Integrand size = 50, antiderivative size = 21 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{(2+x) \log (x (-5+i \pi +\log (3)))} \] Output:
x/(2+x)/ln(x*(ln(3)+I*Pi-5))
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{(2+x) \log (x (-5+i \pi +\log (3)))} \] Input:
Integrate[(-2 - x + 2*Log[-5*x + x*(I*Pi + Log[3])])/((4 + 4*x + x^2)*Log[ -5*x + x*(I*Pi + Log[3])]^2),x]
Output:
x/((2 + x)*Log[x*(-5 + I*Pi + Log[3])])
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.
\(\displaystyle \int \frac {-x+2 \log (-5 x+x (\log (3)+i \pi ))-2}{\left (x^2+4 x+4\right ) \log ^2(-5 x+x (\log (3)+i \pi ))} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-x+2 \log (-5 x+x (\log (3)+i \pi ))-2}{(x+2)^2 \log ^2(-5 x+x (\log (3)+i \pi ))}dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {-x+2 \log (-5 x+x (\log (3)+i \pi ))-2}{(x+2)^2 \log ^2(-x (5-i \pi -\log (3)))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{(-x-2) \log ^2(-x (5-i \pi -\log (3)))}+\frac {2}{(x+2)^2 \log (-x (5-i \pi -\log (3)))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{(-x-2) \log ^2(x (-5+i \pi +\log (3)))}dx+2 \int \frac {1}{(x+2)^2 \log (x (-5+i \pi +\log (3)))}dx\) |
Input:
Int[(-2 - x + 2*Log[-5*x + x*(I*Pi + Log[3])])/((4 + 4*x + x^2)*Log[-5*x + x*(I*Pi + Log[3])]^2),x]
Output:
$Aborted
Time = 2.80 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {x}{\left (2+x \right ) \ln \left (x \left (\ln \left (3\right )+i \pi -5\right )\right )}\) | \(21\) |
norman | \(\frac {x}{\left (2+x \right ) \ln \left (x \left (\ln \left (3\right )+i \pi \right )-5 x \right )}\) | \(24\) |
risch | \(\frac {x}{\left (2+x \right ) \ln \left (x \left (\ln \left (3\right )+i \pi \right )-5 x \right )}\) | \(24\) |
parts | \(\frac {1}{\ln \left (i \pi x +x \ln \left (3\right )-5 x \right )}-\frac {2 \left (\ln \left (3\right )+i \pi -5\right )}{\left (2 i \pi +i \pi x +x \ln \left (3\right )-5 x +2 \ln \left (3\right )-10\right ) \ln \left (i \pi x +x \ln \left (3\right )-5 x \right )}\) | \(68\) |
derivativedivides | \(\frac {\left (\ln \left (3\right )+i \pi -5\right )^{3} x^{2}+\left (2 \ln \left (3\right )^{2}+2 \pi ^{2}-20 \ln \left (3\right )+50\right ) x \left (\ln \left (3\right )+i \pi -5\right )}{\left (\ln \left (3\right )+i \pi -5\right ) \ln \left (x \left (\ln \left (3\right )+i \pi -5\right )\right ) \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right ) x \left (\ln \left (3\right )+i \pi -5\right )+4 \pi ^{2}+x^{2} \left (\ln \left (3\right )+i \pi -5\right )^{2}-40 \ln \left (3\right )-20 x \left (\ln \left (3\right )+i \pi -5\right )+100\right )}\) | \(124\) |
default | \(\frac {\left (\ln \left (3\right )+i \pi -5\right )^{3} x^{2}+\left (2 \ln \left (3\right )^{2}+2 \pi ^{2}-20 \ln \left (3\right )+50\right ) x \left (\ln \left (3\right )+i \pi -5\right )}{\left (\ln \left (3\right )+i \pi -5\right ) \ln \left (x \left (\ln \left (3\right )+i \pi -5\right )\right ) \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right ) x \left (\ln \left (3\right )+i \pi -5\right )+4 \pi ^{2}+x^{2} \left (\ln \left (3\right )+i \pi -5\right )^{2}-40 \ln \left (3\right )-20 x \left (\ln \left (3\right )+i \pi -5\right )+100\right )}\) | \(124\) |
Input:
int((2*ln(x*(ln(3)+I*Pi)-5*x)-x-2)/(x^2+4*x+4)/ln(x*(ln(3)+I*Pi)-5*x)^2,x, method=_RETURNVERBOSE)
Output:
x/(2+x)/ln(x*(ln(3)+I*Pi-5))
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{{\left (x + 2\right )} \log \left ({\left (i \, \pi - 5\right )} x + x \log \left (3\right )\right )} \] Input:
integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi) -5*x)^2,x, algorithm="fricas")
Output:
x/((x + 2)*log((I*pi - 5)*x + x*log(3)))
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{\left (x + 2\right ) \log {\left (- 5 x + x \log {\left (3 \right )} + i \pi x \right )}} \] Input:
integrate((2*ln(x*(ln(3)+I*pi)-5*x)-x-2)/(x**2+4*x+4)/ln(x*(ln(3)+I*pi)-5* x)**2,x)
Output:
x/((x + 2)*log(-5*x + x*log(3) + I*pi*x))
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{x \log \left (i \, \pi + \log \left (3\right ) - 5\right ) + {\left (x + 2\right )} \log \left (x\right ) + 2 \, \log \left (i \, \pi + \log \left (3\right ) - 5\right )} \] Input:
integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi) -5*x)^2,x, algorithm="maxima")
Output:
x/(x*log(I*pi + log(3) - 5) + (x + 2)*log(x) + 2*log(I*pi + log(3) - 5))
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{x \log \left (i \, \pi x + x \log \left (3\right ) - 5 \, x\right ) + 2 \, \log \left (i \, \pi x + x \log \left (3\right ) - 5 \, x\right )} \] Input:
integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi) -5*x)^2,x, algorithm="giac")
Output:
x/(x*log(I*pi*x + x*log(3) - 5*x) + 2*log(I*pi*x + x*log(3) - 5*x))
Time = 8.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{\ln \left (x\,\left (\ln \left (3\right )-5+\Pi \,1{}\mathrm {i}\right )\right )\,\left (x+2\right )} \] Input:
int(-(x - 2*log(x*(Pi*1i + log(3)) - 5*x) + 2)/(log(x*(Pi*1i + log(3)) - 5 *x)^2*(4*x + x^2 + 4)),x)
Output:
x/(log(x*(Pi*1i + log(3) - 5))*(x + 2))
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx=\frac {x}{\mathrm {log}\left (\mathrm {log}\left (3\right ) x +i \pi x -5 x \right ) \left (x +2\right )} \] Input:
int((2*log(x*(log(3)+I*Pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*Pi)-5*x)^ 2,x)
Output:
x/(log(log(3)*x + i*pi*x - 5*x)*(x + 2))