\(\int \frac {e^x (-20-12 x+4 x^2)+e^x (12+8 x) \log (x)+e^{e^{-x} x} (-e^x+x^2-x^3+(e^x+x-x^2) \log (x))+(4 e^x-4 e^x \log (x)) \log (2 x^2)}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx\) [9847]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 112, antiderivative size = 32 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {x \left (5+\frac {1}{4} e^{e^{-x} x}+x-\log \left (2 x^2\right )\right )}{x+\log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx \]

[In]

Int[(E^x*(-20 - 12*x + 4*x^2) + E^x*(12 + 8*x)*Log[x] + E^(x/E^x)*(-E^x + x^2 - x^3 + (E^x + x - x^2)*Log[x])
+ (4*E^x - 4*E^x*Log[x])*Log[2*x^2])/(4*E^x*x^2 + 8*E^x*x*Log[x] + 4*E^x*Log[x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )\right )}{4 (x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{-x} \left (e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )\right )}{(x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \left (-\frac {e^{-x+e^{-x} x} \left (e^x-x^2+x^3-e^x \log (x)-x \log (x)+x^2 \log (x)\right )}{(x+\log (x))^2}+\frac {4 \left (-5-3 x+x^2+3 \log (x)+2 x \log (x)+\log \left (2 x^2\right )-\log (x) \log \left (2 x^2\right )\right )}{(x+\log (x))^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{-x+e^{-x} x} \left (e^x-x^2+x^3-e^x \log (x)-x \log (x)+x^2 \log (x)\right )}{(x+\log (x))^2} \, dx\right )+\int \frac {-5-3 x+x^2+3 \log (x)+2 x \log (x)+\log \left (2 x^2\right )-\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} \left (e^x-x^2+x^3-e^x \log (x)-x \log (x)+x^2 \log (x)\right )}{(x+\log (x))^2} \, dx\right )+\int \left (-\frac {5}{(x+\log (x))^2}-\frac {3 x}{(x+\log (x))^2}+\frac {x^2}{(x+\log (x))^2}+\frac {3 \log (x)}{(x+\log (x))^2}+\frac {2 x \log (x)}{(x+\log (x))^2}-\frac {(-1+\log (x)) \log \left (2 x^2\right )}{(x+\log (x))^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int \left (-\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{(x+\log (x))^2}+\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^3}{(x+\log (x))^2}-\frac {e^{x-e^{-x} \left (-1+e^x\right ) x} (-1+\log (x))}{(x+\log (x))^2}-\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x \log (x)}{(x+\log (x))^2}+\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2 \log (x)}{(x+\log (x))^2}\right ) \, dx\right )+2 \int \frac {x \log (x)}{(x+\log (x))^2} \, dx-3 \int \frac {x}{(x+\log (x))^2} \, dx+3 \int \frac {\log (x)}{(x+\log (x))^2} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx-\int \frac {(-1+\log (x)) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{(x+\log (x))^2} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^3}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{x-e^{-x} \left (-1+e^x\right ) x} (-1+\log (x))}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x \log (x)}{(x+\log (x))^2} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2 \log (x)}{(x+\log (x))^2} \, dx+2 \int \left (-\frac {x^2}{(x+\log (x))^2}+\frac {x}{x+\log (x)}\right ) \, dx-3 \int \frac {x}{(x+\log (x))^2} \, dx+3 \int \left (-\frac {x}{(x+\log (x))^2}+\frac {1}{x+\log (x)}\right ) \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx-\int \left (-\frac {\log \left (2 x^2\right )}{(x+\log (x))^2}+\frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2}\right ) \, dx \\ & = \frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{(x+\log (x))^2} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^3}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{e^{-x} x} (-1+\log (x))}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \left (-\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{(x+\log (x))^2}+\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)}\right ) \, dx-\frac {1}{4} \int \left (-\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^3}{(x+\log (x))^2}+\frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)}\right ) \, dx-2 \int \frac {x^2}{(x+\log (x))^2} \, dx+2 \int \frac {x}{x+\log (x)} \, dx-2 \left (3 \int \frac {x}{(x+\log (x))^2} \, dx\right )+3 \int \frac {1}{x+\log (x)} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx+\int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2} \, dx-\int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)} \, dx+\frac {1}{4} \int \left (\frac {e^{e^{-x} x} (-1-x)}{(x+\log (x))^2}+\frac {e^{e^{-x} x}}{x+\log (x)}\right ) \, dx-2 \int \frac {x^2}{(x+\log (x))^2} \, dx+2 \int \frac {x}{x+\log (x)} \, dx-2 \left (3 \int \frac {x}{(x+\log (x))^2} \, dx\right )+3 \int \frac {1}{x+\log (x)} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx+\int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2} \, dx-\int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{e^{-x} x} (-1-x)}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{e^{-x} x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)} \, dx-2 \int \frac {x^2}{(x+\log (x))^2} \, dx+2 \int \frac {x}{x+\log (x)} \, dx-2 \left (3 \int \frac {x}{(x+\log (x))^2} \, dx\right )+3 \int \frac {1}{x+\log (x)} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx+\int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2} \, dx-\int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{e^{-x} x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)} \, dx+\frac {1}{4} \int \left (-\frac {e^{e^{-x} x}}{(x+\log (x))^2}-\frac {e^{e^{-x} x} x}{(x+\log (x))^2}\right ) \, dx-2 \int \frac {x^2}{(x+\log (x))^2} \, dx+2 \int \frac {x}{x+\log (x)} \, dx-2 \left (3 \int \frac {x}{(x+\log (x))^2} \, dx\right )+3 \int \frac {1}{x+\log (x)} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx+\int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2} \, dx-\int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{e^{-x} x}}{(x+\log (x))^2} \, dx\right )-\frac {1}{4} \int \frac {e^{e^{-x} x} x}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{e^{-x} x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)} \, dx-\frac {1}{4} \int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)} \, dx-2 \int \frac {x^2}{(x+\log (x))^2} \, dx+2 \int \frac {x}{x+\log (x)} \, dx-2 \left (3 \int \frac {x}{(x+\log (x))^2} \, dx\right )+3 \int \frac {1}{x+\log (x)} \, dx-5 \int \frac {1}{(x+\log (x))^2} \, dx+\int \frac {x^2}{(x+\log (x))^2} \, dx+\int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2} \, dx-\int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {x \left (e^{e^{-x} x}+4 (5+x)-4 \log \left (2 x^2\right )\right )}{4 (x+\log (x))} \]

