exp(x*log(log(log(x))^2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x) ),x, algorithm="fricas")
Output:
e^(5*x^2 + x*log(log(log(x))^2) + x - e^x - 1)
Time = 2.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=e^{5 x^{2} + x \log {\left (\log {\left (\log {\left (x \right )} \right )}^{2} \right )} + x - e^{x} - 1} \] Input:
integrate((ln(x)*ln(ln(x))*ln(ln(ln(x))**2)+(-exp(x)+10*x+1)*ln(x)*ln(ln(x ))+2)*exp(x*ln(ln(ln(x))**2)-exp(x)+5*x**2+x-1)/ln(x)/ln(ln(x)),x)
Output:
exp(5*x**2 + x*log(log(log(x))**2) + x - exp(x) - 1)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=e^{\left (5 \, x^{2} + 2 \, x \log \left (\log \left (\log \left (x\right )\right )\right ) + x - e^{x} - 1\right )} \] Input:
integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*l og(log(x))+2)*exp(x*log(log(log(x))^2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x) ),x, algorithm="maxima")
Output:
e^(5*x^2 + 2*x*log(log(log(x))) + x - e^x - 1)
\[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=\int { \frac {{\left (\log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )^{2}\right ) \log \left (\log \left (x\right )\right ) + {\left (10 \, x - e^{x} + 1\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right ) + 2\right )} e^{\left (5 \, x^{2} + x \log \left (\log \left (\log \left (x\right )\right )^{2}\right ) + x - e^{x} - 1\right )}}{\log \left (x\right ) \log \left (\log \left (x\right )\right )} \,d x } \] Input:
integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*l og(log(x))+2)*exp(x*log(log(log(x))^2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x) ),x, algorithm="giac")
Output:
undef
Time = 2.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx={\mathrm {e}}^{-1}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x\,{\left ({\ln \left (\ln \left (x\right )\right )}^2\right )}^x \] Input:
int((exp(x - exp(x) + 5*x^2 + x*log(log(log(x))^2) - 1)*(log(log(x))*log(x )*(10*x - exp(x) + 1) + log(log(log(x))^2)*log(log(x))*log(x) + 2))/(log(l og(x))*log(x)),x)
Output:
exp(-1)*exp(5*x^2)*exp(-exp(x))*exp(x)*(log(log(x))^2)^x
\[ \int \frac {e^{-1-e^x+x+5 x^2+x \log \left (\log ^2(\log (x))\right )} \left (2+\left (1-e^x+10 x\right ) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{\log (x) \log (\log (x))} \, dx=\int \frac {\left (\mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}\right )+\left (-{\mathrm e}^{x}+10 x +1\right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+2\right ) {\mathrm e}^{x \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}\right )-{\mathrm e}^{x}+5 x^{2}+x -1}}{\mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )}d x \] Input:
int((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*log(log (x))+2)*exp(x*log(log(log(x))^2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x)),x)
Output:
int((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*log(log (x))+2)*exp(x*log(log(log(x))^2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x)),x)