Integrand size = 53, antiderivative size = 23 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4 x}{5 \left (x+x \left (-21 x+\frac {x^2}{\log (x)}\right )\right )} \] Output:
4/5/(x+(x^2/ln(x)-21*x)*x)*x
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4 \log (x)}{5 \left (x^2+\log (x)-21 x \log (x)\right )} \] Input:
Integrate[(4*x - 8*x*Log[x] + 84*Log[x]^2)/(5*x^4 + (10*x^2 - 210*x^3)*Log [x] + (5 - 210*x + 2205*x^2)*Log[x]^2),x]
Output:
(4*Log[x])/(5*(x^2 + Log[x] - 21*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.
\(\displaystyle \int \frac {4 x+84 \log ^2(x)-8 x \log (x)}{5 x^4+\left (2205 x^2-210 x+5\right ) \log ^2(x)+\left (10 x^2-210 x^3\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \left (x+21 \log ^2(x)-2 x \log (x)\right )}{5 \left (x^2-21 x \log (x)+\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{5} \int \frac {21 \log ^2(x)-2 x \log (x)+x}{\left (x^2-21 \log (x) x+\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4}{5} \int \left (-\frac {2 x}{(21 x-1)^2 \left (x^2-21 \log (x) x+\log (x)\right )}+\frac {\left (-21 x^3+443 x^2-42 x+1\right ) x}{(1-21 x)^2 \left (x^2-21 \log (x) x+\log (x)\right )^2}+\frac {21}{(21 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{5} \left (\frac {\int \frac {1}{\left (x^2-21 \log (x) x+\log (x)\right )^2}dx}{9261}+\int \frac {x}{\left (x^2-21 \log (x) x+\log (x)\right )^2}dx-\frac {1}{21} \int \frac {x^2}{\left (x^2-21 \log (x) x+\log (x)\right )^2}dx+\frac {\int \frac {1}{(21 x-1)^2 \left (x^2-21 \log (x) x+\log (x)\right )^2}dx}{9261}+\frac {2 \int \frac {1}{(21 x-1) \left (x^2-21 \log (x) x+\log (x)\right )^2}dx}{9261}-\frac {2}{21} \int \frac {1}{(21 x-1)^2 \left (x^2-21 \log (x) x+\log (x)\right )}dx-\frac {2}{21} \int \frac {1}{(21 x-1) \left (x^2-21 \log (x) x+\log (x)\right )}dx+\frac {1}{1-21 x}\right )\) |
Input:
Int[(4*x - 8*x*Log[x] + 84*Log[x]^2)/(5*x^4 + (10*x^2 - 210*x^3)*Log[x] + (5 - 210*x + 2205*x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
norman | \(\frac {4 \ln \left (x \right )}{5 \left (x^{2}-21 x \ln \left (x \right )+\ln \left (x \right )\right )}\) | \(18\) |
parallelrisch | \(\frac {4 \ln \left (x \right )}{5 \left (x^{2}-21 x \ln \left (x \right )+\ln \left (x \right )\right )}\) | \(18\) |
default | \(-\frac {4 \ln \left (x \right )}{5 \left (21 x \ln \left (x \right )-x^{2}-\ln \left (x \right )\right )}\) | \(22\) |
risch | \(-\frac {4}{5 \left (21 x -1\right )}+\frac {4 x^{2}}{5 \left (21 x -1\right ) \left (x^{2}-21 x \ln \left (x \right )+\ln \left (x \right )\right )}\) | \(36\) |
Input:
int((84*ln(x)^2-8*x*ln(x)+4*x)/((2205*x^2-210*x+5)*ln(x)^2+(-210*x^3+10*x^ 2)*ln(x)+5*x^4),x,method=_RETURNVERBOSE)
Output:
4/5*ln(x)/(x^2-21*x*ln(x)+ln(x))
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, \log \left (x\right )}{5 \, {\left (x^{2} - {\left (21 \, x - 1\right )} \log \left (x\right )\right )}} \] Input:
integrate((84*log(x)^2-8*x*log(x)+4*x)/((2205*x^2-210*x+5)*log(x)^2+(-210* x^3+10*x^2)*log(x)+5*x^4),x, algorithm="fricas")
Output:
4/5*log(x)/(x^2 - (21*x - 1)*log(x))
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=- \frac {4 x^{2}}{- 105 x^{3} + 5 x^{2} + \left (2205 x^{2} - 210 x + 5\right ) \log {\left (x \right )}} - \frac {84}{2205 x - 105} \] Input:
integrate((84*ln(x)**2-8*x*ln(x)+4*x)/((2205*x**2-210*x+5)*ln(x)**2+(-210* x**3+10*x**2)*ln(x)+5*x**4),x)
Output:
-4*x**2/(-105*x**3 + 5*x**2 + (2205*x**2 - 210*x + 5)*log(x)) - 84/(2205*x - 105)
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, \log \left (x\right )}{5 \, {\left (x^{2} - {\left (21 \, x - 1\right )} \log \left (x\right )\right )}} \] Input:
integrate((84*log(x)^2-8*x*log(x)+4*x)/((2205*x^2-210*x+5)*log(x)^2+(-210* x^3+10*x^2)*log(x)+5*x^4),x, algorithm="maxima")
Output:
4/5*log(x)/(x^2 - (21*x - 1)*log(x))
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, x^{2}}{5 \, {\left (21 \, x^{3} - 441 \, x^{2} \log \left (x\right ) - x^{2} + 42 \, x \log \left (x\right ) - \log \left (x\right )\right )}} - \frac {4}{5 \, {\left (21 \, x - 1\right )}} \] Input:
integrate((84*log(x)^2-8*x*log(x)+4*x)/((2205*x^2-210*x+5)*log(x)^2+(-210* x^3+10*x^2)*log(x)+5*x^4),x, algorithm="giac")
Output:
4/5*x^2/(21*x^3 - 441*x^2*log(x) - x^2 + 42*x*log(x) - log(x)) - 4/5/(21*x - 1)
Time = 2.91 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=\frac {4\,\ln \left (x\right )}{5\,\left (\ln \left (x\right )-21\,x\,\ln \left (x\right )+x^2\right )} \] Input:
int((4*x + 84*log(x)^2 - 8*x*log(x))/(log(x)*(10*x^2 - 210*x^3) + log(x)^2 *(2205*x^2 - 210*x + 5) + 5*x^4),x)
Output:
(4*log(x))/(5*(log(x) - 21*x*log(x) + x^2))
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-8 x \log (x)+84 \log ^2(x)}{5 x^4+\left (10 x^2-210 x^3\right ) \log (x)+\left (5-210 x+2205 x^2\right ) \log ^2(x)} \, dx=-\frac {4 \,\mathrm {log}\left (x \right )}{105 \,\mathrm {log}\left (x \right ) x -5 \,\mathrm {log}\left (x \right )-5 x^{2}} \] Input:
int((84*log(x)^2-8*x*log(x)+4*x)/((2205*x^2-210*x+5)*log(x)^2+(-210*x^3+10 *x^2)*log(x)+5*x^4),x)
Output:
( - 4*log(x))/(5*(21*log(x)*x - log(x) - x**2))