Integrand size = 76, antiderivative size = 24 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{2+\frac {-1+x+\frac {\log (x)}{2 x-\log (x)}}{x}} \] Output:
exp(2+(x+ln(x)/(2*x-ln(x))-1)/x)
Time = 2.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{3-\frac {2}{x}-\frac {2}{-2 x+\log (x)}} \] Input:
Integrate[(E^((2*x - 6*x^2 + (-2 + 3*x)*Log[x])/(-2*x^2 + x*Log[x]))*(2*x + 4*x^2 - 8*x*Log[x] + 2*Log[x]^2))/(4*x^4 - 4*x^3*Log[x] + x^2*Log[x]^2), x]
Output:
E^(3 - 2/x - 2/(-2*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.
\(\displaystyle \int \frac {\left (4 x^2+2 x+2 \log ^2(x)-8 x \log (x)\right ) \exp \left (\frac {-6 x^2+2 x+(3 x-2) \log (x)}{x \log (x)-2 x^2}\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (2 x^2+x+\log ^2(x)-4 x \log (x)\right ) \exp \left (\frac {-6 x^2+2 x+(3 x-2) \log (x)}{x \log (x)-2 x^2}\right )}{x^2 (2 x-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{-\frac {2 \left (x-3 x^2\right )}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}-2} \left (2 x^2-4 \log (x) x+x+\log ^2(x)\right )}{(2 x-\log (x))^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle 2 \int \frac {e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}-2} \left (2 x^2-4 \log (x) x+x+\log ^2(x)\right )}{(2 x-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}-2}+\frac {e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} (1-2 x) x^{\frac {2-3 x}{2 x^2-x \log (x)}-1}}{(2 x-\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\int e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}-2}dx+\int \frac {e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}-1}}{(2 x-\log (x))^2}dx-2 \int \frac {e^{-\frac {2 (1-3 x) x}{2 x^2-x \log (x)}} x^{\frac {2-3 x}{2 x^2-x \log (x)}}}{(2 x-\log (x))^2}dx\right )\) |
Input:
Int[(E^((2*x - 6*x^2 + (-2 + 3*x)*Log[x])/(-2*x^2 + x*Log[x]))*(2*x + 4*x^ 2 - 8*x*Log[x] + 2*Log[x]^2))/(4*x^4 - 4*x^3*Log[x] + x^2*Log[x]^2),x]
Output:
$Aborted
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-2+3 x \right ) \ln \left (x \right )-6 x^{2}+2 x}{x \left (\ln \left (x \right )-2 x \right )}}\) | \(31\) |
risch | \({\mathrm e}^{\frac {3 x \ln \left (x \right )-6 x^{2}-2 \ln \left (x \right )+2 x}{x \left (\ln \left (x \right )-2 x \right )}}\) | \(32\) |
Input:
int((2*ln(x)^2-8*x*ln(x)+4*x^2+2*x)*exp(((-2+3*x)*ln(x)-6*x^2+2*x)/(x*ln(x )-2*x^2))/(x^2*ln(x)^2-4*x^3*ln(x)+4*x^4),x,method=_RETURNVERBOSE)
Output:
exp(((-2+3*x)*ln(x)-6*x^2+2*x)/x/(ln(x)-2*x))
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {6 \, x^{2} - {\left (3 \, x - 2\right )} \log \left (x\right ) - 2 \, x}{2 \, x^{2} - x \log \left (x\right )}\right )} \] Input:
integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x )/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*x^3*log(x)+4*x^4),x, algorithm="fricas ")
Output:
e^((6*x^2 - (3*x - 2)*log(x) - 2*x)/(2*x^2 - x*log(x)))
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\frac {- 6 x^{2} + 2 x + \left (3 x - 2\right ) \log {\left (x \right )}}{- 2 x^{2} + x \log {\left (x \right )}}} \] Input:
integrate((2*ln(x)**2-8*x*ln(x)+4*x**2+2*x)*exp(((-2+3*x)*ln(x)-6*x**2+2*x )/(x*ln(x)-2*x**2))/(x**2*ln(x)**2-4*x**3*ln(x)+4*x**4),x)
Output:
exp((-6*x**2 + 2*x + (3*x - 2)*log(x))/(-2*x**2 + x*log(x)))
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {2}{2 \, x - \log \left (x\right )} - \frac {2}{x} + 3\right )} \] Input:
integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x )/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*x^3*log(x)+4*x^4),x, algorithm="maxima ")
Output:
e^(2/(2*x - log(x)) - 2/x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (23) = 46\).
Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {6 \, x^{2}}{2 \, x^{2} - x \log \left (x\right )} - \frac {3 \, x \log \left (x\right )}{2 \, x^{2} - x \log \left (x\right )} - \frac {2 \, x}{2 \, x^{2} - x \log \left (x\right )} + \frac {2 \, \log \left (x\right )}{2 \, x^{2} - x \log \left (x\right )}\right )} \] Input:
integrate((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x )/(x*log(x)-2*x^2))/(x^2*log(x)^2-4*x^3*log(x)+4*x^4),x, algorithm="giac")
Output:
e^(6*x^2/(2*x^2 - x*log(x)) - 3*x*log(x)/(2*x^2 - x*log(x)) - 2*x/(2*x^2 - x*log(x)) + 2*log(x)/(2*x^2 - x*log(x)))
Time = 3.45 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=x^{\frac {3\,x-2}{x\,\ln \left (x\right )-2\,x^2}}\,{\mathrm {e}}^{-\frac {6\,x^2}{x\,\ln \left (x\right )-2\,x^2}}\,{\mathrm {e}}^{\frac {2\,x}{x\,\ln \left (x\right )-2\,x^2}} \] Input:
int((exp((2*x + log(x)*(3*x - 2) - 6*x^2)/(x*log(x) - 2*x^2))*(2*x + 2*log (x)^2 - 8*x*log(x) + 4*x^2))/(x^2*log(x)^2 - 4*x^3*log(x) + 4*x^4),x)
Output:
x^((3*x - 2)/(x*log(x) - 2*x^2))*exp(-(6*x^2)/(x*log(x) - 2*x^2))*exp((2*x )/(x*log(x) - 2*x^2))
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {2 x-6 x^2+(-2+3 x) \log (x)}{-2 x^2+x \log (x)}} \left (2 x+4 x^2-8 x \log (x)+2 \log ^2(x)\right )}{4 x^4-4 x^3 \log (x)+x^2 \log ^2(x)} \, dx=\frac {e^{\frac {2}{\mathrm {log}\left (x \right )-2 x}} e^{3}}{e^{\frac {2 \,\mathrm {log}\left (x \right )}{\mathrm {log}\left (x \right ) x -2 x^{2}}}} \] Input:
int((2*log(x)^2-8*x*log(x)+4*x^2+2*x)*exp(((-2+3*x)*log(x)-6*x^2+2*x)/(x*l og(x)-2*x^2))/(x^2*log(x)^2-4*x^3*log(x)+4*x^4),x)
Output:
(e**(2/(log(x) - 2*x))*e**3)/e**((2*log(x))/(log(x)*x - 2*x**2))