Integrand size = 74, antiderivative size = 26 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=-2+e^5-x+\left (-5+\frac {x^2}{4}+\log (x-\log (x))\right )^2 \] Output:
(1/4*x^2-5+ln(x-ln(x)))^2+exp(5)-x-2
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {1}{4} \left (-4 x-10 x^2+\frac {x^4}{4}-40 \log (x-\log (x))+2 x^2 \log (x-\log (x))+4 \log ^2(x-\log (x))\right ) \] Input:
Integrate[(-40 + 40*x + 6*x^2 + 18*x^3 - x^5 + (-4*x - 20*x^2 + x^4)*Log[x ] + (8 - 8*x - 4*x^3 + 4*x^2*Log[x])*Log[x - Log[x]])/(-4*x^2 + 4*x*Log[x] ),x]
Output:
(-4*x - 10*x^2 + x^4/4 - 40*Log[x - Log[x]] + 2*x^2*Log[x - Log[x]] + 4*Lo g[x - Log[x]]^2)/4
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.
\(\displaystyle \int \frac {-x^5+18 x^3+6 x^2+\left (x^4-20 x^2-4 x\right ) \log (x)+\left (-4 x^3+4 x^2 \log (x)-8 x+8\right ) \log (x-\log (x))+40 x-40}{4 x \log (x)-4 x^2} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {-x^5+18 x^3+6 x^2+\left (x^4-20 x^2-4 x\right ) \log (x)+\left (-4 x^3+4 x^2 \log (x)-8 x+8\right ) \log (x-\log (x))+40 x-40}{x (4 \log (x)-4 x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^5-18 x^3-6 x^2-\left (x^4-20 x^2-4 x\right ) \log (x)-\left (-4 x^3+4 x^2 \log (x)-8 x+8\right ) \log (x-\log (x))-40 x+40}{4 x (x-\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {x^5-18 x^3-6 x^2-40 x+\left (-x^4+20 x^2+4 x\right ) \log (x)-4 \left (-x^3+\log (x) x^2-2 x+2\right ) \log (x-\log (x))+40}{x (x-\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {x^5-\log (x) x^4-18 x^3+20 \log (x) x^2-6 x^2+4 \log (x) x-40 x+40}{x (x-\log (x))}+\frac {4 \left (x^3-\log (x) x^2+2 x-2\right ) \log (x-\log (x))}{x (x-\log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (2 \int \frac {x^2}{x-\log (x)}dx+4 \int \frac {x^2 \log (x-\log (x))}{x-\log (x)}dx-40 \int \frac {1}{x-\log (x)}dx+40 \int \frac {1}{x (x-\log (x))}dx-2 \int \frac {x}{x-\log (x)}dx+8 \int \frac {\log (x-\log (x))}{x-\log (x)}dx-8 \int \frac {\log (x-\log (x))}{x (x-\log (x))}dx-4 \int \frac {x \log (x) \log (x-\log (x))}{x-\log (x)}dx+\frac {x^4}{4}-10 x^2-4 x\right )\) |
Input:
Int[(-40 + 40*x + 6*x^2 + 18*x^3 - x^5 + (-4*x - 20*x^2 + x^4)*Log[x] + (8 - 8*x - 4*x^3 + 4*x^2*Log[x])*Log[x - Log[x]])/(-4*x^2 + 4*x*Log[x]),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73
method | result | size |
risch | \(\ln \left (x -\ln \left (x \right )\right )^{2}+\frac {\ln \left (x -\ln \left (x \right )\right ) x^{2}}{2}+\frac {x^{4}}{16}-\frac {5 x^{2}}{2}-x -10 \ln \left (\ln \left (x \right )-x \right )\) | \(45\) |
parallelrisch | \(\frac {x^{4}}{16}+\frac {\ln \left (x -\ln \left (x \right )\right ) x^{2}}{2}-\frac {5 x^{2}}{2}+\ln \left (x -\ln \left (x \right )\right )^{2}-x -10 \ln \left (x -\ln \left (x \right )\right )\) | \(45\) |
Input:
int(((4*x^2*ln(x)-4*x^3-8*x+8)*ln(x-ln(x))+(x^4-20*x^2-4*x)*ln(x)-x^5+18*x ^3+6*x^2+40*x-40)/(4*x*ln(x)-4*x^2),x,method=_RETURNVERBOSE)
Output:
ln(x-ln(x))^2+1/2*ln(x-ln(x))*x^2+1/16*x^4-5/2*x^2-x-10*ln(ln(x)-x)
