Integrand size = 72, antiderivative size = 22 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {12}{5 \left (-x+\frac {\log \left (\frac {x}{5}\right )}{4+x}\right )} \] Output:
12/5/(ln(1/5*x)/(4+x)-x)
Time = 5.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 (4+x)}{5 \left (4 x+x^2+\log (5)-\log (x)\right )} \] Input:
Integrate[(-48 + 180*x + 96*x^2 + 12*x^3 + 12*x*Log[x/5])/(80*x^3 + 40*x^4 + 5*x^5 + (-40*x^2 - 10*x^3)*Log[x/5] + 5*x*Log[x/5]^2),x]
Output:
(-12*(4 + x))/(5*(4*x + x^2 + Log[5] - 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.
\(\displaystyle \int \frac {12 x^3+96 x^2+180 x+12 x \log \left (\frac {x}{5}\right )-48}{5 x^5+40 x^4+80 x^3+\left (-10 x^3-40 x^2\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {12 \left (x^3+8 x^2+15 x+x \log \left (\frac {x}{5}\right )-4\right )}{5 x \left (x^2+4 x-\log (x)+\log (5)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {12}{5} \int -\frac {-x^3-8 x^2-\log \left (\frac {x}{5}\right ) x-15 x+4}{x \left (x^2+4 x-\log (x)+\log (5)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {12}{5} \int \frac {-x^3-8 x^2-\log \left (\frac {x}{5}\right ) x-15 x+4}{x \left (x^2+4 x-\log (x)+\log (5)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {12}{5} \int \left (-\frac {x^2}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}-\frac {8 x}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}-\frac {\log \left (\frac {x}{5}\right )}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}-\frac {15}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}+\frac {4}{\left (x^2+4 x-\log (x)+\log (5)\right )^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {12}{5} \left (-5 \text {Subst}\left (\int \frac {\log (x)}{\left (25 x^2+20 x-\log (5 x)+\log (5)\right )^2}dx,x,\frac {x}{5}\right )-15 \int \frac {1}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}dx+4 \int \frac {1}{x \left (x^2+4 x-\log (x)+\log (5)\right )^2}dx-8 \int \frac {x}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}dx-\int \frac {x^2}{\left (x^2+4 x-\log (x)+\log (5)\right )^2}dx\right )\) |
Input:
Int[(-48 + 180*x + 96*x^2 + 12*x^3 + 12*x*Log[x/5])/(80*x^3 + 40*x^4 + 5*x ^5 + (-40*x^2 - 10*x^3)*Log[x/5] + 5*x*Log[x/5]^2),x]
Output:
$Aborted
Time = 0.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {48}{5}+\frac {12 x}{5}}{-x^{2}+\ln \left (\frac {x}{5}\right )-4 x}\) | \(21\) |
default | \(\frac {\frac {48}{5}+\frac {12 x}{5}}{-x^{2}+\ln \left (\frac {x}{5}\right )-4 x}\) | \(21\) |
risch | \(-\frac {12 \left (4+x \right )}{5 \left (x^{2}+4 x -\ln \left (\frac {x}{5}\right )\right )}\) | \(21\) |
norman | \(\frac {-\frac {48}{5}-\frac {12 x}{5}}{x^{2}+4 x -\ln \left (\frac {x}{5}\right )}\) | \(22\) |
parallelrisch | \(\frac {-48-12 x}{5 x^{2}+20 x -5 \ln \left (\frac {x}{5}\right )}\) | \(23\) |
Input:
int((12*x*ln(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*ln(1/5*x)^2+(-10*x^3-40*x ^2)*ln(1/5*x)+5*x^5+40*x^4+80*x^3),x,method=_RETURNVERBOSE)
Output:
12/5*(4+x)/(-x^2+ln(1/5*x)-4*x)
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \] Input:
integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10* x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+80*x^3),x, algorithm="fricas")
Output:
-12/5*(x + 4)/(x^2 + 4*x - log(1/5*x))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {12 x + 48}{- 5 x^{2} - 20 x + 5 \log {\left (\frac {x}{5} \right )}} \] Input:
integrate((12*x*ln(1/5*x)+12*x**3+96*x**2+180*x-48)/(5*x*ln(1/5*x)**2+(-10 *x**3-40*x**2)*ln(1/5*x)+5*x**5+40*x**4+80*x**3),x)
Output:
(12*x + 48)/(-5*x**2 - 20*x + 5*log(x/5))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x + \log \left (5\right ) - \log \left (x\right )\right )}} \] Input:
integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10* x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+80*x^3),x, algorithm="maxima")
Output:
-12/5*(x + 4)/(x^2 + 4*x + log(5) - log(x))
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \] Input:
integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10* x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+80*x^3),x, algorithm="giac")
Output:
-12/5*(x + 4)/(x^2 + 4*x - log(1/5*x))
Time = 1.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {\frac {12\,x}{5}+\frac {48}{5}}{4\,x-\ln \left (\frac {x}{5}\right )+x^2} \] Input:
int((180*x + 12*x*log(x/5) + 96*x^2 + 12*x^3 - 48)/(5*x*log(x/5)^2 - log(x /5)*(40*x^2 + 10*x^3) + 80*x^3 + 40*x^4 + 5*x^5),x)
Output:
-((12*x)/5 + 48/5)/(4*x - log(x/5) + x^2)
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {12 x +48}{5 \,\mathrm {log}\left (\frac {x}{5}\right )-5 x^{2}-20 x} \] Input:
int((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10*x^3-40 *x^2)*log(1/5*x)+5*x^5+40*x^4+80*x^3),x)
Output:
(12*(x + 4))/(5*(log(x/5) - x**2 - 4*x))