Integrand size = 51, antiderivative size = 25 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 \left (x+\log \left (5+\frac {-3-2 (7+\log (x))-\log (3 x)}{x}\right )\right ) \] Output:
5*ln((-17-ln(3*x)-2*ln(x))/x+5)+5*x
Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=-5 (-x+\log (x)-\log (17-5 x+\log (3)+3 \log (x))) \] Input:
Integrate[(-70 + 85*x - 25*x^2 + (-10 + 10*x)*Log[x] + (-5 + 5*x)*Log[3*x] )/(17*x - 5*x^2 + 2*x*Log[x] + x*Log[3*x]),x]
Output:
-5*(-x + Log[x] - Log[17 - 5*x + Log[3] + 3*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.
\(\displaystyle \int \frac {-25 x^2+85 x+(10 x-10) \log (x)+(5 x-5) \log (3 x)-70}{-5 x^2+17 x+2 x \log (x)+x \log (3 x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 \left (-5 x^2+3 x \log (x)+17 x \left (1+\frac {\log (3)}{17}\right )-3 \log (x)-14 \left (1+\frac {\log (3)}{14}\right )\right )}{-5 x^2+17 x+2 x \log (x)+x \log (3 x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {5 x^2-3 \log (x) x-(17+\log (3)) x+3 \log (x)+\log (3)+14}{-5 x^2+2 \log (x) x+\log (3 x) x+17 x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {5 x^2-3 \log (x) x-(17+\log (3)) x+3 \log (x)+\log (3)+14}{-5 x^2+2 \log (x) x+\log (3 x) x+17 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -5 \int \frac {5 x^2-3 \log (x) x-(17+\log (3)) x+3 \log (x)+14 \left (1+\frac {\log (3)}{14}\right )}{-5 x^2+2 \log (x) x+\log (3 x) x+17 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {5 x}{-5 x+3 \log (x)+17 \left (1+\frac {\log (3)}{17}\right )}+\frac {3 \log (x)}{5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )}+\frac {17+\log (3)}{5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )}+\frac {-14-\log (3)}{\left (5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )\right ) x}+\frac {3 \log (x)}{\left (-5 x+3 \log (x)+17 \left (1+\frac {\log (3)}{17}\right )\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left ((17+\log (3)) \int \frac {1}{5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )}dx+(17+\log (3)) \int \frac {1}{x \left (5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )\right )}dx-(14+\log (3)) \int \frac {1}{x \left (5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )\right )}dx+5 \int \frac {x}{5 x-3 \log (x)-17 \left (1+\frac {\log (3)}{17}\right )}dx+(17+\log (3)) \int \frac {1}{-5 x+3 \log (x)+17 \left (1+\frac {\log (3)}{17}\right )}dx+5 \int \frac {1}{-5 x+3 \log (x)+17 \left (1+\frac {\log (3)}{17}\right )}dx+5 \int \frac {x}{-5 x+3 \log (x)+17 \left (1+\frac {\log (3)}{17}\right )}dx-x+\log (x)\right )\) |
Input:
Int[(-70 + 85*x - 25*x^2 + (-10 + 10*x)*Log[x] + (-5 + 5*x)*Log[3*x])/(17* x - 5*x^2 + 2*x*Log[x] + x*Log[3*x]),x]
Output:
$Aborted
Time = 0.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
default | \(-5 \ln \left (x \right )+5 x +5 \ln \left (3 \ln \left (x \right )+\ln \left (3\right )-5 x +17\right )\) | \(23\) |
