\(\int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx\) [9050]

Optimal result

Integrand size = 63, antiderivative size = 22 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (-x+4 (-16+\log (x)) \left (\log (x)+5 \log ^2(x)\right )^2\right ) \]

[Out]

ln(4*(ln(x)-16)*(ln(x)+5*ln(x)^2)^2-x)

Rubi [F]

\[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx \]

[In]

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]

[Out]

Defer[Int][(x + 64*Log[x]^2 + 636*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5)^(-1), x] + 128*Defer[Int][Log[x]/(x
*(x + 64*Log[x]^2 + 636*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5)), x] + 1908*Defer[Int][Log[x]^2/(x*(x + 64*Lo
g[x]^2 + 636*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5)), x] + 6240*Defer[Int][Log[x]^3/(x*(x + 64*Log[x]^2 + 63
6*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5)), x] - 500*Defer[Int][Log[x]^4/(x*(x + 64*Log[x]^2 + 636*Log[x]^3 +
1560*Log[x]^4 - 100*Log[x]^5)), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)}+\frac {128 \log (x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )}+\frac {1908 \log ^2(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )}+\frac {6240 \log ^3(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )}-\frac {500 \log ^4(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )}\right ) \, dx \\ & = 128 \int \frac {\log (x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )} \, dx-500 \int \frac {\log ^4(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )} \, dx+1908 \int \frac {\log ^2(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )} \, dx+6240 \int \frac {\log ^3(x)}{x \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right )} \, dx+\int \frac {1}{x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (x+64 \log ^2(x)+636 \log ^3(x)+1560 \log ^4(x)-100 \log ^5(x)\right ) \]

[In]

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]

[Out]

Log[x + 64*Log[x]^2 + 636*Log[x]^3 + 1560*Log[x]^4 - 100*Log[x]^5]

Maple [A] (verified)

Time = 0.10 (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	\(\ln \left (100 \ln \left (x \right )^{5}-1560 \ln \left (x \right )^{4}-636 \ln \left (x \right )^{3}-64 \ln \left (x \right )^{2}-x \right )\)	\(30\)

[In]

int((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,method=_RETURNVERBOSE)

[Out]

ln(-100*ln(x)^5+1560*ln(x)^4+636*ln(x)^3+64*ln(x)^2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (100 \, \log \left (x\right )^{5} - 1560 \, \log \left (x\right )^{4} - 636 \, \log \left (x\right )^{3} - 64 \, \log \left (x\right )^{2} - x\right ) \]

[In]

integrate((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, algorithm="fricas")

[Out]

log(100*log(x)^5 - 1560*log(x)^4 - 636*log(x)^3 - 64*log(x)^2 - x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log {\left (- \frac {x}{100} + \log {\left (x \right )}^{5} - \frac {78 \log {\left (x \right )}^{4}}{5} - \frac {159 \log {\left (x \right )}^{3}}{25} - \frac {16 \log {\left (x \right )}^{2}}{25} \right )} \]

[In]

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)

[Out]

log(-x/100 + log(x)**5 - 78*log(x)**4/5 - 159*log(x)**3/25 - 16*log(x)**2/25)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (\log \left (x\right )^{5} - \frac {78}{5} \, \log \left (x\right )^{4} - \frac {159}{25} \, \log \left (x\right )^{3} - \frac {16}{25} \, \log \left (x\right )^{2} - \frac {1}{100} \, x\right ) \]

[In]

integrate((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, algorithm="maxima")

[Out]

log(log(x)^5 - 78/5*log(x)^4 - 159/25*log(x)^3 - 16/25*log(x)^2 - 1/100*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\log \left (100 \, \log \left (x\right )^{5} - 1560 \, \log \left (x\right )^{4} - 636 \, \log \left (x\right )^{3} - 64 \, \log \left (x\right )^{2} - x\right ) \]

[In]

integrate((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, algorithm="giac")

[Out]

log(100*log(x)^5 - 1560*log(x)^4 - 636*log(x)^3 - 64*log(x)^2 - x)

Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x-128 \log (x)-1908 \log ^2(x)-6240 \log ^3(x)+500 \log ^4(x)}{-x^2-64 x \log ^2(x)-636 x \log ^3(x)-1560 x \log ^4(x)+100 x \log ^5(x)} \, dx=\ln \left (-100\,{\ln \left (x\right )}^5+1560\,{\ln \left (x\right )}^4+636\,{\ln \left (x\right )}^3+64\,{\ln \left (x\right )}^2+x\right ) \]

[In]

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*l
og(x)^4 - 100*x*log(x)^5 + x^2),x)

[Out]

log(x + 64*log(x)^2 + 636*log(x)^3 + 1560*log(x)^4 - 100*log(x)^5)