Integrand size = 67, antiderivative size = 26 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \left (x^2+\log ^2(x)-\log \left (5 \log \left (\frac {-1+x}{2 x}\right )\right )\right ) \] Output:
5*ln(x)^2+5*x^2-5*ln(5*ln(1/2*(-1+x)/x))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \left (x^2+\log ^2(x)-\log \left (\log \left (\frac {-1+x}{2 x}\right )\right )\right ) \] Input:
Integrate[(-5 + (-10*x^2 + 10*x^3)*Log[(-1 + x)/(2*x)] + (-10 + 10*x)*Log[ (-1 + x)/(2*x)]*Log[x])/((-x + x^2)*Log[(-1 + x)/(2*x)]),x]
Output:
5*(x^2 + Log[x]^2 - Log[Log[(-1 + x)/(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.
\(\displaystyle \int \frac {\left (10 x^3-10 x^2\right ) \log \left (\frac {x-1}{2 x}\right )+(10 x-10) \log (x) \log \left (\frac {x-1}{2 x}\right )-5}{\left (x^2-x\right ) \log \left (\frac {x-1}{2 x}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (10 x^3-10 x^2\right ) \log \left (\frac {x-1}{2 x}\right )+(10 x-10) \log (x) \log \left (\frac {x-1}{2 x}\right )-5}{(x-1) x \log \left (\frac {x-1}{2 x}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (2 \left (x^2+\log (x)\right )-\frac {1}{(x-1) \log \left (\frac {x-1}{2 x}\right )}\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {2 \left (x^2+\log (x)\right )+\frac {1}{(1-x) \log \left (-\frac {1-x}{2 x}\right )}}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle 5 \int \left (\frac {-2 \log \left (\frac {x-1}{2 x}\right ) x^3+2 \log \left (\frac {x-1}{2 x}\right ) x^2+1}{(1-x) x \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}+\frac {2 \log (x)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (-\int \frac {1}{(x-1) \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}dx+\int \frac {1}{x \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}dx+x^2+\log ^2(x)\right )\) |
Input:
Int[(-5 + (-10*x^2 + 10*x^3)*Log[(-1 + x)/(2*x)] + (-10 + 10*x)*Log[(-1 + x)/(2*x)]*Log[x])/((-x + x^2)*Log[(-1 + x)/(2*x)]),x]
Output:
$Aborted
Time = 0.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (\frac {-1+x}{2 x}\right )\right )\) | \(25\) |
parts | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (\frac {-1+x}{2 x}\right )\right )\) | \(25\) |
default | \(5 \ln \left (x \right )^{2}+5 x^{2}-5 \ln \left (\ln \left (2\right )-\ln \left (1-\frac {1}{x}\right )\right )\) | \(29\) |
risch | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (-1+x \right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}-2 i \ln \left (2\right )-2 i \ln \left (x \right )\right )}{2}\right )\) | \(119\) |
Input:
int(((10*x-10)*ln(1/2*(-1+x)/x)*ln(x)+(10*x^3-10*x^2)*ln(1/2*(-1+x)/x)-5)/ (x^2-x)/ln(1/2*(-1+x)/x),x,method=_RETURNVERBOSE)
Output:
5*x^2+5*ln(x)^2-5*ln(ln(1/2*(-1+x)/x))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (\log \left (\frac {x - 1}{2 \, x}\right )\right ) \] Input:
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm="fricas")
Output:
5*x^2 + 5*log(x)^2 - 5*log(log(1/2*(x - 1)/x))
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 x^{2} + 5 \log {\left (x \right )}^{2} - 5 \log {\left (\log {\left (\frac {\frac {x}{2} - \frac {1}{2}}{x} \right )} \right )} \] Input:
integrate(((10*x-10)*ln(1/2*(-1+x)/x)*ln(x)+(10*x**3-10*x**2)*ln(1/2*(-1+x )/x)-5)/(x**2-x)/ln(1/2*(-1+x)/x),x)
Output:
5*x**2 + 5*log(x)**2 - 5*log(log((x/2 - 1/2)/x))
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (-\log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right )\right ) \] Input:
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm="maxima")
Output:
5*x^2 + 5*log(x)^2 - 5*log(-log(2) + log(x - 1) - log(x))
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (-\log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right )\right ) \] Input:
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm="giac")
Output:
5*x^2 + 5*log(x)^2 - 5*log(-log(2) + log(x - 1) - log(x))
Time = 3.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5\,{\ln \left (x\right )}^2-5\,\ln \left (\ln \left (\frac {\frac {x}{2}-\frac {1}{2}}{x}\right )\right )+5\,x^2 \] Input:
int((log((x/2 - 1/2)/x)*(10*x^2 - 10*x^3) - log((x/2 - 1/2)/x)*log(x)*(10* x - 10) + 5)/(log((x/2 - 1/2)/x)*(x - x^2)),x)
Output:
5*log(x)^2 - 5*log(log((x/2 - 1/2)/x)) + 5*x^2
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=-5 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x -1}{2 x}\right )\right )+5 \mathrm {log}\left (x \right )^{2}+5 x^{2} \] Input:
int(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+x)/x)- 5)/(x^2-x)/log(1/2*(-1+x)/x),x)
Output:
5*( - log(log((x - 1)/(2*x))) + log(x)**2 + x**2)