Integrand size = 72, antiderivative size = 25 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=-\frac {1}{3} \log \left (\frac {(-3+x)^2 x}{e^{\frac {e^x}{x}}+x}\right ) \] Output:
-1/3*ln(x*(-3+x)^2/(x+exp(exp(x)/x)))
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=\frac {1}{3} \left (-2 \log (3-x)-\log (x)+\log \left (e^{\frac {e^x}{x}}+x\right )\right ) \] Input:
Integrate[(-2*x^3 + E^(E^x/x)*(3*x - 3*x^2 + E^x*(3 - 4*x + x^2)))/(-9*x^3 + 3*x^4 + E^(E^x/x)*(-9*x^2 + 3*x^3)),x]
Output:
(-2*Log[3 - x] - Log[x] + Log[E^(E^x/x) + x])/3
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.
\(\displaystyle \int \frac {e^{\frac {e^x}{x}} \left (-3 x^2+e^x \left (x^2-4 x+3\right )+3 x\right )-2 x^3}{3 x^4-9 x^3+e^{\frac {e^x}{x}} \left (3 x^3-9 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^3-e^{\frac {e^x}{x}} \left (-3 x^2+e^x \left (x^2-4 x+3\right )+3 x\right )}{3 (3-x) x^2 \left (x+e^{\frac {e^x}{x}}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {2 x^3-e^{\frac {e^x}{x}} \left (-3 x^2+3 x+e^x \left (x^2-4 x+3\right )\right )}{(3-x) x^2 \left (x+e^{\frac {e^x}{x}}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {e^{x+\frac {e^x}{x}} (x-1)}{x^2 \left (x+e^{\frac {e^x}{x}}\right )}-\frac {2 x^2+3 e^{\frac {e^x}{x}} x-3 e^{\frac {e^x}{x}}}{(x-3) x \left (x+e^{\frac {e^x}{x}}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {e^{x+\frac {e^x}{x}}}{x^2 \left (x+e^{\frac {e^x}{x}}\right )}dx+\int \frac {1}{x+e^{\frac {e^x}{x}}}dx+\int \frac {e^{x+\frac {e^x}{x}}}{x \left (x+e^{\frac {e^x}{x}}\right )}dx-2 \log (3-x)-\log (x)\right )\) |
Input:
Int[(-2*x^3 + E^(E^x/x)*(3*x - 3*x^2 + E^x*(3 - 4*x + x^2)))/(-9*x^3 + 3*x ^4 + E^(E^x/x)*(-9*x^2 + 3*x^3)),x]
Output:
$Aborted
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
norman | \(-\frac {\ln \left (x \right )}{3}-\frac {2 \ln \left (-3+x \right )}{3}+\frac {\ln \left (x +{\mathrm e}^{\frac {{\mathrm e}^{x}}{x}}\right )}{3}\) | \(24\) |
risch | \(-\frac {\ln \left (x \right )}{3}-\frac {2 \ln \left (-3+x \right )}{3}+\frac {\ln \left (x +{\mathrm e}^{\frac {{\mathrm e}^{x}}{x}}\right )}{3}\) | \(24\) |
parallelrisch | \(-\frac {\ln \left (x \right )}{3}-\frac {2 \ln \left (-3+x \right )}{3}+\frac {\ln \left (x +{\mathrm e}^{\frac {{\mathrm e}^{x}}{x}}\right )}{3}\) | \(24\) |
Input:
int((((x^2-4*x+3)*exp(x)-3*x^2+3*x)*exp(exp(x)/x)-2*x^3)/((3*x^3-9*x^2)*ex p(exp(x)/x)+3*x^4-9*x^3),x,method=_RETURNVERBOSE)
Output:
-1/3*ln(x)-2/3*ln(-3+x)+1/3*ln(x+exp(exp(x)/x))
Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=\frac {1}{3} \, \log \left (x + e^{\left (\frac {e^{x}}{x}\right )}\right ) - \frac {2}{3} \, \log \left (x - 3\right ) - \frac {1}{3} \, \log \left (x\right ) \] Input:
integrate((((x^2-4*x+3)*exp(x)-3*x^2+3*x)*exp(exp(x)/x)-2*x^3)/((3*x^3-9*x ^2)*exp(exp(x)/x)+3*x^4-9*x^3),x, algorithm="fricas")
Output:
1/3*log(x + e^(e^x/x)) - 2/3*log(x - 3) - 1/3*log(x)
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=- \frac {\log {\left (x \right )}}{3} - \frac {2 \log {\left (x - 3 \right )}}{3} + \frac {\log {\left (x + e^{\frac {e^{x}}{x}} \right )}}{3} \] Input:
integrate((((x**2-4*x+3)*exp(x)-3*x**2+3*x)*exp(exp(x)/x)-2*x**3)/((3*x**3 -9*x**2)*exp(exp(x)/x)+3*x**4-9*x**3),x)
Output:
-log(x)/3 - 2*log(x - 3)/3 + log(x + exp(exp(x)/x))/3
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=\frac {1}{3} \, \log \left (x + e^{\left (\frac {e^{x}}{x}\right )}\right ) - \frac {2}{3} \, \log \left (x - 3\right ) - \frac {1}{3} \, \log \left (x\right ) \] Input:
integrate((((x^2-4*x+3)*exp(x)-3*x^2+3*x)*exp(exp(x)/x)-2*x^3)/((3*x^3-9*x ^2)*exp(exp(x)/x)+3*x^4-9*x^3),x, algorithm="maxima")
Output:
1/3*log(x + e^(e^x/x)) - 2/3*log(x - 3) - 1/3*log(x)
Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=-\frac {1}{3} \, x + \frac {1}{3} \, \log \left (x e^{x} + e^{\left (\frac {x^{2} + e^{x}}{x}\right )}\right ) - \frac {2}{3} \, \log \left (x - 3\right ) - \frac {1}{3} \, \log \left (x\right ) \] Input:
integrate((((x^2-4*x+3)*exp(x)-3*x^2+3*x)*exp(exp(x)/x)-2*x^3)/((3*x^3-9*x ^2)*exp(exp(x)/x)+3*x^4-9*x^3),x, algorithm="giac")
Output:
-1/3*x + 1/3*log(x*e^x + e^((x^2 + e^x)/x)) - 2/3*log(x - 3) - 1/3*log(x)
Time = 7.47 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=\frac {\ln \left (x+{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}\right )}{3}-\frac {2\,\ln \left (x-3\right )}{3}-\frac {\ln \left (x\right )}{3} \] Input:
int(-(exp(exp(x)/x)*(3*x + exp(x)*(x^2 - 4*x + 3) - 3*x^2) - 2*x^3)/(exp(e xp(x)/x)*(9*x^2 - 3*x^3) + 9*x^3 - 3*x^4),x)
Output:
log(x + exp(exp(x)/x))/3 - (2*log(x - 3))/3 - log(x)/3
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^3+e^{\frac {e^x}{x}} \left (3 x-3 x^2+e^x \left (3-4 x+x^2\right )\right )}{-9 x^3+3 x^4+e^{\frac {e^x}{x}} \left (-9 x^2+3 x^3\right )} \, dx=\frac {\mathrm {log}\left (e^{\frac {e^{x}}{x}}+x \right )}{3}-\frac {2 \,\mathrm {log}\left (x -3\right )}{3}-\frac {\mathrm {log}\left (x \right )}{3} \] Input:
int((((x^2-4*x+3)*exp(x)-3*x^2+3*x)*exp(exp(x)/x)-2*x^3)/((3*x^3-9*x^2)*ex p(exp(x)/x)+3*x^4-9*x^3),x)
Output:
(log(e**(e**x/x) + x) - 2*log(x - 3) - log(x))/3