Integrand size = 75, antiderivative size = 28 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=2+e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \] Output:
2+exp((ln(exp(2)*x)-x*exp(2*x)-x)/(-4+x))
Time = 5.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=e^{-\frac {-2+x+e^{2 x} x}{-4+x}} x^{\frac {1}{-4+x}} \] Input:
Integrate[(E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))*(-4 + 5*x + E^(2*x)* (4*x + 8*x^2 - 2*x^3) - x*Log[E^2*x]))/(16*x - 8*x^2 + x^3),x]
Output:
x^(-4 + x)^(-1)/E^((-2 + x + E^(2*x)*x)/(-4 + 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.
\(\displaystyle \int \frac {e^{\frac {-e^{2 x} x-x+\log \left (e^2 x\right )}{x-4}} \left (e^{2 x} \left (-2 x^3+8 x^2+4 x\right )+5 x+x \left (-\log \left (e^2 x\right )\right )-4\right )}{x^3-8 x^2+16 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {-e^{2 x} x-x+\log \left (e^2 x\right )}{x-4}} \left (e^{2 x} \left (-2 x^3+8 x^2+4 x\right )+5 x+x \left (-\log \left (e^2 x\right )\right )-4\right )}{x \left (x^2-8 x+16\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {e^{\frac {e^{2 x} x+x}{4-x}-\frac {2}{4-x}} x^{\frac {1}{x-4}-1} \left (\log \left (e^2 x\right ) x-5 x-2 e^{2 x} \left (-x^3+4 x^2+2 x\right )+4\right )}{4 (4-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {e^{\frac {e^{2 x} x+x}{4-x}-\frac {2}{4-x}} x^{\frac {1}{x-4}-1} \left (\log \left (e^2 x\right ) x-5 x-2 e^{2 x} \left (-x^3+4 x^2+2 x\right )+4\right )}{(4-x)^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\int \frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}-1} \left (\log \left (e^2 x\right ) x-5 x-2 e^{2 x} \left (-x^3+4 x^2+2 x\right )+4\right )}{(4-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} (\log (x) x-3 x+4) x^{\frac {1}{x-4}-1}}{(x-4)^2}+\frac {2 e^{2 x-\frac {e^{2 x} x+x-2}{x-4}} \left (x^2-4 x-2\right ) x^{\frac {1}{x-4}}}{(x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \int \frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}-1}}{(x-4)^2}dx+3 \int \frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}-1}}{x-4}dx-2 \int e^{2 x-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}}dx+4 \int \frac {e^{2 x-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}}}{(x-4)^2}dx-8 \int \frac {e^{2 x-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}}}{x-4}dx+\int \frac {\int \frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}}}{(x-4)^2}dx}{x}dx-\log (x) \int \frac {e^{-\frac {e^{2 x} x+x-2}{x-4}} x^{\frac {1}{x-4}}}{(x-4)^2}dx\) |
Input:
Int[(E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))*(-4 + 5*x + E^(2*x)*(4*x + 8*x^2 - 2*x^3) - x*Log[E^2*x]))/(16*x - 8*x^2 + x^3),x]
Output:
$Aborted
Time = 1.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86
method | result | size |
risch | \({\mathrm e}^{-\frac {-\ln \left ({\mathrm e}^{2} x \right )+x \,{\mathrm e}^{2 x}+x}{x -4}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{2} x \right )-x \,{\mathrm e}^{2 x}-x}{x -4}}\) | \(24\) |
Input:
int((-x*ln(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((ln(exp(2)*x)- x*exp(2*x)-x)/(x-4))/(x^3-8*x^2+16*x),x,method=_RETURNVERBOSE)
Output:
exp(-(-ln(exp(2)*x)+x*exp(2*x)+x)/(x-4))
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=e^{\left (-\frac {x e^{\left (2 \, x\right )} + x - \log \left (x e^{2}\right )}{x - 4}\right )} \] Input:
integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(ex p(2)*x)-x*exp(2*x)-x)/(-4+x))/(x^3-8*x^2+16*x),x, algorithm="fricas")
Output:
e^(-(x*e^(2*x) + x - log(x*e^2))/(x - 4))
Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=e^{\frac {- x e^{2 x} - x + \log {\left (x e^{2} \right )}}{x - 4}} \] Input:
integrate((-x*ln(exp(2)*x)+(-2*x**3+8*x**2+4*x)*exp(2*x)+5*x-4)*exp((ln(ex p(2)*x)-x*exp(2*x)-x)/(-4+x))/(x**3-8*x**2+16*x),x)
Output:
exp((-x*exp(2*x) - x + log(x*exp(2)))/(x - 4))
Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=e^{\left (-\frac {4 \, e^{\left (2 \, x\right )}}{x - 4} + \frac {\log \left (x\right )}{x - 4} - \frac {2}{x - 4} - e^{\left (2 \, x\right )} - 1\right )} \] Input:
integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(ex p(2)*x)-x*exp(2*x)-x)/(-4+x))/(x^3-8*x^2+16*x),x, algorithm="maxima")
Output:
e^(-4*e^(2*x)/(x - 4) + log(x)/(x - 4) - 2/(x - 4) - e^(2*x) - 1)
\[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=\int { -\frac {{\left (2 \, {\left (x^{3} - 4 \, x^{2} - 2 \, x\right )} e^{\left (2 \, x\right )} + x \log \left (x e^{2}\right ) - 5 \, x + 4\right )} e^{\left (-\frac {x e^{\left (2 \, x\right )} + x - \log \left (x e^{2}\right )}{x - 4}\right )}}{x^{3} - 8 \, x^{2} + 16 \, x} \,d x } \] Input:
integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(ex p(2)*x)-x*exp(2*x)-x)/(-4+x))/(x^3-8*x^2+16*x),x, algorithm="giac")
Output:
integrate(-(2*(x^3 - 4*x^2 - 2*x)*e^(2*x) + x*log(x*e^2) - 5*x + 4)*e^(-(x *e^(2*x) + x - log(x*e^2))/(x - 4))/(x^3 - 8*x^2 + 16*x), x)
Time = 4.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=x^{\frac {1}{x-4}}\,{\mathrm {e}}^{-\frac {x}{x-4}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{2\,x}}{x-4}}\,{\mathrm {e}}^{\frac {2}{x-4}} \] Input:
int((exp(-(x - log(x*exp(2)) + x*exp(2*x))/(x - 4))*(5*x + exp(2*x)*(4*x + 8*x^2 - 2*x^3) - x*log(x*exp(2)) - 4))/(16*x - 8*x^2 + x^3),x)
Output:
x^(1/(x - 4))*exp(-x/(x - 4))*exp(-(x*exp(2*x))/(x - 4))*exp(2/(x - 4))
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx=\frac {e^{\frac {\mathrm {log}\left (e^{2} x \right )}{x -4}}}{e^{\frac {e^{2 x} x +4}{x -4}} e} \] Input:
int((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(exp(2)*x )-x*exp(2*x)-x)/(-4+x))/(x^3-8*x^2+16*x),x)
Output:
e**(log(e**2*x)/(x - 4))/(e**((e**(2*x)*x + 4)/(x - 4))*e)