\(\int \frac {(-e^x+x) \log (e^x-x)+(2 e^x-2 x) \log ^2(e^x-x)+(-2+2 e^x) \log (\frac {8-16 \log (e^x-x)}{\log (e^x-x)})}{(-e^x+x) \log (e^x-x)+(2 e^x-2 x) \log ^2(e^x-x)} \, dx\) [3598]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [C] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 109, antiderivative size = 21 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (4 \left (-4+\frac {2}{\log \left (e^x-x\right )}\right )\right ) \]

[Out]

ln(8/ln(exp(x)-x)-16)^2+x

Rubi [F]

\[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=\int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx \]

[In]

Int[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x)*Log[(8 - 16*Log[E^x - x])/Log[E^x -
 x]])/((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2),x]

[Out]

x + 2*Defer[Int][Log[8*(-2 + Log[E^x - x]^(-1))]/(Log[E^x - x]*(-1 + 2*Log[E^x - x])), x] - 2*Defer[Int][Log[8
*(-2 + Log[E^x - x]^(-1))]/((E^x - x)*Log[E^x - x]*(-1 + 2*Log[E^x - x])), x] + 2*Defer[Int][(x*Log[8*(-2 + Lo
g[E^x - x]^(-1))])/((E^x - x)*Log[E^x - x]*(-1 + 2*Log[E^x - x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-\left (\left (-e^x+x\right ) \log \left (e^x-x\right )\right )-\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )-\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (e^x-x\right ) \left (1-2 \log \left (e^x-x\right )\right ) \log \left (e^x-x\right )} \, dx \\ & = \int \left (\frac {2 (-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx \\ & = 2 \int \frac {(-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+\int \frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx \\ & = 2 \int \left (-\frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx+\int \left (1+\frac {2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx \\ & = x+2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx-2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+2 \int \frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right ) \]

[In]

Integrate[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x)*Log[(8 - 16*Log[E^x - x])/Log
[E^x - x]])/((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2),x]

[Out]

x + Log[8*(-2 + Log[E^x - x]^(-1))]^2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 4.97 (sec) , antiderivative size = 567, normalized size of antiderivative = 27.00

method result size
risch \(\text {Expression too large to display}\) \(567\)

[In]

int(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2*x)*ln(exp(x)-x)^2+(x-exp(x))*ln(exp(x)-x))
/((2*exp(x)-2*x)*ln(exp(x)-x)^2+(x-exp(x))*ln(exp(x)-x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(x)-x)-1/2)^2-2*ln(ln(exp(x)-x))*ln(ln(exp(x)-x)-1/2)+ln(ln(exp(x)-x))^2+I*Pi*ln(ln(exp(x)-x)-1/2)*cs
gn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2+I*Pi*ln(ln(exp(x)-x))*csgn(I*(ln(exp(x)-x)-
1/2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))-I*Pi*ln(ln(exp(x)-x))*csgn(I/ln(exp(x)-x))*
csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-I*Pi*ln(ln(exp(x)-x))*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x)
*(ln(exp(x)-x)-1/2))^2-2*I*Pi*ln(ln(exp(x)-x))+I*Pi*ln(ln(exp(x)-x)-1/2)*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)
-x)*(ln(exp(x)-x)-1/2))^2-I*Pi*ln(ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^3+2*I*Pi*ln(ln(exp(x)-
x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2+I*Pi*ln(ln(exp(x)-x)-1/2)*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2)
)^3+2*I*Pi*ln(ln(exp(x)-x)-1/2)-2*I*Pi*ln(ln(exp(x)-x)-1/2)*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-I*Pi*ln(
ln(exp(x)-x)-1/2)*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))-8*ln
(2)*ln(ln(exp(x)-x))+8*ln(2)*ln(ln(exp(x)-x)-1/2)+x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=\log \left (-\frac {8 \, {\left (2 \, \log \left (-x + e^{x}\right ) - 1\right )}}{\log \left (-x + e^{x}\right )}\right )^{2} + x \]

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="fricas")

[Out]

log(-8*(2*log(-x + e^x) - 1)/log(-x + e^x))^2 + x

Sympy [A] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x + \log {\left (\frac {8 - 16 \log {\left (- x + e^{x} \right )}}{\log {\left (- x + e^{x} \right )}} \right )}^{2} \]

[In]

integrate(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2*x)*ln(exp(x)-x)**2+(x-exp(x))*ln(exp
(x)-x))/((2*exp(x)-2*x)*ln(exp(x)-x)**2+(x-exp(x))*ln(exp(x)-x)),x)

[Out]

x + log((8 - 16*log(-x + exp(x)))/log(-x + exp(x)))**2

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.52 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-2 \, {\left (-i \, \pi - 3 \, \log \left (2\right ) + \log \left (\log \left (-x + e^{x}\right )\right )\right )} \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) + \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} - 2 \, {\left (i \, \pi + 3 \, \log \left (2\right )\right )} \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="maxima")

[Out]

-2*(-I*pi - 3*log(2) + log(log(-x + e^x)))*log(2*log(-x + e^x) - 1) + log(2*log(-x + e^x) - 1)^2 - 2*(I*pi + 3
*log(2))*log(log(-x + e^x)) + log(log(-x + e^x))^2 + x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (18) = 36\).

Time = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.62 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-\log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} + 2 \, \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) - 2 \, \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="giac")

[Out]

-log(2*log(-x + e^x) - 1)^2 + 2*log(2*log(-x + e^x) - 1)*log(-16*log(-x + e^x) + 8) - 2*log(-16*log(-x + e^x)
+ 8)*log(log(-x + e^x)) + log(log(-x + e^x))^2 + x

Mupad [B] (verification not implemented)

Time = 9.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx={\ln \left (\frac {8}{\ln \left ({\mathrm {e}}^x-x\right )}-16\right )}^2+x \]

[In]

int(-(log(-(16*log(exp(x) - x) - 8)/log(exp(x) - x))*(2*exp(x) - 2) - log(exp(x) - x)^2*(2*x - 2*exp(x)) + log
(exp(x) - x)*(x - exp(x)))/(log(exp(x) - x)^2*(2*x - 2*exp(x)) - log(exp(x) - x)*(x - exp(x))),x)

[Out]

x + log(8/log(exp(x) - x) - 16)^2