\(\int \frac {(-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} (1-2 x+x^2)) \log (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}})+e^{\frac {2 e^e x}{-1+x}} (e^e (-16-8 x)+e^e (-8-4 x) \log (\frac {2+x}{5}))}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} (2-3 x+x^3)} \, dx\) [6051]

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

Optimal result

Integrand size = 141, antiderivative size = 31 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=\log \left (\left (3-e^{\frac {2 e^e x}{-1+x}}\right )^2\right ) \left (2+\log \left (\frac {2+x}{5}\right )\right ) \]

[Out]

(2+ln(2/5+1/5*x))*ln((3-exp(x*exp(exp(1))/(-1+x))^2)^2)

Rubi [F]

\[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=\int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx \]

[In]

Int[((-3 + 6*x - 3*x^2 + E^((2*E^E*x)/(-1 + x))*(1 - 2*x + x^2))*Log[9 - 6*E^((2*E^E*x)/(-1 + x)) + E^((4*E^E*
x)/(-1 + x))] + E^((2*E^E*x)/(-1 + x))*(E^E*(-16 - 8*x) + E^E*(-8 - 4*x)*Log[(2 + x)/5]))/(-6 + 9*x - 3*x^3 +
E^((2*E^E*x)/(-1 + x))*(2 - 3*x + x^3)),x]

[Out]

4*Log[3 - E^((-2*E^E*x)/(1 - x))] + Defer[Int][Log[(-3 + E^((2*E^E*x)/(-1 + x)))^2]/(2 + x), x] - 4*Defer[Int]
[(E^(E + (2*E^E*x)/(-1 + x))*Log[2/5 + x/5])/((-3 + E^((2*E^E*x)/(-1 + x)))*(1 - x)^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\log \left (\left (-3+e^{\frac {2 e^e x}{-1+x}}\right )^2\right )}{2+x}-\frac {4 e^{e+\frac {2 e^e x}{-1+x}} \left (2+\log \left (\frac {2+x}{5}\right )\right )}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (-1+x)^2}\right ) \, dx \\ & = -\left (4 \int \frac {e^{e+\frac {2 e^e x}{-1+x}} \left (2+\log \left (\frac {2+x}{5}\right )\right )}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (-1+x)^2} \, dx\right )+\int \frac {\log \left (\left (-3+e^{\frac {2 e^e x}{-1+x}}\right )^2\right )}{2+x} \, dx \\ & = -\left (4 \int \left (\frac {2 e^{e+\frac {2 e^e x}{-1+x}}}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (-1+x)^2}+\frac {e^{e+\frac {2 e^e x}{-1+x}} \log \left (\frac {2}{5}+\frac {x}{5}\right )}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (1-x)^2}\right ) \, dx\right )+\int \frac {\log \left (\left (-3+e^{\frac {2 e^e x}{-1+x}}\right )^2\right )}{2+x} \, dx \\ & = -\left (4 \int \frac {e^{e+\frac {2 e^e x}{-1+x}} \log \left (\frac {2}{5}+\frac {x}{5}\right )}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (1-x)^2} \, dx\right )-8 \int \frac {e^{e+\frac {2 e^e x}{-1+x}}}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (-1+x)^2} \, dx+\int \frac {\log \left (\left (-3+e^{\frac {2 e^e x}{-1+x}}\right )^2\right )}{2+x} \, dx \\ & = 4 \log \left (3-e^{-\frac {2 e^e x}{1-x}}\right )-4 \int \frac {e^{e+\frac {2 e^e x}{-1+x}} \log \left (\frac {2}{5}+\frac {x}{5}\right )}{\left (-3+e^{\frac {2 e^e x}{-1+x}}\right ) (1-x)^2} \, dx+\int \frac {\log \left (\left (-3+e^{\frac {2 e^e x}{-1+x}}\right )^2\right )}{2+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=4 \log \left (3-e^{2 e^e+\frac {2 e^e}{-1+x}}\right )+\log \left (\left (-3+e^{2 e^e+\frac {2 e^e}{-1+x}}\right )^2\right ) \log \left (\frac {2+x}{5}\right ) \]

[In]

Integrate[((-3 + 6*x - 3*x^2 + E^((2*E^E*x)/(-1 + x))*(1 - 2*x + x^2))*Log[9 - 6*E^((2*E^E*x)/(-1 + x)) + E^((
4*E^E*x)/(-1 + x))] + E^((2*E^E*x)/(-1 + x))*(E^E*(-16 - 8*x) + E^E*(-8 - 4*x)*Log[(2 + x)/5]))/(-6 + 9*x - 3*
x^3 + E^((2*E^E*x)/(-1 + x))*(2 - 3*x + x^3)),x]

[Out]

4*Log[3 - E^(2*E^E + (2*E^E)/(-1 + x))] + Log[(-3 + E^(2*E^E + (2*E^E)/(-1 + x)))^2]*Log[(2 + x)/5]

