\(\int \frac {-1-x+e^{3 x} (-3 x-3 x^2)+e^x (1+2 x+x^2+x^3)+(1+3 x+2 x^2+x^3+e^{2 x} (-3 x-3 x^2)) \log (x)+(e^x (x+x^2)+(x+x^2) \log (x)) \log (\frac {x+x^2}{e^x+\log (x)})}{e^{3 x} (-x-x^2)+e^x (x^2+x^3)+(x^2+x^3+e^{2 x} (-x-x^2)) \log (x)+(e^x (x+x^2)+(x+x^2) \log (x)) \log (\frac {x+x^2}{e^x+\log (x)})} \, dx\) [7988]

Optimal result

Integrand size = 189, antiderivative size = 28 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=\log \left (e^x \left (-e^{2 x}+x+\log \left (\frac {x (1+x)}{e^x+\log (x)}\right )\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=\int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx \]

[In]

Int[(-1 - x + E^(3*x)*(-3*x - 3*x^2) + E^x*(1 + 2*x + x^2 + x^3) + (1 + 3*x + 2*x^2 + x^3 + E^(2*x)*(-3*x - 3*
x^2))*Log[x] + (E^x*(x + x^2) + (x + x^2)*Log[x])*Log[(x + x^2)/(E^x + Log[x])])/(E^(3*x)*(-x - x^2) + E^x*(x^
2 + x^3) + (x^2 + x^3 + E^(2*x)*(-x - x^2))*Log[x] + (E^x*(x + x^2) + (x + x^2)*Log[x])*Log[(x + x^2)/(E^x + L
og[x])]),x]

[Out]

3*x - Defer[Int][Log[x]/((E^x + Log[x])*(-x + Log[x]^2 - Log[(x*(1 + x))/(E^x + Log[x])])), x] + Defer[Int][(E
^x*Log[x])/((E^(2*x) - x - Log[(x*(1 + x))/(E^x + Log[x])])*(-x + Log[x]^2 - Log[(x*(1 + x))/(E^x + Log[x])]))
, x] - Defer[Int][Log[x]^2/((E^(2*x) - x - Log[(x*(1 + x))/(E^x + Log[x])])*(-x + Log[x]^2 - Log[(x*(1 + x))/(
E^x + Log[x])])), x] + 2*Defer[Int][(Log[x]^2*Log[(x*(1 + x))/(E^x + Log[x])])/((E^(2*x) - x - Log[(x*(1 + x))
/(E^x + Log[x])])*(-x + Log[x]^2 - Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][1/(x*(E^x + Log[x])*(x -
Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - 2*Defer[Int][1/((E^(2*x) - x - Log[(x*(1 + x))/(E^x + Log[
x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] + Defer[Int][E^x/(x*(E^(2*x) - x - Log[(x*(1 + x))
/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][E^x/((1 + x)*(E^(2*x) - x
- Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][(E^x*x*
Log[x])/((E^(2*x) - x - Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x]
- 2*Defer[Int][Log[(x*(1 + x))/(E^x + Log[x])]/((E^(2*x) - x - Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2
+ Log[(x*(1 + x))/(E^x + Log[x])])), x] + 2*Defer[Int][Log[(x*(1 + x))/(E^x + Log[x])]^2/((E^(2*x) - x - Log[
(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - 2*Defer[Int][x^2/((-E^(2*
x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][1
/((1 + x)*(-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])),
x] - Defer[Int][E^x/((1 + x)*(-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/
(E^x + Log[x])])), x] + Defer[Int][Log[x]/(x*(-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 +
Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][(E^x*x*Log[x])/((-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[
x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[Int][Log[x]^2/(x*(-E^(2*x) + x + Log[(x*(1
+ x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] + 2*Defer[Int][(x*Log[x]^2)/((-E
^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - Defer[In
t][Log[x]^2/((1 + x)*(-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + L
og[x])])), x] - 2*Defer[Int][Log[(x*(1 + x))/(E^x + Log[x])]/((-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])
*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] + Defer[Int][Log[(x*(1 + x))/(E^x + Log[x])]/(x*(-E^(2*
x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x] - 4*Defer[Int]
[(x*Log[(x*(1 + x))/(E^x + Log[x])])/((-E^(2*x) + x + Log[(x*(1 + x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*
(1 + x))/(E^x + Log[x])])), x] + Defer[Int][Log[(x*(1 + x))/(E^x + Log[x])]/((1 + x)*(-E^(2*x) + x + Log[(x*(1
+ x))/(E^x + Log[x])])*(x - Log[x]^2 + Log[(x*(1 + x))/(E^x + Log[x])])), x]

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x+\log \left (e^{2 x}-x-\log \left (\frac {x (1+x)}{e^x+\log (x)}\right )\right ) \]

[In]

Integrate[(-1 - x + E^(3*x)*(-3*x - 3*x^2) + E^x*(1 + 2*x + x^2 + x^3) + (1 + 3*x + 2*x^2 + x^3 + E^(2*x)*(-3*
x - 3*x^2))*Log[x] + (E^x*(x + x^2) + (x + x^2)*Log[x])*Log[(x + x^2)/(E^x + Log[x])])/(E^(3*x)*(-x - x^2) + E
^x*(x^2 + x^3) + (x^2 + x^3 + E^(2*x)*(-x - x^2))*Log[x] + (E^x*(x + x^2) + (x + x^2)*Log[x])*Log[(x + x^2)/(E
^x + Log[x])]),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 228.46 (sec) , antiderivative size = 260, normalized size of antiderivative = 9.29

method	result	size
risch	\(x +\ln \left (\ln \left (\ln \left (x \right )+{\mathrm e}^{x}\right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i x \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )-\pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i x \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i x \left (1+x \right )}{\ln \left (x \right )+{\mathrm e}^{x}}\right )^{3}-2 i {\mathrm e}^{2 x}+2 i x +2 i \ln \left (x \right )+2 i \ln \left (1+x \right )\right )}{2}\right )\)	\(260\)

