Integrand size = 85, antiderivative size = 24 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {1}{2 x-\frac {\left (3-x-3 e^2 x\right ) \log (x)}{x}} \] Output:
1/(2*x-(-3*exp(2)*x-x+3)*ln(x)/x)
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{2 x^2+\left (-3+x+3 e^2 x\right ) \log (x)} \] Input:
Integrate[(3 - x - 3*E^2*x - 2*x^2 - 3*Log[x])/(4*x^4 + (-12*x^2 + 4*x^3 + 12*E^2*x^3)*Log[x] + (9 - 6*x + x^2 + 9*E^4*x^2 + E^2*(-18*x + 6*x^2))*Lo g[x]^2),x]
Output:
x/(2*x^2 + (-3 + x + 3*E^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.
\(\displaystyle \int \frac {-2 x^2-3 e^2 x-x-3 \log (x)+3}{4 x^4+\left (9 e^4 x^2+x^2+e^2 \left (6 x^2-18 x\right )-6 x+9\right ) \log ^2(x)+\left (12 e^2 x^3+4 x^3-12 x^2\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-2 x^2+\left (-1-3 e^2\right ) x-3 \log (x)+3}{4 x^4+\left (9 e^4 x^2+x^2+e^2 \left (6 x^2-18 x\right )-6 x+9\right ) \log ^2(x)+\left (12 e^2 x^3+4 x^3-12 x^2\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-2 x^2-\left (1+3 e^2\right ) x-3 \log (x)+3}{\left (2 x^2+\left (3 e^2 x+x-3\right ) \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2+\left (1+3 e^2\right ) x \log (x)-3 \log (x)\right )}+\frac {2 \left (1+3 e^2\right ) x^3-\left (11-6 e^2-9 e^4\right ) x^2-6 \left (1+3 e^2\right ) x+9}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2+\left (1+3 e^2\right ) x \log (x)-3 \log (x)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \left (7+6 e^2+9 e^4\right ) \int \frac {1}{\left (2 x^2+\left (1+3 e^2\right ) \log (x) x-3 \log (x)\right )^2}dx}{\left (1+3 e^2\right )^2}+\frac {\left (5-6 e^2-9 e^4\right ) \int \frac {x}{\left (2 x^2+\left (1+3 e^2\right ) \log (x) x-3 \log (x)\right )^2}dx}{1+3 e^2}-2 \int \frac {x^2}{\left (2 x^2+\left (1+3 e^2\right ) \log (x) x-3 \log (x)\right )^2}dx+\frac {54 \int \frac {1}{\left (\left (1+3 e^2\right ) x-3\right ) \left (2 x^2+\left (1+3 e^2\right ) \log (x) x-3 \log (x)\right )^2}dx}{\left (1+3 e^2\right )^2}+3 \int \frac {1}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2+\left (1+3 e^2\right ) \log (x) x-3 \log (x)\right )}dx\) |
Input:
Int[(3 - x - 3*E^2*x - 2*x^2 - 3*Log[x])/(4*x^4 + (-12*x^2 + 4*x^3 + 12*E^ 2*x^3)*Log[x] + (9 - 6*x + x^2 + 9*E^4*x^2 + E^2*(-18*x + 6*x^2))*Log[x]^2 ),x]
Output:
$Aborted
Time = 0.56 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {x}{3 x \,{\mathrm e}^{2} \ln \left (x \right )+x \ln \left (x \right )+2 x^{2}-3 \ln \left (x \right )}\) | \(26\) |
norman | \(\frac {x}{3 x \,{\mathrm e}^{2} \ln \left (x \right )+x \ln \left (x \right )+2 x^{2}-3 \ln \left (x \right )}\) | \(26\) |
risch | \(\frac {x}{3 x \,{\mathrm e}^{2} \ln \left (x \right )+x \ln \left (x \right )+2 x^{2}-3 \ln \left (x \right )}\) | \(26\) |
parallelrisch | \(\frac {x}{3 x \,{\mathrm e}^{2} \ln \left (x \right )+x \ln \left (x \right )+2 x^{2}-3 \ln \left (x \right )}\) | \(26\) |
Input:
int((-3*ln(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+x ^2-6*x+9)*ln(x)^2+(12*x^3*exp(2)+4*x^3-12*x^2)*ln(x)+4*x^4),x,method=_RETU RNVERBOSE)
Output:
x/(3*x*exp(2)*ln(x)+x*ln(x)+2*x^2-3*ln(x))
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{2 \, x^{2} + {\left (3 \, x e^{2} + x - 3\right )} \log \left (x\right )} \] Input:
integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*e xp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*exp(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, al gorithm="fricas")
Output:
x/(2*x^2 + (3*x*e^2 + x - 3)*log(x))
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{2 x^{2} + \left (x + 3 x e^{2} - 3\right ) \log {\left (x \right )}} \] Input:
integrate((-3*ln(x)-3*exp(2)*x-2*x**2-x+3)/((9*x**2*exp(2)**2+(6*x**2-18*x )*exp(2)+x**2-6*x+9)*ln(x)**2+(12*x**3*exp(2)+4*x**3-12*x**2)*ln(x)+4*x**4 ),x)
Output:
x/(2*x**2 + (x + 3*x*exp(2) - 3)*log(x))
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{2 \, x^{2} + {\left (x {\left (3 \, e^{2} + 1\right )} - 3\right )} \log \left (x\right )} \] Input:
integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*e xp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*exp(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, al gorithm="maxima")
Output:
x/(2*x^2 + (x*(3*e^2 + 1) - 3)*log(x))
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{3 \, x e^{2} \log \left (x\right ) + 2 \, x^{2} + x \log \left (x\right ) - 3 \, \log \left (x\right )} \] Input:
integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*e xp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*exp(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, al gorithm="giac")
Output:
x/(3*x*e^2*log(x) + 2*x^2 + x*log(x) - 3*log(x))
Time = 2.56 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{\ln \left (x\right )\,\left (x+3\,x\,{\mathrm {e}}^2-3\right )+2\,x^2} \] Input:
int(-(x + 3*log(x) + 3*x*exp(2) + 2*x^2 - 3)/(log(x)^2*(9*x^2*exp(4) - exp (2)*(18*x - 6*x^2) - 6*x + x^2 + 9) + log(x)*(12*x^3*exp(2) - 12*x^2 + 4*x ^3) + 4*x^4),x)
Output:
x/(log(x)*(x + 3*x*exp(2) - 3) + 2*x^2)
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx=\frac {x}{3 \,\mathrm {log}\left (x \right ) e^{2} x +\mathrm {log}\left (x \right ) x -3 \,\mathrm {log}\left (x \right )+2 x^{2}} \] Input:
int((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+ x^2-6*x+9)*log(x)^2+(12*x^3*exp(2)+4*x^3-12*x^2)*log(x)+4*x^4),x)
Output:
x/(3*log(x)*e**2*x + log(x)*x - 3*log(x) + 2*x**2)