Integrand size = 89, antiderivative size = 27 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {1}{5} \left (x+\frac {5 x \left (x+x^4+\log (4)\right )}{(1+x) \log (-x)}\right ) \] Output:
x/(1+x)/ln(-x)*(2*ln(2)+x^4+x)+1/5*x
Time = 0.93 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {x}{5}+\frac {x \left (x+x^4+\log (4)\right )}{(1+x) \log (-x)} \] Input:
Integrate[(-5*x - 5*x^2 - 5*x^4 - 5*x^5 + (-5 - 5*x)*Log[4] + (10*x + 5*x^ 2 + 25*x^4 + 20*x^5 + 5*Log[4])*Log[-x] + (1 + 2*x + x^2)*Log[-x]^2)/((5 + 10*x + 5*x^2)*Log[-x]^2),x]
Output:
x/5 + (x*(x + x^4 + Log[4]))/((1 + 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.
\(\displaystyle \int \frac {-5 x^5-5 x^4-5 x^2+\left (x^2+2 x+1\right ) \log ^2(-x)+\left (20 x^5+25 x^4+5 x^2+10 x+5 \log (4)\right ) \log (-x)-5 x+(-5 x-5) \log (4)}{\left (5 x^2+10 x+5\right ) \log ^2(-x)} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-5 x^5-5 x^4-5 x^2+\left (x^2+2 x+1\right ) \log ^2(-x)+\left (20 x^5+25 x^4+5 x^2+10 x+5 \log (4)\right ) \log (-x)-5 x+(-5 x-5) \log (4)}{\left (\sqrt {5} x+\sqrt {5}\right )^2 \log ^2(-x)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {x^4+x+\log (4)}{(x+1) \log ^2(-x)}+\frac {4 x^5+5 x^4+x^2+2 x+\log (4)}{(x+1)^2 \log (-x)}+\frac {1}{5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {x^4+x+\log (4)}{(x+1) \log ^2(-x)}dx+\int \frac {4 x^5+5 x^4+x^2+2 x+\log (4)}{(x+1)^2 \log (-x)}dx+\frac {x}{5}\) |
Input:
Int[(-5*x - 5*x^2 - 5*x^4 - 5*x^5 + (-5 - 5*x)*Log[4] + (10*x + 5*x^2 + 25 *x^4 + 20*x^5 + 5*Log[4])*Log[-x] + (1 + 2*x + x^2)*Log[-x]^2)/((5 + 10*x + 5*x^2)*Log[-x]^2),x]
Output:
$Aborted
Time = 13.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x \left (2 \ln \left (2\right )+x^{4}+x \right )}{\left (1+x \right ) \ln \left (-x \right )}+\frac {x}{5}\) | \(27\) |
norman | \(\frac {x^{2}+x^{5}-\frac {\ln \left (-x \right )}{5}+2 x \ln \left (2\right )+\frac {\ln \left (-x \right ) x^{2}}{5}}{\left (1+x \right ) \ln \left (-x \right )}\) | \(40\) |
parallelrisch | \(\frac {5 x^{5}+\ln \left (-x \right ) x^{2}+10 x \ln \left (2\right )+5 x^{2}-\ln \left (-x \right )}{5 \ln \left (-x \right ) \left (1+x \right )}\) | \(44\) |
derivativedivides | \(\frac {x}{5}-2 \ln \left (2\right ) \left (-\frac {1}{\ln \left (-x \right )}-\frac {1}{\left (-1-x \right ) \ln \left (-x \right )}\right )+\frac {x^{4}}{\ln \left (-x \right )}-\frac {x^{3}}{\ln \left (-x \right )}+\frac {x^{2}}{\ln \left (-x \right )}\) | \(64\) |
default | \(\frac {x}{5}-2 \ln \left (2\right ) \left (-\frac {1}{\ln \left (-x \right )}-\frac {1}{\left (-1-x \right ) \ln \left (-x \right )}\right )+\frac {x^{4}}{\ln \left (-x \right )}-\frac {x^{3}}{\ln \left (-x \right )}+\frac {x^{2}}{\ln \left (-x \right )}\) | \(64\) |
Input:
int(((x^2+2*x+1)*ln(-x)^2+(10*ln(2)+20*x^5+25*x^4+5*x^2+10*x)*ln(-x)+2*(-5 *x-5)*ln(2)-5*x^5-5*x^4-5*x^2-5*x)/(5*x^2+10*x+5)/ln(-x)^2,x,method=_RETUR NVERBOSE)
