\(\int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} (1-3 x-x^2+8 x^3-4 x^4+e^x (-3 x-x^2+2 x^3))}{1-3 x+4 x^3} \, dx\) [8179]

Optimal result

Integrand size = 84, antiderivative size = 23 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right )}{(1+x) (-1+2 x)^2}-\frac {3 \exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x}{(1+x) (-1+2 x)^2}-\frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^2}{(1+x) (-1+2 x)^2}+\frac {8 \exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^3}{(1+x) (-1+2 x)^2}-\frac {4 \exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^4}{(1+x) (-1+2 x)^2}+\frac {\exp \left (x+\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x (-3+2 x)}{(-1+2 x)^2}\right ) \, dx \\ & = -\left (3 \int \frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x}{(1+x) (-1+2 x)^2} \, dx\right )-4 \int \frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^4}{(1+x) (-1+2 x)^2} \, dx+8 \int \frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^3}{(1+x) (-1+2 x)^2} \, dx+\int \frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right )}{(1+x) (-1+2 x)^2} \, dx-\int \frac {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x^2}{(1+x) (-1+2 x)^2} \, dx+\int \frac {\exp \left (x+\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) x (-3+2 x)}{(-1+2 x)^2} \, dx \\ & = -\left (3 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x}{(1-2 x)^2} \, dx\right )-4 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^4}{(1-2 x)^2} \, dx+8 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^3}{(1-2 x)^2} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2}{(1-2 x)^2} \, dx+\int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right ) x (-3+2 x)}{(1-2 x)^2} \, dx \\ & = -\left (3 \int \left (\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{2 (-1+2 x)^2}+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{2 (-1+2 x)}\right ) \, dx\right )-4 \int \left (\frac {3}{16} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}+\frac {1}{4} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x+\frac {1}{4} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{16 (-1+2 x)^2}+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{4 (-1+2 x)}\right ) \, dx+8 \int \left (\frac {1}{4} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}+\frac {1}{4} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{8 (-1+2 x)^2}+\frac {3 e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{8 (-1+2 x)}\right ) \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int \left (\frac {1}{4} e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{4 (-1+2 x)^2}+\frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{2 (-1+2 x)}\right ) \, dx+\int \left (\frac {1}{2} \exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )-\frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{(-1+2 x)^2}-\frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{2 (-1+2 x)}\right ) \, dx \\ & = -\left (\frac {1}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx\right )-2 \left (\frac {1}{4} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx\right )+\frac {1}{2} \int \exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right ) \, dx-\frac {1}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx-\frac {1}{2} \int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{-1+2 x} \, dx-\frac {3}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx+3 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2 \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{(-1+2 x)^2} \, dx-\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx \\ & = -\left (\frac {1}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx\right )-2 \left (\frac {1}{4} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx\right )+\frac {1}{2} \int e^{\frac {1+e^x-2 x}{-1+2 x}} (1+x) \, dx-\frac {1}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx-\frac {1}{2} \int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{-1+2 x} \, dx-\frac {3}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx+3 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2 \, dx-\int \frac {e^{\frac {1+e^x-2 x}{-1+2 x}} (1+x)}{(1-2 x)^2} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx \\ & = -\left (\frac {1}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx\right )-2 \left (\frac {1}{4} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx\right )-\frac {1}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx-\frac {1}{2} \int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{-1+2 x} \, dx+\frac {1}{2} \int \left (e^{\frac {1+e^x-2 x}{-1+2 x}}+e^{\frac {1+e^x-2 x}{-1+2 x}} x\right ) \, dx-\frac {3}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx+3 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2 \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx-\int \left (\frac {3 e^{\frac {1+e^x-2 x}{-1+2 x}}}{2 (-1+2 x)^2}+\frac {e^{\frac {1+e^x-2 x}{-1+2 x}}}{2 (-1+2 x)}\right ) \, dx \\ & = -\left (\frac {1}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx\right )-2 \left (\frac {1}{4} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx\right )+\frac {1}{2} \int e^{\frac {1+e^x-2 x}{-1+2 x}} \, dx+\frac {1}{2} \int e^{\frac {1+e^x-2 x}{-1+2 x}} x \, dx-\frac {1}{2} \int \frac {e^{\frac {1+e^x-2 x}{-1+2 x}}}{-1+2 x} \, dx-\frac {1}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx-\frac {1}{2} \int \frac {\exp \left (\frac {1+e^x-2 x-\log (1+x)+2 x \log (1+x)}{-1+2 x}\right )}{-1+2 x} \, dx-\frac {3}{4} \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-2 x}{-1+2 x}}}{(-1+2 x)^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} \, dx+2 \int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx+3 \int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(1-2 x)^2} \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x \, dx-\int e^{\frac {1+e^x-x-2 x^2}{-1+2 x}} x^2 \, dx+\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{(-1+2 x)^2} \, dx-\int \frac {e^{\frac {1+e^x-x-2 x^2}{-1+2 x}}}{-1+2 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \]

