Integrand size = 106, antiderivative size = 24 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^2}{4+x+x \left (3+x-\log \left (\frac {x}{1+x}\right )\right )} \] Output:
x^2/(4+(3+x-ln(x/(1+x)))*x+x)
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^2}{(2+x)^2-x \log \left (\frac {x}{1+x}\right )} \] Input:
Integrate[(8*x + 13*x^2 + 4*x^3 + (-x^2 - x^3)*Log[x/(1 + x)])/(16 + 48*x + 56*x^2 + 32*x^3 + 9*x^4 + x^5 + (-8*x - 16*x^2 - 10*x^3 - 2*x^4)*Log[x/( 1 + x)] + (x^2 + x^3)*Log[x/(1 + x)]^2),x]
Output:
x^2/((2 + x)^2 - x*Log[x/(1 + 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.
\(\displaystyle \int \frac {4 x^3+13 x^2+\left (-x^3-x^2\right ) \log \left (\frac {x}{x+1}\right )+8 x}{x^5+9 x^4+32 x^3+56 x^2+\left (x^3+x^2\right ) \log ^2\left (\frac {x}{x+1}\right )+\left (-2 x^4-10 x^3-16 x^2-8 x\right ) \log \left (\frac {x}{x+1}\right )+48 x+16} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x \left (4 x^2+13 x-(x+1) x \log \left (\frac {x}{x+1}\right )+8\right )}{(x+1) \left ((x+2)^2-x \log \left (\frac {x}{x+1}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x}{x^2+4 x-x \log \left (\frac {x}{x+1}\right )+4}-\frac {x \left (x^3+x^2-5 x-4\right )}{(x+1) \left (x^2+4 x-x \log \left (\frac {x}{x+1}\right )+4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {1}{\left (x^2-\log \left (\frac {x}{x+1}\right ) x+4 x+4\right )^2}dx+5 \int \frac {x}{\left (x^2-\log \left (\frac {x}{x+1}\right ) x+4 x+4\right )^2}dx+\int \frac {1}{(x+1) \left (x^2-\log \left (\frac {x}{x+1}\right ) x+4 x+4\right )^2}dx+\int \frac {x}{x^2-\log \left (\frac {x}{x+1}\right ) x+4 x+4}dx-\int \frac {x^3}{\left (x^2-\log \left (\frac {x}{x+1}\right ) x+4 x+4\right )^2}dx\) |
Input:
Int[(8*x + 13*x^2 + 4*x^3 + (-x^2 - x^3)*Log[x/(1 + x)])/(16 + 48*x + 56*x ^2 + 32*x^3 + 9*x^4 + x^5 + (-8*x - 16*x^2 - 10*x^3 - 2*x^4)*Log[x/(1 + x) ] + (x^2 + x^3)*Log[x/(1 + x)]^2),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {x^{2}}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\) | \(26\) |
parallelrisch | \(\frac {x^{2}}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\) | \(26\) |
norman | \(\frac {-4 x +x \ln \left (\frac {x}{1+x}\right )-4}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\) | \(38\) |
derivativedivides | \(\frac {\left (1-\frac {1}{1+x}\right )^{2}}{\left (1-\frac {1}{1+x}\right )^{2} \ln \left (1-\frac {1}{1+x}\right )+\left (1-\frac {1}{1+x}\right )^{2}-\left (1-\frac {1}{1+x}\right ) \ln \left (1-\frac {1}{1+x}\right )+\frac {4}{1+x}}\) | \(77\) |
default | \(\frac {\left (1-\frac {1}{1+x}\right )^{2}}{\left (1-\frac {1}{1+x}\right )^{2} \ln \left (1-\frac {1}{1+x}\right )+\left (1-\frac {1}{1+x}\right )^{2}-\left (1-\frac {1}{1+x}\right ) \ln \left (1-\frac {1}{1+x}\right )+\frac {4}{1+x}}\) | \(77\) |
Input:
int(((-x^3-x^2)*ln(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*ln(x/(1+x))^2+(-2 *x^4-10*x^3-16*x^2-8*x)*ln(x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+16),x,met hod=_RETURNVERBOSE)
