Integrand size = 97, antiderivative size = 36 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {5+x^2+\frac {1}{5} (4+x) \log \left (x-\frac {x}{2 \log (5)}\right )}{x (x-\log (x))} \] Output:
(1/5*(4+x)*ln(x-1/2*x/ln(5))+x^2+5)/(x-ln(x))/x
Time = 0.79 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=-\frac {-25-6 x^2+x \log (x)-(4+x) \log \left (x-\frac {x}{\log (25)}\right )}{5 x (x-\log (x))} \] Input:
Integrate[(25 - 46*x + 6*x^2 + (4 - 7*x - x^2)*Log[(-x + 2*x*Log[5])/(2*Lo g[5])] + Log[x]*(21 - x - 5*x^2 + 4*Log[(-x + 2*x*Log[5])/(2*Log[5])]))/(5 *x^4 - 10*x^3*Log[x] + 5*x^2*Log[x]^2),x]
Output:
-1/5*(-25 - 6*x^2 + x*Log[x] - (4 + x)*Log[x - x/Log[25]])/(x*(x - Log[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.
| \(\displaystyle \int \frac {6 x^2+\left (-x^2-7 x+4\right ) \log \left (\frac {2 x \log (5)-x}{2 \log (5)}\right )+\log (x) \left (-5 x^2-x+4 \log \left (\frac {2 x \log (5)-x}{2 \log (5)}\right )+21\right )-46 x+25}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {6 x^2+\left (-x^2-7 x+4\right ) \log \left (\frac {2 x \log (5)-x}{2 \log (5)}\right )+\log (x) \left (-5 x^2-x+4 \log \left (\frac {2 x \log (5)-x}{2 \log (5)}\right )+21\right )-46 x+25}{5 x^2 (x-\log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {6 x^2-46 x+\left (-x^2-7 x+4\right ) \log \left (-\frac {x (1-\log (25))}{2 \log (5)}\right )+\log (x) \left (-5 x^2-x+4 \log \left (-\frac {x (1-\log (25))}{2 \log (5)}\right )+21\right )+25}{x^2 (x-\log (x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{5} \int \left (-\frac {\log (x)}{x (x-\log (x))^2}+\frac {21 \log (x)}{x^2 (x-\log (x))^2}-\frac {5 \log (x)}{(x-\log (x))^2}+\frac {\left (-x^2-7 x+4 \log (x)+4\right ) \log \left (x \left (1-\frac {1}{\log (25)}\right )\right )}{x^2 (x-\log (x))^2}-\frac {46}{x (x-\log (x))^2}+\frac {25}{x^2 (x-\log (x))^2}+\frac {6}{(x-\log (x))^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (25 \int \frac {1}{x^2 (x-\log (x))^2}dx-21 \int \frac {1}{x^2 (x-\log (x))}dx+4 \int \frac {\log \left (x \left (1-\frac {1}{\log (25)}\right )\right )}{x^2 (x-\log (x))^2}dx+4 \int \frac {\log (x) \log \left (x \left (1-\frac {1}{\log (25)}\right )\right )}{x^2 (x-\log (x))^2}dx+5 \int \frac {1}{(x-\log (x))^2}dx-25 \int \frac {1}{x (x-\log (x))^2}dx-5 \int \frac {x}{(x-\log (x))^2}dx+\int \frac {1}{x (x-\log (x))}dx-5 \int \frac {1}{\log (x)-x}dx-\int \frac {\log \left (x \left (1-\frac {1}{\log (25)}\right )\right )}{(x-\log (x))^2}dx-7 \int \frac {\log \left (x \left (1-\frac {1}{\log (25)}\right )\right )}{x (x-\log (x))^2}dx\right )\) |
Input:
Int[(25 - 46*x + 6*x^2 + (4 - 7*x - x^2)*Log[(-x + 2*x*Log[5])/(2*Log[5])] + Log[x]*(21 - x - 5*x^2 + 4*Log[(-x + 2*x*Log[5])/(2*Log[5])]))/(5*x^4 - 10*x^3*Log[x] + 5*x^2*Log[x]^2),x]
Output:
$Aborted
Time = 0.83 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.47
| method | result | size |
| parallelrisch | \(\frac {25+5 x^{2}+\ln \left (\frac {x \left (2 \ln \left (5\right )-1\right )}{2 \ln \left (5\right )}\right ) x +4 \ln \left (\frac {x \left (2 \ln \left (5\right )-1\right )}{2 \ln \left (5\right )}\right )}{5 x \left (x -\ln \left (x \right )\right )}\) | \(53\) |
