\(\int \frac {-2 e^{2 x} x^5+(-5 x^5+5 x^4 \log (4 e^{e^{2 x}+x})) \log (-x+\log (4 e^{e^{2 x}+x}))+(-x-x \log (x)+\log (4 e^{e^{2 x}+x}) (1+\log (x))) \log ^2(-x+\log (4 e^{e^{2 x}+x}))}{(-x+\log (4 e^{e^{2 x}+x})) \log ^2(-x+\log (4 e^{e^{2 x}+x}))} \, dx\) [1028]

Optimal result

Integrand size = 136, antiderivative size = 28 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \left (\log (x)+\frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \]

[Out]

(x^4/ln(ln(4*exp(exp(2*x)+x))-x)+ln(x))*x

Rubi [F]

\[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx \]

[In]

Int[(-2*E^(2*x)*x^5 + (-5*x^5 + 5*x^4*Log[4*E^(E^(2*x) + x)])*Log[-x + Log[4*E^(E^(2*x) + x)]] + (-x - x*Log[x
] + Log[4*E^(E^(2*x) + x)]*(1 + Log[x]))*Log[-x + Log[4*E^(E^(2*x) + x)]]^2)/((-x + Log[4*E^(E^(2*x) + x)])*Lo
g[-x + Log[4*E^(E^(2*x) + x)]]^2),x]

[Out]

x*Log[x] + 2*Defer[Int][(E^(2*x)*x^5)/((x - Log[4*E^(E^(2*x) + x)])*Log[-x + Log[4*E^(E^(2*x) + x)]]^2), x] +
5*Defer[Int][x^4/Log[-x + Log[4*E^(E^(2*x) + x)]], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\log (x)+\frac {2 e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}+\frac {5 x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \, dx \\ & = x+2 \int \frac {e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+5 \int \frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+\int \log (x) \, dx \\ & = x \log (x)+2 \int \frac {e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+5 \int \frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \log (x)+\frac {x^5}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \]

[In]

Integrate[(-2*E^(2*x)*x^5 + (-5*x^5 + 5*x^4*Log[4*E^(E^(2*x) + x)])*Log[-x + Log[4*E^(E^(2*x) + x)]] + (-x - x
*Log[x] + Log[4*E^(E^(2*x) + x)]*(1 + Log[x]))*Log[-x + Log[4*E^(E^(2*x) + x)]]^2)/((-x + Log[4*E^(E^(2*x) + x
)])*Log[-x + Log[4*E^(E^(2*x) + x)]]^2),x]

[Out]

x*Log[x] + x^5/Log[-x + Log[4*E^(E^(2*x) + x)]]

Maple [A] (verified)

Time = 26.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(x \ln \left (x \right )+\frac {x^{5}}{\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\)	\(29\)
parallelrisch	\(\frac {120 x^{5}+120 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right ) x \ln \left (x \right )}{120 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\)	\(46\)

[In]

int((((ln(x)+1)*ln(4*exp(exp(2*x)+x))-x*ln(x)-x)*ln(ln(4*exp(exp(2*x)+x))-x)^2+(5*x^4*ln(4*exp(exp(2*x)+x))-5*
x^5)*ln(ln(4*exp(exp(2*x)+x))-x)-2*x^5*exp(2*x))/(ln(4*exp(exp(2*x)+x))-x)/ln(ln(4*exp(exp(2*x)+x))-x)^2,x,met
hod=_RETURNVERBOSE)

[Out]

x*ln(x)+x^5/ln(2*ln(2)+ln(exp(exp(2*x)+x))-x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]

[In]

integrate((((log(x)+1)*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(exp(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp
(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*
x)+x))-x)^2,x, algorithm="fricas")

[Out]

(x^5 + x*log(x)*log(e^(2*x) + 2*log(2)))/log(e^(2*x) + 2*log(2))

Sympy [F(-1)]

Timed out. \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate((((ln(x)+1)*ln(4*exp(exp(2*x)+x))-x*ln(x)-x)*ln(ln(4*exp(exp(2*x)+x))-x)**2+(5*x**4*ln(4*exp(exp(2*x
)+x))-5*x**5)*ln(ln(4*exp(exp(2*x)+x))-x)-2*x**5*exp(2*x))/(ln(4*exp(exp(2*x)+x))-x)/ln(ln(4*exp(exp(2*x)+x))-
x)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]

[In]

integrate((((log(x)+1)*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(exp(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp
(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*
x)+x))-x)^2,x, algorithm="maxima")

[Out]

(x^5 + x*log(x)*log(e^(2*x) + 2*log(2)))/log(e^(2*x) + 2*log(2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]

[In]

integrate((((log(x)+1)*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(exp(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp
(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*
x)+x))-x)^2,x, algorithm="giac")

[Out]

(x^5 + x*log(x)*log(e^(2*x) + 2*log(2)))/log(e^(2*x) + 2*log(2))

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^5}{\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (4\right )\right )}+x\,\ln \left (x\right ) \]

[In]

int((log(log(4*exp(x + exp(2*x))) - x)^2*(x + x*log(x) - log(4*exp(x + exp(2*x)))*(log(x) + 1)) - log(log(4*ex
p(x + exp(2*x))) - x)*(5*x^4*log(4*exp(x + exp(2*x))) - 5*x^5) + 2*x^5*exp(2*x))/(log(log(4*exp(x + exp(2*x)))
- x)^2*(x - log(4*exp(x + exp(2*x))))),x)

[Out]

x^5/log(exp(2*x) + log(4)) + x*log(x)