Output:
x^2/(x^2-x*ln(x/(1+x))+4*x+4)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^{2}}{x^{2} - x \log \left (\frac {x}{x + 1}\right ) + 4 \, x + 4} \] Input:
integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x ))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+ 16),x, algorithm="fricas")
Output:
x^2/(x^2 - x*log(x/(x + 1)) + 4*x + 4)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=- \frac {x^{2}}{- x^{2} + x \log {\left (\frac {x}{x + 1} \right )} - 4 x - 4} \] Input:
integrate(((-x**3-x**2)*ln(x/(1+x))+4*x**3+13*x**2+8*x)/((x**3+x**2)*ln(x/ (1+x))**2+(-2*x**4-10*x**3-16*x**2-8*x)*ln(x/(1+x))+x**5+9*x**4+32*x**3+56 *x**2+48*x+16),x)
Output:
-x**2/(-x**2 + x*log(x/(x + 1)) - 4*x - 4)
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^{2}}{x^{2} + x \log \left (x + 1\right ) - x \log \left (x\right ) + 4 \, x + 4} \] Input:
integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x ))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+ 16),x, algorithm="maxima")
Output:
x^2/(x^2 + x*log(x + 1) - x*log(x) + 4*x + 4)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=-\frac {x^{2}}{{\left (x + 1\right )}^{2} {\left (\frac {x \log \left (\frac {x}{x + 1}\right )}{x + 1} - \frac {x^{2} \log \left (\frac {x}{x + 1}\right )}{{\left (x + 1\right )}^{2}} + \frac {4 \, x}{x + 1} - \frac {x^{2}}{{\left (x + 1\right )}^{2}} - 4\right )}} \] Input:
integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x ))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+ 16),x, algorithm="giac")
Output:
-x^2/((x + 1)^2*(x*log(x/(x + 1))/(x + 1) - x^2*log(x/(x + 1))/(x + 1)^2 + 4*x/(x + 1) - x^2/(x + 1)^2 - 4))
Timed out. \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\int \frac {8\,x+13\,x^2+4\,x^3-\ln \left (\frac {x}{x+1}\right )\,\left (x^3+x^2\right )}{48\,x+{\ln \left (\frac {x}{x+1}\right )}^2\,\left (x^3+x^2\right )-\ln \left (\frac {x}{x+1}\right )\,\left (2\,x^4+10\,x^3+16\,x^2+8\,x\right )+56\,x^2+32\,x^3+9\,x^4+x^5+16} \,d x \] Input:
int((8*x + 13*x^2 + 4*x^3 - log(x/(x + 1))*(x^2 + x^3))/(48*x + log(x/(x + 1))^2*(x^2 + x^3) - log(x/(x + 1))*(8*x + 16*x^2 + 10*x^3 + 2*x^4) + 56*x ^2 + 32*x^3 + 9*x^4 + x^5 + 16),x)
Output:
int((8*x + 13*x^2 + 4*x^3 - log(x/(x + 1))*(x^2 + x^3))/(48*x + log(x/(x + 1))^2*(x^2 + x^3) - log(x/(x + 1))*(8*x + 16*x^2 + 10*x^3 + 2*x^4) + 56*x ^2 + 32*x^3 + 9*x^4 + x^5 + 16), x)
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {-\mathrm {log}\left (\frac {x}{x +1}\right ) x +4 x +4}{\mathrm {log}\left (\frac {x}{x +1}\right ) x -x^{2}-4 x -4} \] Input:
int(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x))^2+( -2*x^4-10*x^3-16*x^2-8*x)*log(x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+16),x)
Output:
( - log(x/(x + 1))*x + 4*x + 4)/(log(x/(x + 1))*x - x**2 - 4*x - 4)