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {1}{16} \, x^{4} - \frac {5}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{2} - 20\right )} \log \left (x - \log \left (x\right )\right ) + \log \left (x - \log \left (x\right )\right )^{2} - x \] Input:
integrate(((4*x^2*log(x)-4*x^3-8*x+8)*log(x-log(x))+(x^4-20*x^2-4*x)*log(x )-x^5+18*x^3+6*x^2+40*x-40)/(4*x*log(x)-4*x^2),x, algorithm="fricas")
Output:
1/16*x^4 - 5/2*x^2 + 1/2*(x^2 - 20)*log(x - log(x)) + log(x - log(x))^2 - x
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {x^{4}}{16} + \frac {x^{2} \log {\left (x - \log {\left (x \right )} \right )}}{2} - \frac {5 x^{2}}{2} - x - 10 \log {\left (- x + \log {\left (x \right )} \right )} + \log {\left (x - \log {\left (x \right )} \right )}^{2} \] Input:
integrate(((4*x**2*ln(x)-4*x**3-8*x+8)*ln(x-ln(x))+(x**4-20*x**2-4*x)*ln(x )-x**5+18*x**3+6*x**2+40*x-40)/(4*x*ln(x)-4*x**2),x)
Output:
x**4/16 + x**2*log(x - log(x))/2 - 5*x**2/2 - x - 10*log(-x + log(x)) + lo g(x - log(x))**2
Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {1}{16} \, x^{4} - \frac {5}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{2} - 20\right )} \log \left (x - \log \left (x\right )\right ) + \log \left (x - \log \left (x\right )\right )^{2} - x \] Input:
integrate(((4*x^2*log(x)-4*x^3-8*x+8)*log(x-log(x))+(x^4-20*x^2-4*x)*log(x )-x^5+18*x^3+6*x^2+40*x-40)/(4*x*log(x)-4*x^2),x, algorithm="maxima")
Output:
1/16*x^4 - 5/2*x^2 + 1/2*(x^2 - 20)*log(x - log(x)) + log(x - log(x))^2 - x
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {1}{16} \, x^{4} + \frac {1}{2} \, x^{2} \log \left (x - \log \left (x\right )\right ) - \frac {5}{2} \, x^{2} + \log \left (x - \log \left (x\right )\right )^{2} - x - 10 \, \log \left (x - \log \left (x\right )\right ) \] Input:
integrate(((4*x^2*log(x)-4*x^3-8*x+8)*log(x-log(x))+(x^4-20*x^2-4*x)*log(x )-x^5+18*x^3+6*x^2+40*x-40)/(4*x*log(x)-4*x^2),x, algorithm="giac")
Output:
1/16*x^4 + 1/2*x^2*log(x - log(x)) - 5/2*x^2 + log(x - log(x))^2 - x - 10* log(x - log(x))
Time = 3.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=\frac {x^2\,\ln \left (x-\ln \left (x\right )\right )}{2}-10\,\ln \left (\ln \left (x\right )-x\right )-x-\frac {5\,x^2}{2}+\frac {x^4}{16}+{\ln \left (x-\ln \left (x\right )\right )}^2 \] Input:
int(-(log(x - log(x))*(8*x - 4*x^2*log(x) + 4*x^3 - 8) - 40*x - 6*x^2 - 18 *x^3 + x^5 + log(x)*(4*x + 20*x^2 - x^4) + 40)/(4*x*log(x) - 4*x^2),x)
Output:
(x^2*log(x - log(x)))/2 - 10*log(log(x) - x) - x - (5*x^2)/2 + x^4/16 + lo g(x - log(x))^2
Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {-40+40 x+6 x^2+18 x^3-x^5+\left (-4 x-20 x^2+x^4\right ) \log (x)+\left (8-8 x-4 x^3+4 x^2 \log (x)\right ) \log (x-\log (x))}{-4 x^2+4 x \log (x)} \, dx=-10 \,\mathrm {log}\left (\mathrm {log}\left (x \right )-x \right )+\mathrm {log}\left (x -\mathrm {log}\left (x \right )\right )^{2}+\frac {\mathrm {log}\left (x -\mathrm {log}\left (x \right )\right ) x^{2}}{2}+\frac {x^{4}}{16}-\frac {5 x^{2}}{2}-x \] Input:
int(((4*x^2*log(x)-4*x^3-8*x+8)*log(x-log(x))+(x^4-20*x^2-4*x)*log(x)-x^5+ 18*x^3+6*x^2+40*x-40)/(4*x*log(x)-4*x^2),x)
Output:
( - 160*log(log(x) - x) + 16*log( - log(x) + x)**2 + 8*log( - log(x) + x)* x**2 + x**4 - 40*x**2 - 16*x)/16