Integrand size = 63, antiderivative size = 22 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (-x+4 (-16+\log (x)) \left (\log (x)+5 \log ^2(x)\right )^2\right ) \] Output:
ln(4*(ln(x)-16)*(ln(x)+5*ln(x)^2)^2-x)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right ) \] Input:
Integrate[(-x - 128*Log[x] - 1908*Log[x]^2 - 6240*Log[x]^3 + 500*Log[x]^4) /(-x^2 - 64*x*Log[x]^2 - 636*x*Log[x]^3 - 1560*x*Log[x]^4 + 100*x*Log[x]^5 ),x]
Output:
Log[x + 64*Log[x]^2 + 636*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5]
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.
\(\displaystyle \int \frac {-x+500 \log ^4(x)-6240 \log ^3(x)-1908 \log ^2(x)-128 \log (x)}{-x^2+100 x \log ^5(x)-1560 x \log ^4(x)-636 x \log ^3(x)-64 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {500 \log ^4(x)}{x \left (x-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)\right )}+\frac {6240 \log ^3(x)}{x \left (x-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)\right )}+\frac {1908 \log ^2(x)}{x \left (x-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)\right )}+\frac {128 \log (x)}{x \left (x-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)\right )}+\frac {1}{x-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)+x}dx+128 \int \frac {\log (x)}{x \left (-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)+x\right )}dx+1908 \int \frac {\log ^2(x)}{x \left (-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)+x\right )}dx+6240 \int \frac {\log ^3(x)}{x \left (-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)+x\right )}dx-500 \int \frac {\log ^4(x)}{x \left (-100 \log ^5(x)+1560 \log ^4(x)+636 \log ^3(x)+64 \log ^2(x)+x\right )}dx\) |
Input:
Int[(-x - 128*Log[x] - 1908*Log[x]^2 - 6240*Log[x]^3 + 500*Log[x]^4)/(-x^2 - 64*x*Log[x]^2 - 636*x*Log[x]^3 - 1560*x*Log[x]^4 + 100*x*Log[x]^5),x]
Output:
$Aborted
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
norman | \(\ln \left (-100 \ln \left (x \right )^{5}+1560 \ln \left (x \right )^{4}+636 \ln \left (x \right )^{3}+64 \ln \left (x \right )^{2}+x \right )\) | \(28\) |
risch | \(\ln \left (\ln \left (x \right )^{5}-\frac {78 \ln \left (x \right )^{4}}{5}-\frac {159 \ln \left (x \right )^{3}}{25}-\frac {16 \ln \left (x \right )^{2}}{25}-\frac {x}{100}\right )\) | \(28\) |
parallelrisch | \(\ln \left (-100 \ln \left (x \right )^{5}+1560 \ln \left (x \right )^{4}+636 \ln \left (x \right )^{3}+64 \ln \left (x \right )^{2}+x \right )\) | \(28\) |
default | \(\ln \left (100 \ln \left (x \right )^{5}-1560 \ln \left (x \right )^{4}-636 \ln \left (x \right )^{3}-64 \ln \left (x \right )^{2}-x \right )\) | \(30\) |
Input:
int((500*ln(x)^4-6240*ln(x)^3-1908*ln(x)^2-128*ln(x)-x)/(100*x*ln(x)^5-156 0*x*ln(x)^4-636*x*ln(x)^3-64*x*ln(x)^2-x^2),x,method=_RETURNVERBOSE)
Output:
ln(-100*ln(x)^5+1560*ln(x)^4+636*ln(x)^3+64*ln(x)^2+x)
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (100 \, \log \left (x\right )^{5} - 1560 \, \log \left (x\right )^{4} - 636 \, \log \left (x\right )^{3} - 64 \, \log \left (x\right )^{2} - x\right ) \] Input:
integrate((500*log(x)^4-6240*log(x)^3-1908*log(x)^2-128*log(x)-x)/(100*x*l og(x)^5-1560*x*log(x)^4-636*x*log(x)^3-64*x*log(x)^2-x^2),x, algorithm="fr icas")
Output:
log(100*log(x)^5 - 1560*log(x)^4 - 636*log(x)^3 - 64*log(x)^2 - x)
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log {\left (- \frac {x}{100} + \log {\left (x \right )}^{5} - \frac {78 \log {\left (x \right )}^{4}}{5} - \frac {159 \log {\left (x \right )}^{3}}{25} - \frac {16 \log {\left (x \right )}^{2}}{25} \right )} \] Input:
integrate((500*ln(x)**4-6240*ln(x)**3-1908*ln(x)**2-128*ln(x)-x)/(100*x*ln (x)**5-1560*x*ln(x)**4-636*x*ln(x)**3-64*x*ln(x)**2-x**2),x)
Output:
log(-x/100 + log(x)**5 - 78*log(x)**4/5 - 159*log(x)**3/25 - 16*log(x)**2/ 25)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (\log \left (x\right )^{5} - \frac {78}{5} \, \log \left (x\right )^{4} - \frac {159}{25} \, \log \left (x\right )^{3} - \frac {16}{25} \, \log \left (x\right )^{2} - \frac {1}{100} \, x\right ) \] Input:
integrate((500*log(x)^4-6240*log(x)^3-1908*log(x)^2-128*log(x)-x)/(100*x*l og(x)^5-1560*x*log(x)^4-636*x*log(x)^3-64*x*log(x)^2-x^2),x, algorithm="ma xima")
Output:
log(log(x)^5 - 78/5*log(x)^4 - 159/25*log(x)^3 - 16/25*log(x)^2 - 1/100*x)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (100 \, \log \left (x\right )^{5} - 1560 \, \log \left (x\right )^{4} - 636 \, \log \left (x\right )^{3} - 64 \, \log \left (x\right )^{2} - x\right ) \] Input:
integrate((500*log(x)^4-6240*log(x)^3-1908*log(x)^2-128*log(x)-x)/(100*x*l og(x)^5-1560*x*log(x)^4-636*x*log(x)^3-64*x*log(x)^2-x^2),x, algorithm="gi ac")
Output:
log(100*log(x)^5 - 1560*log(x)^4 - 636*log(x)^3 - 64*log(x)^2 - x)
Time = 4.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\ln \left (-100\,{\ln \left (x\right )}^5+1560\,{\ln \left (x\right )}^4+636\,{\ln \left (x\right )}^3+64\,{\ln \left (x\right )}^2+x\right ) \] Input:
int((x + 128*log(x) + 1908*log(x)^2 + 6240*log(x)^3 - 500*log(x)^4)/(64*x* log(x)^2 + 636*x*log(x)^3 + 1560*x*log(x)^4 - 100*x*log(x)^5 + x^2),x)
Output:
log(x + 64*log(x)^2 + 636*log(x)^3 + 1560*log(x)^4 - 100*log(x)^5)
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\mathrm {log}\left (100 \mathrm {log}\left (x \right )^{5}-1560 \mathrm {log}\left (x \right )^{4}-636 \mathrm {log}\left (x \right )^{3}-64 \mathrm {log}\left (x \right )^{2}-x \right ) \] Input:
int((500*log(x)^4-6240*log(x)^3-1908*log(x)^2-128*log(x)-x)/(100*x*log(x)^ 5-1560*x*log(x)^4-636*x*log(x)^3-64*x*log(x)^2-x^2),x)
Output:
log(100*log(x)**5 - 1560*log(x)**4 - 636*log(x)**3 - 64*log(x)**2 - x)