Integrand size = 84, antiderivative size = 23 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \] Output:
exp(ln(1+x)+exp(x)/(-1+2*x)-x-1)*x
Time = 5.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \] Input:
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]
Output:
E^(-1 - x + E^x/(-1 + 2*x))*x*(1 + x)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (-4 x^4+8 x^3-x^2+e^x \left (2 x^3-x^2-3 x\right )-3 x+1\right ) \exp \left (\frac {-2 x^2-x+e^x+(2 x-1) \log (x+1)+1}{2 x-1}\right )}{4 x^3-3 x+1} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-4 x^4+8 x^3-x^2+e^x \left (2 x^3-x^2-3 x\right )-3 x+1\right ) \exp \left (\frac {-2 x^2-x+e^x+(2 x-1) \log (x+1)+1}{2 x-1}\right )}{9 (x+1)}-\frac {2 \left (-4 x^4+8 x^3-x^2+e^x \left (2 x^3-x^2-3 x\right )-3 x+1\right ) \exp \left (\frac {-2 x^2-x+e^x+(2 x-1) \log (x+1)+1}{2 x-1}\right )}{9 (2 x-1)}+\frac {2 \left (-4 x^4+8 x^3-x^2+e^x \left (2 x^3-x^2-3 x\right )-3 x+1\right ) \exp \left (\frac {-2 x^2-x+e^x+(2 x-1) \log (x+1)+1}{2 x-1}\right )}{3 (2 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} \left (-4 x^4+2 \left (e^x+4\right ) x^3-\left (e^x+1\right ) x^2-3 \left (e^x+1\right ) x+1\right )}{(1-2 x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} x^2}{(2 x-1)^2}+\frac {e^{\frac {-2 x^2-x+e^x+1}{2 x-1}+x} \left (2 x^2-x-3\right ) x}{(2 x-1)^2}-\frac {3 e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} x}{(2 x-1)^2}+\frac {e^{\frac {-2 x^2-x+e^x+1}{2 x-1}}}{(2 x-1)^2}-\frac {4 e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} x^4}{(2 x-1)^2}+\frac {8 e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} x^3}{(2 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{\frac {-2 x^2-x+e^x+1}{2 x-1}}dx+\int e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} xdx-\int e^{\frac {-2 x^2-x+e^x+1}{2 x-1}} x^2dx+\frac {1}{4} \int e^{\frac {-2 x+e^x+1}{2 x-1}}dx+\frac {1}{2} \int e^{\frac {-2 x+e^x+1}{2 x-1}} xdx-\frac {3}{2} \int \frac {e^{\frac {-2 x+e^x+1}{2 x-1}}}{(2 x-1)^2}dx-\frac {5}{4} \int \frac {e^{\frac {-2 x+e^x+1}{2 x-1}}}{2 x-1}dx\) |
Input:
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]
Output:
$Aborted
Time = 1.70 (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 | \(x \,{\mathrm e}^{\frac {2 \ln \left (1+x \right ) x -2 x^{2}+{\mathrm e}^{x}-\ln \left (1+x \right )-x +1}{-1+2 x}}\) | \(37\) |
Input:
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)+e xp(x)-2*x^2-x+1)/(-1+2*x))/(4*x^3-3*x+1),x,method=_RETURNVERBOSE)
Output:
x*exp(((-1+2*x)*ln(1+x)+exp(x)-2*x^2-x+1)/(-1+2*x))
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=x e^{\left (-\frac {2 \, x^{2} - {\left (2 \, x - 1\right )} \log \left (x + 1\right ) + x - e^{x} - 1}{2 \, x - 1}\right )} \] Input:
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")
Output:
x*e^(-(2*x^2 - (2*x - 1)*log(x + 1) + x - e^x - 1)/(2*x - 1))
Time = 4.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=x e^{\frac {- 2 x^{2} - x + \left (2 x - 1\right ) \log {\left (x + 1 \right )} + e^{x} + 1}{2 x - 1}} \] Input:
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)
Output:
x*exp((-2*x**2 - x + (2*x - 1)*log(x + 1) + exp(x) + 1)/(2*x - 1))
\[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=\int { -\frac {{\left (4 \, x^{4} - 8 \, x^{3} + x^{2} - {\left (2 \, x^{3} - x^{2} - 3 \, x\right )} e^{x} + 3 \, x - 1\right )} e^{\left (-\frac {2 \, x^{2} - {\left (2 \, x - 1\right )} \log \left (x + 1\right ) + x - e^{x} - 1}{2 \, x - 1}\right )}}{4 \, x^{3} - 3 \, x + 1} \,d x } \] Input:
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")
Output:
-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)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx={\left (x^{2} e^{\left (\frac {e^{x}}{2 \, x - 1}\right )} + x e^{\left (\frac {e^{x}}{2 \, x - 1}\right )}\right )} e^{\left (-x - 1\right )} \] Input:
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")
Output:
(x^2*e^(e^x/(2*x - 1)) + x*e^(e^x/(2*x - 1)))*e^(-x - 1)
Timed out. \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=\int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^x-x-2\,x^2+\ln \left (x+1\right )\,\left (2\,x-1\right )+1}{2\,x-1}}\,\left (3\,x+{\mathrm {e}}^x\,\left (-2\,x^3+x^2+3\,x\right )+x^2-8\,x^3+4\,x^4-1\right )}{4\,x^3-3\,x+1} \,d x \] Input:
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)
Output:
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)
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx=\frac {e^{\frac {e^{x}}{2 x -1}} x \left (x +1\right )}{e^{x} e} \] Input:
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)*log(1+x)+ exp(x)-2*x^2-x+1)/(-1+2*x))/(4*x^3-3*x+1),x)
Output:
(e**(e**x/(2*x - 1))*x*(x + 1))/(e**x*e)