Maple [C] (warning: unable to verify)

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

Time = 26.06 (sec) , antiderivative size = 179, normalized size of antiderivative = 5.77

method result size
risch \(2 \ln \left (\frac {2}{5}+\frac {x}{5}\right ) \ln \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )-\frac {i \pi \ln \left (-2-x \right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )^{2}\right )}{2}+i \pi \ln \left (-2-x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )^{2}\right )}^{2}-\frac {i \pi \ln \left (-2-x \right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )^{2}\right )}^{3}}{2}-8 \,{\mathrm e}^{{\mathrm e}}+4 \ln \left ({\mathrm e}^{\frac {2 x \,{\mathrm e}^{{\mathrm e}}}{-1+x}}-3\right )\) \(179\)

[In]

int((((x^2-2*x+1)*exp(x*exp(exp(1))/(-1+x))^2-3*x^2+6*x-3)*ln(exp(x*exp(exp(1))/(-1+x))^4-6*exp(x*exp(exp(1))/
(-1+x))^2+9)+((-4*x-8)*exp(exp(1))*ln(2/5+1/5*x)+(-8*x-16)*exp(exp(1)))*exp(x*exp(exp(1))/(-1+x))^2)/((x^3-3*x
+2)*exp(x*exp(exp(1))/(-1+x))^2-3*x^3+9*x-6),x,method=_RETURNVERBOSE)

[Out]

2*ln(2/5+1/5*x)*ln(exp(2*x*exp(exp(1))/(-1+x))-3)-1/2*I*Pi*ln(-2-x)*csgn(I*(exp(2*x*exp(exp(1))/(-1+x))-3))^2*
csgn(I*(exp(2*x*exp(exp(1))/(-1+x))-3)^2)+I*Pi*ln(-2-x)*csgn(I*(exp(2*x*exp(exp(1))/(-1+x))-3))*csgn(I*(exp(2*
x*exp(exp(1))/(-1+x))-3)^2)^2-1/2*I*Pi*ln(-2-x)*csgn(I*(exp(2*x*exp(exp(1))/(-1+x))-3)^2)^3-8*exp(exp(1))+4*ln
(exp(2*x*exp(exp(1))/(-1+x))-3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx={\left (\log \left (\frac {1}{5} \, x + \frac {2}{5}\right ) + 2\right )} \log \left (e^{\left (\frac {4 \, x e^{e}}{x - 1}\right )} - 6 \, e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} + 9\right ) \]

[In]

integrate((((x^2-2*x+1)*exp(x*exp(exp(1))/(-1+x))^2-3*x^2+6*x-3)*log(exp(x*exp(exp(1))/(-1+x))^4-6*exp(x*exp(e
xp(1))/(-1+x))^2+9)+((-4*x-8)*exp(exp(1))*log(2/5+1/5*x)+(-8*x-16)*exp(exp(1)))*exp(x*exp(exp(1))/(-1+x))^2)/(
(x^3-3*x+2)*exp(x*exp(exp(1))/(-1+x))^2-3*x^3+9*x-6),x, algorithm="fricas")

[Out]

(log(1/5*x + 2/5) + 2)*log(e^(4*x*e^e/(x - 1)) - 6*e^(2*x*e^e/(x - 1)) + 9)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).

Time = 0.56 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=\log {\left (\frac {x}{5} + \frac {2}{5} \right )} \log {\left (e^{\frac {4 x e^{e}}{x - 1}} - 6 e^{\frac {2 x e^{e}}{x - 1}} + 9 \right )} + 4 \log {\left (e^{\frac {2 x e^{e}}{x - 1}} - 3 \right )} \]

[In]

integrate((((x**2-2*x+1)*exp(x*exp(exp(1))/(-1+x))**2-3*x**2+6*x-3)*ln(exp(x*exp(exp(1))/(-1+x))**4-6*exp(x*ex
p(exp(1))/(-1+x))**2+9)+((-4*x-8)*exp(exp(1))*ln(2/5+1/5*x)+(-8*x-16)*exp(exp(1)))*exp(x*exp(exp(1))/(-1+x))**
2)/((x**3-3*x+2)*exp(x*exp(exp(1))/(-1+x))**2-3*x**3+9*x-6),x)

[Out]

log(x/5 + 2/5)*log(exp(4*x*exp(E)/(x - 1)) - 6*exp(2*x*exp(E)/(x - 1)) + 9) + 4*log(exp(2*x*exp(E)/(x - 1)) -
3)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=-2 \, {\left (\log \left (5\right ) - \log \left (x + 2\right ) - 2\right )} \log \left (e^{\left (\frac {2 \, e^{e}}{x - 1} + 2 \, e^{e}\right )} - 3\right ) \]

[In]

integrate((((x^2-2*x+1)*exp(x*exp(exp(1))/(-1+x))^2-3*x^2+6*x-3)*log(exp(x*exp(exp(1))/(-1+x))^4-6*exp(x*exp(e
xp(1))/(-1+x))^2+9)+((-4*x-8)*exp(exp(1))*log(2/5+1/5*x)+(-8*x-16)*exp(exp(1)))*exp(x*exp(exp(1))/(-1+x))^2)/(
(x^3-3*x+2)*exp(x*exp(exp(1))/(-1+x))^2-3*x^3+9*x-6),x, algorithm="maxima")

[Out]

-2*(log(5) - log(x + 2) - 2)*log(e^(2*e^e/(x - 1) + 2*e^e) - 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (26) = 52\).