parallelrisch | \(5 \ln \left (-\frac {2 \ln \left (x \right )}{5}-\frac {\ln \left (3 x \right )}{5}+x -\frac {17}{5}\right )+5 x -5 \ln \left (x \right )\) | \(25\) |
norman | \(-5 \ln \left (3 x \right )+5 x +5 \ln \left (5 x -2 \ln \left (x \right )-\ln \left (3 x \right )-17\right )\) | \(29\) |
risch | \(5 x -5 \ln \left (x \right )+5 \ln \left (\ln \left (x \right )-\frac {i \left (2 i \ln \left (3\right )-10 i x +34 i\right )}{6}\right )\) | \(30\) |
Input:
int(((5*x-5)*ln(3*x)+(10*x-10)*ln(x)-25*x^2+85*x-70)/(x*ln(3*x)+2*x*ln(x)- 5*x^2+17*x),x,method=_RETURNVERBOSE)
Output:
-5*ln(x)+5*x+5*ln(3*ln(x)+ln(3)-5*x+17)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 \, x - 5 \, \log \left (x\right ) + 5 \, \log \left (-5 \, x + \log \left (3\right ) + 3 \, \log \left (x\right ) + 17\right ) \] Input:
integrate(((5*x-5)*log(3*x)+(10*x-10)*log(x)-25*x^2+85*x-70)/(x*log(3*x)+2 *x*log(x)-5*x^2+17*x),x, algorithm="fricas")
Output:
5*x - 5*log(x) + 5*log(-5*x + log(3) + 3*log(x) + 17)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 x - 5 \log {\left (x \right )} + 5 \log {\left (- \frac {5 x}{3} + \log {\left (x \right )} + \frac {\log {\left (3 \right )}}{3} + \frac {17}{3} \right )} \] Input:
integrate(((5*x-5)*ln(3*x)+(10*x-10)*ln(x)-25*x**2+85*x-70)/(x*ln(3*x)+2*x *ln(x)-5*x**2+17*x),x)
Output:
5*x - 5*log(x) + 5*log(-5*x/3 + log(x) + log(3)/3 + 17/3)
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 \, x - 5 \, \log \left (x\right ) + 5 \, \log \left (-\frac {5}{3} \, x + \frac {1}{3} \, \log \left (3\right ) + \log \left (x\right ) + \frac {17}{3}\right ) \] Input:
integrate(((5*x-5)*log(3*x)+(10*x-10)*log(x)-25*x^2+85*x-70)/(x*log(3*x)+2 *x*log(x)-5*x^2+17*x),x, algorithm="maxima")
Output:
5*x - 5*log(x) + 5*log(-5/3*x + 1/3*log(3) + log(x) + 17/3)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 \, x - 5 \, \log \left (x\right ) + 5 \, \log \left (-5 \, x + \log \left (3\right ) + 3 \, \log \left (x\right ) + 17\right ) \] Input:
integrate(((5*x-5)*log(3*x)+(10*x-10)*log(x)-25*x^2+85*x-70)/(x*log(3*x)+2 *x*log(x)-5*x^2+17*x),x, algorithm="giac")
Output:
5*x - 5*log(x) + 5*log(-5*x + log(3) + 3*log(x) + 17)
Time = 1.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5\,x+5\,\ln \left (\ln \left (3\right )-5\,x+3\,\ln \left (x\right )+17\right )-5\,\ln \left (x\right ) \] Input:
int((85*x + log(x)*(10*x - 10) - 25*x^2 + log(3*x)*(5*x - 5) - 70)/(17*x + x*log(3*x) + 2*x*log(x) - 5*x^2),x)
Output:
5*x + 5*log(log(3) - 5*x + 3*log(x) + 17) - 5*log(x)
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-70+85 x-25 x^2+(-10+10 x) \log (x)+(-5+5 x) \log (3 x)}{17 x-5 x^2+2 x \log (x)+x \log (3 x)} \, dx=5 \,\mathrm {log}\left (\mathrm {log}\left (3 x \right )+2 \,\mathrm {log}\left (x \right )-5 x +17\right )-5 \,\mathrm {log}\left (3 x \right )+5 x \] Input:
int(((5*x-5)*log(3*x)+(10*x-10)*log(x)-25*x^2+85*x-70)/(x*log(3*x)+2*x*log (x)-5*x^2+17*x),x)
Output:
5*(log(log(3*x) + 2*log(x) - 5*x + 17) - log(3*x) + x)