[In]

Integrate[(E^x*(-20 - 12*x + 4*x^2) + E^x*(12 + 8*x)*Log[x] + E^(x/E^x)*(-E^x + x^2 - x^3 + (E^x + x - x^2)*Lo
g[x]) + (4*E^x - 4*E^x*Log[x])*Log[2*x^2])/(4*E^x*x^2 + 8*E^x*x*Log[x] + 4*E^x*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 5.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56

method result size
parallelrisch \(-\frac {-240 x +48 x \ln \left (2 x^{2}\right )+48 x \ln \left (x \right )-48 \ln \left (x \right ) \ln \left ({\mathrm e}^{x}\right )-12 \,{\mathrm e}^{x \,{\mathrm e}^{-x}} x -48 x \ln \left ({\mathrm e}^{x}\right )}{48 \left (x +\ln \left (x \right )\right )}\) \(50\)
risch \(-2 x +\frac {\left (10+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 \ln \left (2\right )+6 x \right ) x}{2 x +2 \ln \left (x \right )}+\frac {{\mathrm e}^{x \,{\mathrm e}^{-x}} x}{4 x +4 \ln \left (x \right )}\) \(88\)

[In]

int((((x+exp(x)-x^2)*ln(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*ln(x)+4*exp(x))*ln(2*x^2)+(8*x+12)*exp(x)*
ln(x)+(4*x^2-12*x-20)*exp(x))/(4*exp(x)*ln(x)^2+8*x*exp(x)*ln(x)+4*exp(x)*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/48*(-240*x+48*x*ln(2*x^2)+48*x*ln(x)-48*ln(x)*ln(exp(x))-12*exp(x/exp(x))*x-48*x*ln(exp(x)))/(x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {4 \, x^{2} + x e^{\left (x e^{\left (-x\right )}\right )} - 4 \, x \log \left (2\right ) - 8 \, x \log \left (x\right ) + 20 \, x}{4 \, {\left (x + \log \left (x\right )\right )}} \]

[In]

integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*log(x)+4*exp(x))*log(2*x^2)+(8*x+12
)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x))/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="frica
s")

[Out]

1/4*(4*x^2 + x*e^(x*e^(-x)) - 4*x*log(2) - 8*x*log(x) + 20*x)/(x + log(x))