Time = 2.41 (sec) , antiderivative size = 178, normalized size of antiderivative = 5.74 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=\frac {x \log \left (x + 2\right ) \log \left (e^{\left (\frac {4 \, x e^{e}}{x - 1}\right )} - 6 \, e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} + 9\right ) - 2 \, x \log \left (5\right ) \log \left (e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} - 3\right ) + 4 \, e^{e} \log \left (5\right ) - 4 \, e^{e} \log \left (x + 2\right ) + 4 \, e^{e} \log \left (\frac {1}{5} \, x + \frac {2}{5}\right ) - \log \left (x + 2\right ) \log \left (e^{\left (\frac {4 \, x e^{e}}{x - 1}\right )} - 6 \, e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} + 9\right ) + 4 \, x \log \left (e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} - 3\right ) + 2 \, \log \left (5\right ) \log \left (e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} - 3\right ) - 4 \, \log \left (e^{\left (\frac {2 \, x e^{e}}{x - 1}\right )} - 3\right )}{x - 1} \]

[In]

integrate((((x^2-2*x+1)*exp(x*exp(exp(1))/(-1+x))^2-3*x^2+6*x-3)*log(exp(x*exp(exp(1))/(-1+x))^4-6*exp(x*exp(e
xp(1))/(-1+x))^2+9)+((-4*x-8)*exp(exp(1))*log(2/5+1/5*x)+(-8*x-16)*exp(exp(1)))*exp(x*exp(exp(1))/(-1+x))^2)/(
(x^3-3*x+2)*exp(x*exp(exp(1))/(-1+x))^2-3*x^3+9*x-6),x, algorithm="giac")

[Out]

(x*log(x + 2)*log(e^(4*x*e^e/(x - 1)) - 6*e^(2*x*e^e/(x - 1)) + 9) - 2*x*log(5)*log(e^(2*x*e^e/(x - 1)) - 3) +
 4*e^e*log(5) - 4*e^e*log(x + 2) + 4*e^e*log(1/5*x + 2/5) - log(x + 2)*log(e^(4*x*e^e/(x - 1)) - 6*e^(2*x*e^e/
(x - 1)) + 9) + 4*x*log(e^(2*x*e^e/(x - 1)) - 3) + 2*log(5)*log(e^(2*x*e^e/(x - 1)) - 3) - 4*log(e^(2*x*e^e/(x
 - 1)) - 3))/(x - 1)

Mupad [B] (verification not implemented)

Time = 11.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-3+6 x-3 x^2+e^{\frac {2 e^e x}{-1+x}} \left (1-2 x+x^2\right )\right ) \log \left (9-6 e^{\frac {2 e^e x}{-1+x}}+e^{\frac {4 e^e x}{-1+x}}\right )+e^{\frac {2 e^e x}{-1+x}} \left (e^e (-16-8 x)+e^e (-8-4 x) \log \left (\frac {2+x}{5}\right )\right )}{-6+9 x-3 x^3+e^{\frac {2 e^e x}{-1+x}} \left (2-3 x+x^3\right )} \, dx=4\,\ln \left ({\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{\mathrm {e}}}{x-1}}-3\right )+\ln \left (\frac {x}{5}+\frac {2}{5}\right )\,\ln \left ({\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{\mathrm {e}}}{x-1}}-6\,{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{\mathrm {e}}}{x-1}}+9\right ) \]

[In]

int((log(exp((4*x*exp(exp(1)))/(x - 1)) - 6*exp((2*x*exp(exp(1)))/(x - 1)) + 9)*(6*x + exp((2*x*exp(exp(1)))/(
x - 1))*(x^2 - 2*x + 1) - 3*x^2 - 3) - exp((2*x*exp(exp(1)))/(x - 1))*(exp(exp(1))*(8*x + 16) + exp(exp(1))*lo
g(x/5 + 2/5)*(4*x + 8)))/(9*x + exp((2*x*exp(exp(1)))/(x - 1))*(x^3 - 3*x + 2) - 3*x^3 - 6),x)

[Out]

4*log(exp((2*x*exp(exp(1)))/(x - 1)) - 3) + log(x/5 + 2/5)*log(exp((4*x*exp(exp(1)))/(x - 1)) - 6*exp((2*x*exp
(exp(1)))/(x - 1)) + 9)