[In]

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

[Out]

E^(-1 - x + E^x/(-1 + 2*x))*x*(1 + x)

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48

method	result	size
parallelrisch	\(x \,{\mathrm e}^{\frac {\left (-1+2 x \right ) \ln \left (1+x \right )+{\mathrm e}^{x}-2 x^{2}-x +1}{-1+2 x}}\)	\(34\)
risch	\(x \,{\mathrm e}^{\frac {2 \ln \left (1+x \right ) x -2 x^{2}+{\mathrm e}^{x}-\ln \left (1+x \right )-x +1}{-1+2 x}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=x e^{\left (-\frac {2 \, x^{2} - {\left (2 \, x - 1\right )} \log \left (x + 1\right ) + x - e^{x} - 1}{2 \, x - 1}\right )} \]

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((-1+2*x)*log(1+x)+exp(x)-2*x^2-x+1)/(-1+2*x))/(4
*x^3-3*x+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 4.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=x e^{\frac {- 2 x^{2} - x + \left (2 x - 1\right ) \log {\left (x + 1 \right )} + e^{x} + 1}{2 x - 1}} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((-1+2*x)*log(1+x)+exp(x)-2*x^2-x+1)/(-1+2*x))/(4
*x^3-3*x+1),x, algorithm="maxima")

[Out]

-integrate((4*x^4 - 8*x^3 + x^2 - (2*x^3 - x^2 - 3*x)*e^x + 3*x - 1)*e^(-(2*x^2 - (2*x - 1)*log(x + 1) + x - e
^x - 1)/(2*x - 1))/(4*x^3 - 3*x + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx={\left (x^{2} e^{\left (\frac {e^{x}}{2 \, x - 1}\right )} + x e^{\left (\frac {e^{x}}{2 \, x - 1}\right )}\right )} e^{\left (-x - 1\right )} \]

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((-1+2*x)*log(1+x)+exp(x)-2*x^2-x+1)/(-1+2*x))/(4
*x^3-3*x+1),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=\int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^x-x-2\,x^2+\ln \left (x+1\right )\,\left (2\,x-1\right )+1}{2\,x-1}}\,\left (3\,x+{\mathrm {e}}^x\,\left (-2\,x^3+x^2+3\,x\right )+x^2-8\,x^3+4\,x^4-1\right )}{4\,x^3-3\,x+1} \,d x \]

[In]

int(-(exp((exp(x) - x - 2*x^2 + log(x + 1)*(2*x - 1) + 1)/(2*x - 1))*(3*x + exp(x)*(3*x + x^2 - 2*x^3) + x^2 -
8*x^3 + 4*x^4 - 1))/(4*x^3 - 3*x + 1),x)

[Out]

int(-(exp((exp(x) - x - 2*x^2 + log(x + 1)*(2*x - 1) + 1)/(2*x - 1))*(3*x + exp(x)*(3*x + x^2 - 2*x^3) + x^2 -
8*x^3 + 4*x^4 - 1))/(4*x^3 - 3*x + 1), x)