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=- 2 x + \frac {x e^{x e^{- x}}}{4 x + 4 \log {\left (x \right )}} + \frac {3 x^{2} - x \log {\left (2 \right )} + 5 x}{x + \log {\left (x \right )}} \]

[In]

integrate((((x+exp(x)-x**2)*ln(x)-exp(x)-x**3+x**2)*exp(x/exp(x))+(-4*exp(x)*ln(x)+4*exp(x))*ln(2*x**2)+(8*x+1
2)*exp(x)*ln(x)+(4*x**2-12*x-20)*exp(x))/(4*exp(x)*ln(x)**2+8*x*exp(x)*ln(x)+4*exp(x)*x**2),x)

[Out]

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

Maxima [F]

\[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\int { \frac {4 \, {\left (2 \, x + 3\right )} e^{x} \log \left (x\right ) - {\left (x^{3} - x^{2} + {\left (x^{2} - x - e^{x}\right )} \log \left (x\right ) + e^{x}\right )} e^{\left (x e^{\left (-x\right )}\right )} + 4 \, {\left (x^{2} - 3 \, x - 5\right )} e^{x} - 4 \, {\left (e^{x} \log \left (x\right ) - e^{x}\right )} \log \left (2 \, x^{2}\right )}{4 \, {\left (x^{2} e^{x} + 2 \, x e^{x} \log \left (x\right ) + e^{x} \log \left (x\right )^{2}\right )}} \,d x } \]

[In]

integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*log(x)+4*exp(x))*log(2*x^2)+(8*x+12
)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x))/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="maxim
a")

[Out]

(x^2 - x*(log(2) - 5) - 2*x*log(x))/(x + log(x)) + 1/4*integrate(-(x^3 - x^2 - (log(x) - 1)*e^x + (x^2 - x)*lo
g(x))*e^(x*e^(-x) - x)/(x^2 + 2*x*log(x) + log(x)^2), x)

Giac [F]

\[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\int { \frac {4 \, {\left (2 \, x + 3\right )} e^{x} \log \left (x\right ) - {\left (x^{3} - x^{2} + {\left (x^{2} - x - e^{x}\right )} \log \left (x\right ) + e^{x}\right )} e^{\left (x e^{\left (-x\right )}\right )} + 4 \, {\left (x^{2} - 3 \, x - 5\right )} e^{x} - 4 \, {\left (e^{x} \log \left (x\right ) - e^{x}\right )} \log \left (2 \, x^{2}\right )}{4 \, {\left (x^{2} e^{x} + 2 \, x e^{x} \log \left (x\right ) + e^{x} \log \left (x\right )^{2}\right )}} \,d x } \]

[In]

integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*log(x)+4*exp(x))*log(2*x^2)+(8*x+12
)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x))/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="giac"
)

[Out]

integrate(1/4*(4*(2*x + 3)*e^x*log(x) - (x^3 - x^2 + (x^2 - x - e^x)*log(x) + e^x)*e^(x*e^(-x)) + 4*(x^2 - 3*x
 - 5)*e^x - 4*(e^x*log(x) - e^x)*log(2*x^2))/(x^2*e^x + 2*x*e^x*log(x) + e^x*log(x)^2), x)

Mupad [B] (verification not implemented)

Time = 13.94 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.28 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=4\,x+\frac {\frac {x\,\left (3\,x-\ln \left (2\,x^2\right )+2\,\ln \left (x\right )-3\,x^2+5\right )}{x+1}-\frac {x\,\ln \left (x\right )\,\left (6\,x-\ln \left (2\,x^2\right )+2\,\ln \left (x\right )+5\right )}{x+1}}{x+\ln \left (x\right )}+\frac {\ln \left (2\,x^2\right )-2\,\ln \left (x\right )+1}{x+1}+\frac {x\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}{4\,\left (x+\ln \left (x\right )\right )} \]

[In]

int(-(exp(x*exp(-x))*(exp(x) - log(x)*(x + exp(x) - x^2) - x^2 + x^3) + exp(x)*(12*x - 4*x^2 + 20) - log(2*x^2
)*(4*exp(x) - 4*exp(x)*log(x)) - exp(x)*log(x)*(8*x + 12))/(4*x^2*exp(x) + 4*exp(x)*log(x)^2 + 8*x*exp(x)*log(
x)),x)

[Out]

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