Integrand size = 60, antiderivative size = 24 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=-2+\frac {e^{-4+x}}{\left (1+\frac {1}{-2+x}\right ) x \log (x)} \] Output:
exp(-1+x)/exp(3)/x/ln(x)/(1/(-2+x)+1)-2
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {e^{-4+x} (-2+x)}{(-1+x) x \log (x)} \] Input:
Integrate[(E^(-1 + x)*(-2 + 3*x - x^2) + E^(-1 + x)*(-2 + 6*x - 4*x^2 + x^ 3)*Log[x])/(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^2),x]
Output:
(E^(-4 + x)*(-2 + x))/((-1 + x)*x*Log[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.
\(\displaystyle \int \frac {e^{x-1} \left (-x^2+3 x-2\right )+e^{x-1} \left (x^3-4 x^2+6 x-2\right ) \log (x)}{e^3 \left (x^4-2 x^3+x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{x-1} \left (x^2-3 x+2\right )+e^{x-1} \left (-x^3+4 x^2-6 x+2\right ) \log (x)}{\left (x^4-2 x^3+x^2\right ) \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{x-1} \left (x^2-3 x+2\right )+e^{x-1} \left (-x^3+4 x^2-6 x+2\right ) \log (x)}{\left (x^4-2 x^3+x^2\right ) \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\frac {\int \frac {e^{x-1} \left (x^2-3 x+2\right )+e^{x-1} \left (-x^3+4 x^2-6 x+2\right ) \log (x)}{x^2 \left (x^2-2 x+1\right ) \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle -\frac {4 \int \frac {e^{x-1} \left (x^2-3 x+2\right )+e^{x-1} \left (-x^3+4 x^2-6 x+2\right ) \log (x)}{4 (1-x)^2 x^2 \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {e^{x-1} \left (x^2-3 x+2\right )+e^{x-1} \left (-x^3+4 x^2-6 x+2\right ) \log (x)}{(1-x)^2 x^2 \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {\int \frac {e^{x-1} \left (-\log (x) x^3+4 \log (x) x^2+x^2-6 \log (x) x-3 x+2 \log (x)+2\right )}{(1-x)^2 x^2 \log ^2(x)}dx}{e^3}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {e^{x-1} (x-2)}{(x-1) x^2 \log ^2(x)}+\frac {e^{x-1} \left (-x^3+4 x^2-6 x+2\right )}{(x-1)^2 x^2 \log (x)}\right )dx}{e^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \int \frac {e^{x-1}}{x^2 \log ^2(x)}dx+2 \int \frac {e^{x-1}}{x^2 \log (x)}dx-\int \frac {e^{x-1}}{(x-1) \log ^2(x)}dx+\int \frac {e^{x-1}}{x \log ^2(x)}dx-\int \frac {e^{x-1}}{(x-1)^2 \log (x)}dx+\int \frac {e^{x-1}}{(x-1) \log (x)}dx-2 \int \frac {e^{x-1}}{x \log (x)}dx}{e^3}\) |
Input:
Int[(E^(-1 + x)*(-2 + 3*x - x^2) + E^(-1 + x)*(-2 + 6*x - 4*x^2 + x^3)*Log [x])/(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^2),x]
Output:
$Aborted
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {\left (-2+x \right ) {\mathrm e}^{x -4}}{\ln \left (x \right ) \left (-1+x \right ) x}\) | \(21\) |
parallelrisch | \(\frac {{\mathrm e}^{-3} \left (x \,{\mathrm e}^{-1+x}-2 \,{\mathrm e}^{-1+x}\right )}{x \ln \left (x \right ) \left (-1+x \right )}\) | \(31\) |
Input:
int(((x^3-4*x^2+6*x-2)*exp(-1+x)*ln(x)+(-x^2+3*x-2)*exp(-1+x))/(x^4-2*x^3+ x^2)/exp(3)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
1/ln(x)/(-1+x)*(-2+x)/x*exp(x-4)
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {{\left (x - 2\right )} e^{\left (x - 4\right )}}{{\left (x^{2} - x\right )} \log \left (x\right )} \] Input:
integrate(((x^3-4*x^2+6*x-2)*exp(-1+x)*log(x)+(-x^2+3*x-2)*exp(-1+x))/(x^4 -2*x^3+x^2)/exp(3)/log(x)^2,x, algorithm="fricas")
Output:
(x - 2)*e^(x - 4)/((x^2 - x)*log(x))
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\left (x - 2\right ) e^{x - 1}}{x^{2} e^{3} \log {\left (x \right )} - x e^{3} \log {\left (x \right )}} \] Input:
integrate(((x**3-4*x**2+6*x-2)*exp(-1+x)*ln(x)+(-x**2+3*x-2)*exp(-1+x))/(x **4-2*x**3+x**2)/exp(3)/ln(x)**2,x)
Output:
(x - 2)*exp(x - 1)/(x**2*exp(3)*log(x) - x*exp(3)*log(x))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {{\left (x - 2\right )} e^{\left (x - 3\right )}}{{\left (x^{2} e - x e\right )} \log \left (x\right )} \] Input:
integrate(((x^3-4*x^2+6*x-2)*exp(-1+x)*log(x)+(-x^2+3*x-2)*exp(-1+x))/(x^4 -2*x^3+x^2)/exp(3)/log(x)^2,x, algorithm="maxima")
Output:
(x - 2)*e^(x - 3)/((x^2*e - x*e)*log(x))
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {{\left (x e^{x} - 2 \, e^{x}\right )} e^{\left (-3\right )}}{x^{2} e \log \left (x\right ) - x e \log \left (x\right )} \] Input:
integrate(((x^3-4*x^2+6*x-2)*exp(-1+x)*log(x)+(-x^2+3*x-2)*exp(-1+x))/(x^4 -2*x^3+x^2)/exp(3)/log(x)^2,x, algorithm="giac")
Output:
(x*e^x - 2*e^x)*e^(-3)/(x^2*e*log(x) - x*e*log(x))
Time = 2.77 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{x-4}\,\left (x-2\right )}{x\,\ln \left (x\right )\,\left (x-1\right )} \] Input:
int(-(exp(-3)*(exp(x - 1)*(x^2 - 3*x + 2) - exp(x - 1)*log(x)*(6*x - 4*x^2 + x^3 - 2)))/(log(x)^2*(x^2 - 2*x^3 + x^4)),x)
Output:
(exp(x - 4)*(x - 2))/(x*log(x)*(x - 1))
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {e^{x} \left (x -2\right )}{\mathrm {log}\left (x \right ) e^{4} x \left (x -1\right )} \] Input:
int(((x^3-4*x^2+6*x-2)*exp(-1+x)*log(x)+(-x^2+3*x-2)*exp(-1+x))/(x^4-2*x^3 +x^2)/exp(3)/log(x)^2,x)
Output:
(e**x*(x - 2))/(log(x)*e**4*x*(x - 1))