\(\int \frac {e^{20 x+x \log (\frac {1-x \log (\frac {e^{-x} \log (2)}{x})}{x})} (-19-x-x^2+20 x \log (\frac {e^{-x} \log (2)}{x})+(-1+x \log (\frac {e^{-x} \log (2)}{x})) \log (\frac {1-x \log (\frac {e^{-x} \log (2)}{x})}{x}))}{-1+x \log (\frac {e^{-x} \log (2)}{x})} \, dx\) [2600]

Optimal result

Integrand size = 113, antiderivative size = 25 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{x \left (20+\log \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )\right )} \] Output:

exp(x*(20+ln(1/x-ln(ln(2)/exp(x)/x))))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{20 x} \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )^x \] Input:

Integrate[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 
+ 20*x*Log[Log[2]/(E^x*x)] + (-1 + x*Log[Log[2]/(E^x*x)])*Log[(1 - x*Log[L 
og[2]/(E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]
 

Output:

E^(20*x)*(x^(-1) - Log[Log[2]/(E^x*x)])^x
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, 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.

\(\displaystyle \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-x^2-x+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (x \log \left (\frac {e^{-x} \log (2)}{x}\right )-1\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )-19\right )}{x \log \left (\frac {e^{-x} \log (2)}{x}\right )-1} \, dx\)

\(\Big \downarrow \) 7257

\(\displaystyle e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x\)

Input:

Int[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 + 20*x 
*Log[Log[2]/(E^x*x)] + (-1 + x*Log[Log[2]/(E^x*x)])*Log[(1 - x*Log[Log[2]/ 
(E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]
 

Output:

E^(20*x)*((1 - x*Log[Log[2]/(E^x*x)])/x)^x
 

Defintions of rubi rules used

rule 7257

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]
 

Maple [A] (verified)

Time = 5.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\({\mathrm e}^{x \left (\ln \left (-\frac {x \ln \left (\frac {\ln \left (2\right ) {\mathrm e}^{-x}}{x}\right )-1}{x}\right )+20\right )}\)	\(27\)

Input:

int(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*ln(ln(2 
)/exp(x)/x)-x^2-x-19)*exp(x*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x)/(x*ln(ln 
(2)/exp(x)/x)-1),x,method=_RETURNVERBOSE)
 

Output:

exp(x*(ln(-(x*ln(ln(2)/exp(x)/x)-1)/x)+20))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )} \] Input:

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 
0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x 
)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="fricas")
 

Output:

e^(x*log(-(x*log(e^(-x)*log(2)/x) - 1)/x) + 20*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\text {Timed out} \] Input:

integrate(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*l 
n(ln(2)/exp(x)/x)-x**2-x-19)*exp(x*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x)/( 
x*ln(ln(2)/exp(x)/x)-1),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (x^{2} + x \log \left (x\right ) - x \log \left (\log \left (2\right )\right ) + 1\right ) - x \log \left (x\right ) + 20 \, x\right )} \] Input:

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 
0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x 
)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="maxima")
 

Output:

e^(x*log(x^2 + x*log(x) - x*log(log(2)) + 1) - x*log(x) + 20*x)
 

Giac [F]

\[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\int { -\frac {{\left (x^{2} - 20 \, x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - {\left (x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1\right )} \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + x + 19\right )} e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )}}{x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1} \,d x } \] Input:

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 
0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x 
)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx={\mathrm {e}}^{20\,x}\,{\left (x-\ln \left (\frac {\ln \left (2\right )}{x}\right )+\frac {1}{x}\right )}^x \] Input:

int(-(exp(20*x + x*log(-(x*log((exp(-x)*log(2))/x) - 1)/x))*(x - log(-(x*l 
og((exp(-x)*log(2))/x) - 1)/x)*(x*log((exp(-x)*log(2))/x) - 1) - 20*x*log( 
(exp(-x)*log(2))/x) + x^2 + 19))/(x*log((exp(-x)*log(2))/x) - 1),x)
 

Output:

exp(20*x)*(x - log(log(2)/x) + 1/x)^x
 

Reduce [F]

\[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\int \frac {\left (\left (x \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{{\mathrm e}^{x} x}\right )-1\right ) \mathrm {log}\left (\frac {-x \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{{\mathrm e}^{x} x}\right )+1}{x}\right )+20 x \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{{\mathrm e}^{x} x}\right )-x^{2}-x -19\right ) {\mathrm e}^{x \,\mathrm {log}\left (\frac {-x \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{{\mathrm e}^{x} x}\right )+1}{x}\right )+20 x}}{x \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{{\mathrm e}^{x} x}\right )-1}d x \] Input:

int(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x*lo 
g(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x 
)/(x*log(log(2)/exp(x)/x)-1),x)
 

Output:

int(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x*lo 
g(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x 
)/(x*log(log(2)/exp(x)/x)-1),x)