Output:
x/(1+x)/ln(-x)*(2*ln(2)+x^4+x)+1/5*x
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {5 \, x^{5} + 5 \, x^{2} + 10 \, x \log \left (2\right ) + {\left (x^{2} + x\right )} \log \left (-x\right )}{5 \, {\left (x + 1\right )} \log \left (-x\right )} \] Input:
integrate(((x^2+2*x+1)*log(-x)^2+(10*log(2)+20*x^5+25*x^4+5*x^2+10*x)*log( -x)+2*(-5*x-5)*log(2)-5*x^5-5*x^4-5*x^2-5*x)/(5*x^2+10*x+5)/log(-x)^2,x, a lgorithm="fricas")
Output:
1/5*(5*x^5 + 5*x^2 + 10*x*log(2) + (x^2 + x)*log(-x))/((x + 1)*log(-x))
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {x}{5} + \frac {x^{5} + x^{2} + 2 x \log {\left (2 \right )}}{\left (x + 1\right ) \log {\left (- x \right )}} \] Input:
integrate(((x**2+2*x+1)*ln(-x)**2+(10*ln(2)+20*x**5+25*x**4+5*x**2+10*x)*l n(-x)+2*(-5*x-5)*ln(2)-5*x**5-5*x**4-5*x**2-5*x)/(5*x**2+10*x+5)/ln(-x)**2 ,x)
Output:
x/5 + (x**5 + x**2 + 2*x*log(2))/((x + 1)*log(-x))
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {5 \, x^{5} + 5 \, x^{2} + 10 \, x \log \left (2\right ) + {\left (x^{2} + x\right )} \log \left (-x\right )}{5 \, {\left (x + 1\right )} \log \left (-x\right )} \] Input:
integrate(((x^2+2*x+1)*log(-x)^2+(10*log(2)+20*x^5+25*x^4+5*x^2+10*x)*log( -x)+2*(-5*x-5)*log(2)-5*x^5-5*x^4-5*x^2-5*x)/(5*x^2+10*x+5)/log(-x)^2,x, a lgorithm="maxima")
Output:
1/5*(5*x^5 + 5*x^2 + 10*x*log(2) + (x^2 + x)*log(-x))/((x + 1)*log(-x))
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {1}{5} \, x + \frac {x^{5} + x^{2} + 2 \, x \log \left (2\right )}{x \log \left (-x\right ) + \log \left (-x\right )} \] Input:
integrate(((x^2+2*x+1)*log(-x)^2+(10*log(2)+20*x^5+25*x^4+5*x^2+10*x)*log( -x)+2*(-5*x-5)*log(2)-5*x^5-5*x^4-5*x^2-5*x)/(5*x^2+10*x+5)/log(-x)^2,x, a lgorithm="giac")
Output:
1/5*x + (x^5 + x^2 + 2*x*log(2))/(x*log(-x) + log(-x))
Time = 2.86 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {x}{5}+\frac {x\,\left (5\,x^4+5\,x+10\,\ln \left (2\right )\right )}{5\,\ln \left (-x\right )\,\left (x+1\right )} \] Input:
int(-(5*x + 2*log(2)*(5*x + 5) - log(-x)^2*(2*x + x^2 + 1) - log(-x)*(10*x + 10*log(2) + 5*x^2 + 25*x^4 + 20*x^5) + 5*x^2 + 5*x^4 + 5*x^5)/(log(-x)^ 2*(10*x + 5*x^2 + 5)),x)
Output:
x/5 + (x*(5*x + 10*log(2) + 5*x^4))/(5*log(-x)*(x + 1))
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-5 x-5 x^2-5 x^4-5 x^5+(-5-5 x) \log (4)+\left (10 x+5 x^2+25 x^4+20 x^5+5 \log (4)\right ) \log (-x)+\left (1+2 x+x^2\right ) \log ^2(-x)}{\left (5+10 x+5 x^2\right ) \log ^2(-x)} \, dx=\frac {x \left (\mathrm {log}\left (-x \right ) x +\mathrm {log}\left (-x \right )+10 \,\mathrm {log}\left (2\right )+5 x^{4}+5 x \right )}{5 \,\mathrm {log}\left (-x \right ) \left (x +1\right )} \] Input:
int(((x^2+2*x+1)*log(-x)^2+(10*log(2)+20*x^5+25*x^4+5*x^2+10*x)*log(-x)+2* (-5*x-5)*log(2)-5*x^5-5*x^4-5*x^2-5*x)/(5*x^2+10*x+5)/log(-x)^2,x)
Output:
(x*(log( - x)*x + log( - x) + 10*log(2) + 5*x**4 + 5*x))/(5*log( - x)*(x + 1))