Integrand size = 81, antiderivative size = 23 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=\frac {2}{-2+\frac {5}{x}-x+\log (6)-\frac {3 \log (x)}{x}} \] Output:
2/(ln(6)-3*ln(x)/x-2-x+5/x)
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 x}{-5+2 x+x^2-x \log (6)+3 \log (x)} \] Input:
Integrate[(16 + 2*x^2 - 6*Log[x])/(25 - 20*x - 6*x^2 + 4*x^3 + x^4 + (10*x - 4*x^2 - 2*x^3)*Log[6] + x^2*Log[6]^2 + (-30 + 12*x + 6*x^2 - 6*x*Log[6] )*Log[x] + 9*Log[x]^2),x]
Output:
(-2*x)/(-5 + 2*x + x^2 - x*Log[6] + 3*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.
\(\displaystyle \int \frac {2 x^2-6 \log (x)+16}{x^4+4 x^3-6 x^2+x^2 \log ^2(6)+\left (6 x^2+12 x-6 x \log (6)-30\right ) \log (x)+\left (-2 x^3-4 x^2+10 x\right ) \log (6)-20 x+9 \log ^2(x)+25} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {2 x^2-6 \log (x)+16}{x^4+4 x^3+x^2 \left (\log ^2(6)-6\right )+\left (6 x^2+12 x-6 x \log (6)-30\right ) \log (x)+\left (-2 x^3-4 x^2+10 x\right ) \log (6)-20 x+9 \log ^2(x)+25}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (x^2-3 \log (x)+8\right )}{\left (-x^2+x (\log (6)-2)-3 \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {x^2-3 \log (x)+8}{\left (-x^2-(2-\log (6)) x-3 \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {2 x^2+(2-\log (6)) x+3}{\left (-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5\right )^2}+\frac {1}{-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (3 \int \frac {1}{\left (-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5\right )^2}dx+(2-\log (6)) \int \frac {x}{\left (-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5\right )^2}dx+2 \int \frac {x^2}{\left (-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5\right )^2}dx+\int \frac {1}{-x^2-2 \left (1-\frac {\log (6)}{2}\right ) x-3 \log (x)+5}dx\right )\) |
Input:
Int[(16 + 2*x^2 - 6*Log[x])/(25 - 20*x - 6*x^2 + 4*x^3 + x^4 + (10*x - 4*x ^2 - 2*x^3)*Log[6] + x^2*Log[6]^2 + (-30 + 12*x + 6*x^2 - 6*x*Log[6])*Log[ x] + 9*Log[x]^2),x]
Output:
$Aborted
Time = 0.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\) | \(24\) |
norman | \(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\) | \(24\) |
parallelrisch | \(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\) | \(24\) |
risch | \(\frac {2 x}{x \ln \left (2\right )+x \ln \left (3\right )-x^{2}-2 x -3 \ln \left (x \right )+5}\) | \(28\) |
Input:
int((-6*ln(x)+2*x^2+16)/(9*ln(x)^2+(-6*x*ln(6)+6*x^2+12*x-30)*ln(x)+x^2*ln (6)^2+(-2*x^3-4*x^2+10*x)*ln(6)+x^4+4*x^3-6*x^2-20*x+25),x,method=_RETURNV ERBOSE)
Output:
2*x/(x*ln(6)-x^2-3*ln(x)-2*x+5)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x \log \left (6\right ) + 2 \, x + 3 \, \log \left (x\right ) - 5} \] Input:
integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log (x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, al gorithm="fricas")
Output:
-2*x/(x^2 - x*log(6) + 2*x + 3*log(x) - 5)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=- \frac {2 x}{x^{2} - x \log {\left (6 \right )} + 2 x + 3 \log {\left (x \right )} - 5} \] Input:
integrate((-6*ln(x)+2*x**2+16)/(9*ln(x)**2+(-6*x*ln(6)+6*x**2+12*x-30)*ln( x)+x**2*ln(6)**2+(-2*x**3-4*x**2+10*x)*ln(6)+x**4+4*x**3-6*x**2-20*x+25),x )
Output:
-2*x/(x**2 - x*log(6) + 2*x + 3*log(x) - 5)
Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} + 3 \, \log \left (x\right ) - 5} \] Input:
integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log (x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, al gorithm="maxima")
Output:
-2*x/(x^2 - x*(log(3) + log(2) - 2) + 3*log(x) - 5)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x \log \left (6\right ) + 2 \, x + 3 \, \log \left (x\right ) - 5} \] Input:
integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log (x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, al gorithm="giac")
Output:
-2*x/(x^2 - x*log(6) + 2*x + 3*log(x) - 5)
Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2\,x}{3\,\ln \left (x\right )-x\,\left (\ln \left (6\right )-2\right )+x^2-5} \] Input:
int((2*x^2 - 6*log(x) + 16)/(x^2*log(6)^2 - 20*x + 9*log(x)^2 - log(6)*(4* x^2 - 10*x + 2*x^3) - 6*x^2 + 4*x^3 + x^4 + log(x)*(12*x - 6*x*log(6) + 6* x^2 - 30) + 25),x)
Output:
-(2*x)/(3*log(x) - x*(log(6) - 2) + x^2 - 5)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 x}{3 \,\mathrm {log}\left (x \right )-\mathrm {log}\left (6\right ) x +x^{2}+2 x -5} \] Input:
int((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log(x)+x^ 2*log(6)^2+(-2*x^3-4*x^2+10*x)*log(6)+x^4+4*x^3-6*x^2-20*x+25),x)
Output:
( - 2*x)/(3*log(x) - log(6)*x + x**2 + 2*x - 5)