| risch | \(-\frac {4}{5 x}+\frac {50+2 x \ln \left (2 \ln \left (5\right )-1\right )-2 x \ln \left (2\right )-2 x \ln \left (\ln \left (5\right )\right )+12 x^{2}+8 \ln \left (2 \ln \left (5\right )-1\right )-8 \ln \left (2\right )-8 \ln \left (\ln \left (5\right )\right )+8 x}{10 x \left (x -\ln \left (x \right )\right )}\) | \(69\) |
| default | \(\frac {-25-6 x^{2}-4 \ln \left (x \right )}{5 x \left (\ln \left (x \right )-x \right )}+\frac {\ln \left (2 \ln \left (5\right )-1\right ) \left (-4-x \right )}{5 x \left (\ln \left (x \right )-x \right )}-\frac {\ln \left (2\right ) \left (-4-x \right )}{5 x \left (\ln \left (x \right )-x \right )}-\frac {\ln \left (\ln \left (5\right )\right ) \left (-4-x \right )}{5 x \left (\ln \left (x \right )-x \right )}\) | \(92\) |
Input:
int(((4*ln(1/2*(2*x*ln(5)-x)/ln(5))-5*x^2-x+21)*ln(x)+(-x^2-7*x+4)*ln(1/2* (2*x*ln(5)-x)/ln(5))+6*x^2-46*x+25)/(5*x^2*ln(x)^2-10*x^3*ln(x)+5*x^4),x,m ethod=_RETURNVERBOSE)
Output:
1/5/x*(25+5*x^2+ln(1/2*x*(2*ln(5)-1)/ln(5))*x+4*ln(1/2*x*(2*ln(5)-1)/ln(5) ))/(x-ln(x))
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (34) = 68\).
Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.33 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {6 \, x^{2} - x \log \left (\frac {2 \, \log \left (5\right )}{2 \, \log \left (5\right ) - 1}\right ) + 4 \, \log \left (\frac {2 \, x \log \left (5\right ) - x}{2 \, \log \left (5\right )}\right ) + 25}{5 \, {\left (x^{2} - x \log \left (\frac {2 \, x \log \left (5\right ) - x}{2 \, \log \left (5\right )}\right ) - x \log \left (\frac {2 \, \log \left (5\right )}{2 \, \log \left (5\right ) - 1}\right )\right )}} \] Input:
integrate(((4*log(1/2*(2*x*log(5)-x)/log(5))-5*x^2-x+21)*log(x)+(-x^2-7*x+ 4)*log(1/2*(2*x*log(5)-x)/log(5))+6*x^2-46*x+25)/(5*x^2*log(x)^2-10*x^3*lo g(x)+5*x^4),x, algorithm="fricas")
Output:
1/5*(6*x^2 - x*log(2*log(5)/(2*log(5) - 1)) + 4*log(1/2*(2*x*log(5) - x)/l og(5)) + 25)/(x^2 - x*log(1/2*(2*x*log(5) - x)/log(5)) - x*log(2*log(5)/(2 *log(5) - 1)))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (27) = 54\).
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.94 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {- 6 x^{2} - 4 x - x \log {\left (-1 + 2 \log {\left (5 \right )} \right )} + x \log {\left (\log {\left (5 \right )} \right )} + x \log {\left (2 \right )} - 25 - 4 \log {\left (-1 + 2 \log {\left (5 \right )} \right )} + 4 \log {\left (\log {\left (5 \right )} \right )} + 4 \log {\left (2 \right )}}{- 5 x^{2} + 5 x \log {\left (x \right )}} - \frac {4}{5 x} \] Input:
integrate(((4*ln(1/2*(2*x*ln(5)-x)/ln(5))-5*x**2-x+21)*ln(x)+(-x**2-7*x+4) *ln(1/2*(2*x*ln(5)-x)/ln(5))+6*x**2-46*x+25)/(5*x**2*ln(x)**2-10*x**3*ln(x )+5*x**4),x)
Output:
(-6*x**2 - 4*x - x*log(-1 + 2*log(5)) + x*log(log(5)) + x*log(2) - 25 - 4* log(-1 + 2*log(5)) + 4*log(log(5)) + 4*log(2))/(-5*x**2 + 5*x*log(x)) - 4/ (5*x)
Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.67 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {6 \, x^{2} - x {\left (\log \left (2\right ) - \log \left (2 \, \log \left (5\right ) - 1\right ) + \log \left (\log \left (5\right )\right )\right )} - 4 \, \log \left (2\right ) + 4 \, \log \left (x\right ) + 4 \, \log \left (2 \, \log \left (5\right ) - 1\right ) - 4 \, \log \left (\log \left (5\right )\right ) + 25}{5 \, {\left (x^{2} - x \log \left (x\right )\right )}} \] Input:
integrate(((4*log(1/2*(2*x*log(5)-x)/log(5))-5*x^2-x+21)*log(x)+(-x^2-7*x+ 4)*log(1/2*(2*x*log(5)-x)/log(5))+6*x^2-46*x+25)/(5*x^2*log(x)^2-10*x^3*lo g(x)+5*x^4),x, algorithm="maxima")
Output:
1/5*(6*x^2 - x*(log(2) - log(2*log(5) - 1) + log(log(5))) - 4*log(2) + 4*l og(x) + 4*log(2*log(5) - 1) - 4*log(log(5)) + 25)/(x^2 - x*log(x))
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.86 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {6 \, x^{2} - x \log \left (2\right ) + x \log \left (2 \, \log \left (5\right ) - 1\right ) - x \log \left (\log \left (5\right )\right ) + 4 \, x - 4 \, \log \left (2\right ) + 4 \, \log \left (2 \, \log \left (5\right ) - 1\right ) - 4 \, \log \left (\log \left (5\right )\right ) + 25}{5 \, {\left (x^{2} - x \log \left (x\right )\right )}} - \frac {4}{5 \, x} \] Input:
integrate(((4*log(1/2*(2*x*log(5)-x)/log(5))-5*x^2-x+21)*log(x)+(-x^2-7*x+ 4)*log(1/2*(2*x*log(5)-x)/log(5))+6*x^2-46*x+25)/(5*x^2*log(x)^2-10*x^3*lo g(x)+5*x^4),x, algorithm="giac")
Output:
1/5*(6*x^2 - x*log(2) + x*log(2*log(5) - 1) - x*log(log(5)) + 4*x - 4*log( 2) + 4*log(2*log(5) - 1) - 4*log(log(5)) + 25)/(x^2 - x*log(x)) - 4/5/x
Time = 1.95 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.36 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {4\,\ln \left (\ln \left (5\right )-\frac {1}{2}\right )-4\,\ln \left (\ln \left (5\right )\right )+4\,\ln \left (x\right )+x\,\ln \left (\ln \left (5\right )-\frac {1}{2}\right )-x\,\ln \left (\ln \left (5\right )\right )+6\,x^2+25}{5\,x\,\left (x-\ln \left (x\right )\right )} \] Input:
int(-(46*x + log(-(x/2 - x*log(5))/log(5))*(7*x + x^2 - 4) + log(x)*(x - 4 *log(-(x/2 - x*log(5))/log(5)) + 5*x^2 - 21) - 6*x^2 - 25)/(5*x^2*log(x)^2 - 10*x^3*log(x) + 5*x^4),x)
Output:
(4*log(log(5) - 1/2) - 4*log(log(5)) + 4*log(x) + x*log(log(5) - 1/2) - x* log(log(5)) + 6*x^2 + 25)/(5*x*(x - log(x)))
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.58 \[ \int \frac {25-46 x+6 x^2+\left (4-7 x-x^2\right ) \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )+\log (x) \left (21-x-5 x^2+4 \log \left (\frac {-x+2 x \log (5)}{2 \log (5)}\right )\right )}{5 x^4-10 x^3 \log (x)+5 x^2 \log ^2(x)} \, dx=\frac {-\mathrm {log}\left (\frac {2 \,\mathrm {log}\left (5\right ) x -x}{2 \,\mathrm {log}\left (5\right )}\right ) x -4 \,\mathrm {log}\left (\frac {2 \,\mathrm {log}\left (5\right ) x -x}{2 \,\mathrm {log}\left (5\right )}\right )-5 \,\mathrm {log}\left (x \right ) x -25}{5 x \left (\mathrm {log}\left (x \right )-x \right )} \] Input:
int(((4*log(1/2*(2*x*log(5)-x)/log(5))-5*x^2-x+21)*log(x)+(-x^2-7*x+4)*log (1/2*(2*x*log(5)-x)/log(5))+6*x^2-46*x+25)/(5*x^2*log(x)^2-10*x^3*log(x)+5 *x^4),x)
Output:
( - log((2*log(5)*x - x)/(2*log(5)))*x - 4*log((2*log(5)*x - x)/(2*log(5)) ) - 5*log(x)*x - 25)/(5*x*(log(x) - x))