Integrand size = 70, antiderivative size = 28 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {1}{4} \log \left (\log (5)+\frac {1}{2} \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )\right ) \] Output:
1/4*ln(1/2*ln(-8/5*x^2/ln(-5+2*x))+ln(5))
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {1}{4} \log \left (\log \left (-\frac {40 x^2}{\log (-5+2 x)}\right )\right ) \] Input:
Integrate[(-x + (-5 + 2*x)*Log[-5 + 2*x])/((-20*x + 8*x^2)*Log[5]*Log[-5 + 2*x] + (-10*x + 4*x^2)*Log[-5 + 2*x]*Log[(-8*x^2)/(5*Log[-5 + 2*x])]),x]
Output:
Log[Log[(-40*x^2)/Log[-5 + 2*x]]]/4
Time = 0.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {7292, 27, 7235}
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.
\(\displaystyle \int \frac {(2 x-5) \log (2 x-5)-x}{\left (4 x^2-10 x\right ) \log \left (-\frac {8 x^2}{5 \log (2 x-5)}\right ) \log (2 x-5)+\left (8 x^2-20 x\right ) \log (5) \log (2 x-5)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x-(2 x-5) \log (2 x-5)}{2 (5-2 x) x \log (2 x-5) \log \left (-\frac {40 x^2}{\log (2 x-5)}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {x+(5-2 x) \log (2 x-5)}{(5-2 x) x \log (2 x-5) \log \left (-\frac {40 x^2}{\log (2 x-5)}\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \frac {1}{4} \log \left (\log \left (-\frac {40 x^2}{\log (2 x-5)}\right )\right )\) |
Input:
Int[(-x + (-5 + 2*x)*Log[-5 + 2*x])/((-20*x + 8*x^2)*Log[5]*Log[-5 + 2*x] + (-10*x + 4*x^2)*Log[-5 + 2*x]*Log[(-8*x^2)/(5*Log[-5 + 2*x])]),x]
Output:
Log[Log[(-40*x^2)/Log[-5 + 2*x]]]/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 1.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {\ln \left (\ln \left (-\frac {8 x^{2}}{5 \ln \left (-5+2 x \right )}\right )+2 \ln \left (5\right )\right )}{4}\) | \(23\) |
parallelrisch | \(\frac {\ln \left (\ln \left (-\frac {8 x^{2}}{5 \ln \left (-5+2 x \right )}\right )+2 \ln \left (5\right )\right )}{4}\) | \(23\) |
risch | \(\frac {\ln \left (\ln \left (\ln \left (-5+2 x \right )\right )+\frac {i \left (2 \pi \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (-5+2 x \right )}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (-5+2 x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (-5+2 x \right )}\right )-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (-5+2 x \right )}\right )^{2}+\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (-5+2 x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (-5+2 x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (-5+2 x \right )}\right )^{3}+2 i \ln \left (5\right )+6 i \ln \left (2\right )+4 i \ln \left (x \right )-2 \pi \right )}{2}\right )}{4}\) | \(213\) |
Input:
int(((-5+2*x)*ln(-5+2*x)-x)/((4*x^2-10*x)*ln(-5+2*x)*ln(-8/5*x^2/ln(-5+2*x ))+(8*x^2-20*x)*ln(5)*ln(-5+2*x)),x,method=_RETURNVERBOSE)
Output:
1/4*ln(ln(-8/5*x^2/ln(-5+2*x))+2*ln(5))
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {1}{4} \, \log \left (2 \, \log \left (5\right ) + \log \left (-\frac {8 \, x^{2}}{5 \, \log \left (2 \, x - 5\right )}\right )\right ) \] Input:
integrate(((-5+2*x)*log(-5+2*x)-x)/((4*x^2-10*x)*log(-5+2*x)*log(-8/5*x^2/ log(-5+2*x))+(8*x^2-20*x)*log(5)*log(-5+2*x)),x, algorithm="fricas")
Output:
1/4*log(2*log(5) + log(-8/5*x^2/log(2*x - 5)))
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {\log {\left (\log {\left (- \frac {8 x^{2}}{5 \log {\left (2 x - 5 \right )}} \right )} + 2 \log {\left (5 \right )} \right )}}{4} \] Input:
integrate(((-5+2*x)*ln(-5+2*x)-x)/((4*x**2-10*x)*ln(-5+2*x)*ln(-8/5*x**2/l n(-5+2*x))+(8*x**2-20*x)*ln(5)*ln(-5+2*x)),x)
Output:
log(log(-8*x**2/(5*log(2*x - 5))) + 2*log(5))/4
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (5\right ) - 3 \, \log \left (2\right ) - 2 \, \log \left (x\right ) + \log \left (-\log \left (2 \, x - 5\right )\right )\right ) \] Input:
integrate(((-5+2*x)*log(-5+2*x)-x)/((4*x^2-10*x)*log(-5+2*x)*log(-8/5*x^2/ log(-5+2*x))+(8*x^2-20*x)*log(5)*log(-5+2*x)),x, algorithm="maxima")
Output:
1/4*log(-log(5) - 3*log(2) - 2*log(x) + log(-log(2*x - 5)))
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (5\right ) - \log \left (-8 \, x^{2}\right ) + \log \left (\log \left (2 \, x - 5\right )\right )\right ) \] Input:
integrate(((-5+2*x)*log(-5+2*x)-x)/((4*x^2-10*x)*log(-5+2*x)*log(-8/5*x^2/ log(-5+2*x))+(8*x^2-20*x)*log(5)*log(-5+2*x)),x, algorithm="giac")
Output:
1/4*log(-log(5) - log(-8*x^2) + log(log(2*x - 5)))
Time = 4.62 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {\ln \left (\ln \left (-\frac {40\,x^2}{\ln \left (2\,x-5\right )}\right )\right )}{4} \] Input:
int((x - log(2*x - 5)*(2*x - 5))/(log(2*x - 5)*log(-(8*x^2)/(5*log(2*x - 5 )))*(10*x - 4*x^2) + log(5)*log(2*x - 5)*(20*x - 8*x^2)),x)
Output:
log(log(-(40*x^2)/log(2*x - 5)))/4
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-x+(-5+2 x) \log (-5+2 x)}{\left (-20 x+8 x^2\right ) \log (5) \log (-5+2 x)+\left (-10 x+4 x^2\right ) \log (-5+2 x) \log \left (-\frac {8 x^2}{5 \log (-5+2 x)}\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (-\frac {8 x^{2}}{5 \,\mathrm {log}\left (2 x -5\right )}\right )+2 \,\mathrm {log}\left (5\right )\right )}{4} \] Input:
int(((-5+2*x)*log(-5+2*x)-x)/((4*x^2-10*x)*log(-5+2*x)*log(-8/5*x^2/log(-5 +2*x))+(8*x^2-20*x)*log(5)*log(-5+2*x)),x)
Output:
log(log(( - 8*x**2)/(5*log(2*x - 5))) + 2*log(5))/4