Integrand size = 85, antiderivative size = 25 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-7+\frac {\log (2 x)}{x \log \left (\frac {4+x}{2 (1+x)}\right )} \] Output:
ln(2*x)/ln((4+x)/(2+2*x))/x-7
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log (2 x)}{x \log \left (\frac {4+x}{2+2 x}\right )} \] Input:
Integrate[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x)]^2),x]
Output:
Log[2*x]/(x*Log[(4 + x)/(2 + 2*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.
\(\displaystyle \int \frac {\left (x^2+5 x+4\right ) \log \left (\frac {x+4}{2 x+2}\right )+\log (2 x) \left (\left (-x^2-5 x-4\right ) \log \left (\frac {x+4}{2 x+2}\right )+3 x\right )}{\left (x^4+5 x^3+4 x^2\right ) \log ^2\left (\frac {x+4}{2 x+2}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2+5 x+4\right ) \log \left (\frac {x+4}{2 x+2}\right )+\log (2 x) \left (\left (-x^2-5 x-4\right ) \log \left (\frac {x+4}{2 x+2}\right )+3 x\right )}{x^2 \left (x^2+5 x+4\right ) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {1-\log (2 x)}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}+\frac {3 \log (2 x)}{x (x+1) (x+4) \log ^2\left (\frac {x+4}{2 x+2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}dx-\int \frac {\log (2 x)}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}dx+\frac {3}{4} \int \frac {\log (2 x)}{x \log ^2\left (\frac {x+4}{2 x+2}\right )}dx-\int \frac {\log (2 x)}{(x+1) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx+\frac {1}{4} \int \frac {\log (2 x)}{(x+4) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx\) |
Input:
Int[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x) ]^2),x]
Output:
$Aborted
Time = 1.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {\ln \left (2 x \right )}{\ln \left (\frac {4+x}{2+2 x}\right ) x}\) | \(22\) |
risch | \(-\frac {2 \ln \left (2 x \right )}{x \left (i \pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}+i \pi \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{3}+2 \ln \left (2\right )+2 \ln \left (1+x \right )-2 \ln \left (4+x \right )\right )}\) | \(131\) |
Input:
int((((-x^2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x^2+5*x+4)*ln((4+x)/(2+ 2*x)))/(x^4+5*x^3+4*x^2)/ln((4+x)/(2+2*x))^2,x,method=_RETURNVERBOSE)
Output:
1/x*ln(2*x)/ln(1/2*(4+x)/(1+x))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log \left (2 \, x\right )}{x \log \left (\frac {x + 4}{2 \, {\left (x + 1\right )}}\right )} \] Input:
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm="frica s")
Output:
log(2*x)/(x*log(1/2*(x + 4)/(x + 1)))
Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log {\left (2 x \right )}}{x \log {\left (\frac {x + 4}{2 x + 2} \right )}} \] Input:
integrate((((-x**2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x**2+5*x+4)*ln(( 4+x)/(2+2*x)))/(x**4+5*x**3+4*x**2)/ln((4+x)/(2+2*x))**2,x)
Output:
log(2*x)/(x*log((x + 4)/(2*x + 2)))
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \] Input:
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm="maxim a")
Output:
-(log(2) + log(x))/(x*log(2) - x*log(x + 4) + x*log(x + 1))
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \] Input:
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm="giac" )
Output:
-(log(2) + log(x))/(x*log(2) - x*log(x + 4) + x*log(x + 1))
Time = 8.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\ln \left (2\,x\right )}{x\,\ln \left (\frac {x+4}{2\,x+2}\right )} \] Input:
int((log(2*x)*(3*x - log((x + 4)/(2*x + 2))*(5*x + x^2 + 4)) + log((x + 4) /(2*x + 2))*(5*x + x^2 + 4))/(log((x + 4)/(2*x + 2))^2*(4*x^2 + 5*x^3 + x^ 4)),x)
Output:
log(2*x)/(x*log((x + 4)/(2*x + 2)))
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\mathrm {log}\left (2 x \right )}{\mathrm {log}\left (\frac {x +4}{2 x +2}\right ) x} \] Input:
int((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/ (2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x)
Output:
log(2*x)/(log((x + 4)/(2*x + 2))*x)