[In]

int((((x^2+x)*ln(x)+(x^2+x)*exp(x))*ln((x^2+x)/(ln(x)+exp(x)))+((-3*x^2-3*x)*exp(x)^2+x^3+2*x^2+3*x+1)*ln(x)+(
-3*x^2-3*x)*exp(x)^3+(x^3+x^2+2*x+1)*exp(x)-x-1)/(((x^2+x)*ln(x)+(x^2+x)*exp(x))*ln((x^2+x)/(ln(x)+exp(x)))+((
-x^2-x)*exp(x)^2+x^3+x^2)*ln(x)+(-x^2-x)*exp(x)^3+(x^3+x^2)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x + \log \left (x - e^{\left (2 \, x\right )} + \log \left (\frac {x^{2} + x}{e^{x} + \log \left (x\right )}\right )\right ) \]

[In]

integrate((((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x)+exp(x)))+((-3*x^2-3*x)*exp(x)^2+x^3+2*x^2+3*x+1
)*log(x)+(-3*x^2-3*x)*exp(x)^3+(x^3+x^2+2*x+1)*exp(x)-x-1)/(((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x
)+exp(x)))+((-x^2-x)*exp(x)^2+x^3+x^2)*log(x)+(-x^2-x)*exp(x)^3+(x^3+x^2)*exp(x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x + \log {\left (x - e^{2 x} + \log {\left (\frac {x^{2} + x}{e^{x} + \log {\left (x \right )}} \right )} \right )} \]

[In]

integrate((((x**2+x)*ln(x)+(x**2+x)*exp(x))*ln((x**2+x)/(ln(x)+exp(x)))+((-3*x**2-3*x)*exp(x)**2+x**3+2*x**2+3
*x+1)*ln(x)+(-3*x**2-3*x)*exp(x)**3+(x**3+x**2+2*x+1)*exp(x)-x-1)/(((x**2+x)*ln(x)+(x**2+x)*exp(x))*ln((x**2+x
)/(ln(x)+exp(x)))+((-x**2-x)*exp(x)**2+x**3+x**2)*ln(x)+(-x**2-x)*exp(x)**3+(x**3+x**2)*exp(x)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x + \log \left (-x + e^{\left (2 \, x\right )} - \log \left (x + 1\right ) - \log \left (x\right ) + \log \left (e^{x} + \log \left (x\right )\right )\right ) \]

[In]

integrate((((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x)+exp(x)))+((-3*x^2-3*x)*exp(x)^2+x^3+2*x^2+3*x+1
)*log(x)+(-3*x^2-3*x)*exp(x)^3+(x^3+x^2+2*x+1)*exp(x)-x-1)/(((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x
)+exp(x)))+((-x^2-x)*exp(x)^2+x^3+x^2)*log(x)+(-x^2-x)*exp(x)^3+(x^3+x^2)*exp(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x + \log \left (x - e^{\left (2 \, x\right )} + \log \left (x + 1\right ) + \log \left (x\right ) - \log \left (e^{x} + \log \left (x\right )\right )\right ) \]

[In]

integrate((((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x)+exp(x)))+((-3*x^2-3*x)*exp(x)^2+x^3+2*x^2+3*x+1
)*log(x)+(-3*x^2-3*x)*exp(x)^3+(x^3+x^2+2*x+1)*exp(x)-x-1)/(((x^2+x)*log(x)+(x^2+x)*exp(x))*log((x^2+x)/(log(x
)+exp(x)))+((-x^2-x)*exp(x)^2+x^3+x^2)*log(x)+(-x^2-x)*exp(x)^3+(x^3+x^2)*exp(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-1-x+e^{3 x} \left (-3 x-3 x^2\right )+e^x \left (1+2 x+x^2+x^3\right )+\left (1+3 x+2 x^2+x^3+e^{2 x} \left (-3 x-3 x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )}{e^{3 x} \left (-x-x^2\right )+e^x \left (x^2+x^3\right )+\left (x^2+x^3+e^{2 x} \left (-x-x^2\right )\right ) \log (x)+\left (e^x \left (x+x^2\right )+\left (x+x^2\right ) \log (x)\right ) \log \left (\frac {x+x^2}{e^x+\log (x)}\right )} \, dx=x+\ln \left (x-{\mathrm {e}}^{2\,x}+\ln \left (\frac {x\,\left (x+1\right )}{{\mathrm {e}}^x+\ln \left (x\right )}\right )\right ) \]

[In]

int(-(x + exp(3*x)*(3*x + 3*x^2) - exp(x)*(2*x + x^2 + x^3 + 1) - log(x)*(3*x - exp(2*x)*(3*x + 3*x^2) + 2*x^2
+ x^3 + 1) - log((x + x^2)/(exp(x) + log(x)))*(exp(x)*(x + x^2) + log(x)*(x + x^2)) + 1)/(exp(x)*(x^2 + x^3)
- exp(3*x)*(x + x^2) + log((x + x^2)/(exp(x) + log(x)))*(exp(x)*(x + x^2) + log(x)*(x + x^2)) + log(x)*(x^2 -
exp(2*x)*(x + x^2) + x^3)),x)

[Out]

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