Integrand size = 82, antiderivative size = 25 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\left (2+\frac {5}{x}\right )^2+\frac {\log \left (\log \left (x+\frac {1+x}{4}\right )\right )}{x} \] Output:
(5/x+2)^2+ln(ln(1/4+5/4*x))/x
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {25}{x^2}+\frac {20}{x}+\frac {\log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x} \] Input:
Integrate[(5*x^2 + (-50 - 270*x - 100*x^2)*Log[(1 + 5*x)/4] + (-x - 5*x^2) *Log[(1 + 5*x)/4]*Log[Log[(1 + 5*x)/4]])/((x^3 + 5*x^4)*Log[(1 + 5*x)/4]), x]
Output:
25/x^2 + 20/x + Log[Log[(1 + 5*x)/4]]/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.
\(\displaystyle \int \frac {5 x^2+\left (-100 x^2-270 x-50\right ) \log \left (\frac {1}{4} (5 x+1)\right )+\left (-5 x^2-x\right ) \log \left (\frac {1}{4} (5 x+1)\right ) \log \left (\log \left (\frac {1}{4} (5 x+1)\right )\right )}{\left (5 x^4+x^3\right ) \log \left (\frac {1}{4} (5 x+1)\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {5 x^2+\left (-100 x^2-270 x-50\right ) \log \left (\frac {1}{4} (5 x+1)\right )+\left (-5 x^2-x\right ) \log \left (\frac {1}{4} (5 x+1)\right ) \log \left (\log \left (\frac {1}{4} (5 x+1)\right )\right )}{x^3 (5 x+1) \log \left (\frac {1}{4} (5 x+1)\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {5}{x (5 x+1) \log \left (\frac {5 x}{4}+\frac {1}{4}\right )}-\frac {20 x+x \log \left (\log \left (\frac {1}{4} (5 x+1)\right )\right )+50}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log \left (\log \left (\frac {5 x}{4}+\frac {1}{4}\right )\right )}{x^2}dx+5 \int \frac {1}{x \log \left (\frac {5 x}{4}+\frac {1}{4}\right )}dx+\frac {(2 x+5)^2}{x^2}-5 \log \left (\log \left (\frac {1}{4} (5 x+1)\right )\right )\) |
Input:
Int[(5*x^2 + (-50 - 270*x - 100*x^2)*Log[(1 + 5*x)/4] + (-x - 5*x^2)*Log[( 1 + 5*x)/4]*Log[Log[(1 + 5*x)/4]])/((x^3 + 5*x^4)*Log[(1 + 5*x)/4]),x]
Output:
$Aborted
Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {\ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x}+\frac {20 x +25}{x^{2}}\) | \(23\) |
parallelrisch | \(-\frac {-1250+2500 x^{2}-50 x \ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )-1000 x}{50 x^{2}}\) | \(26\) |
Input:
int(((-5*x^2-x)*ln(1/4+5/4*x)*ln(ln(1/4+5/4*x))+(-100*x^2-270*x-50)*ln(1/4 +5/4*x)+5*x^2)/(5*x^4+x^3)/ln(1/4+5/4*x),x,method=_RETURNVERBOSE)
Output:
ln(ln(1/4+5/4*x))/x+5*(4*x+5)/x^2
Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x \log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right ) + 20 \, x + 25}{x^{2}} \] Input:
integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-5 0)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)/log(1/4+5/4*x),x, algorithm="fricas")
Output:
(x*log(log(5/4*x + 1/4)) + 20*x + 25)/x^2
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {\log {\left (\log {\left (\frac {5 x}{4} + \frac {1}{4} \right )} \right )}}{x} - \frac {- 20 x - 25}{x^{2}} \] Input:
integrate(((-5*x**2-x)*ln(1/4+5/4*x)*ln(ln(1/4+5/4*x))+(-100*x**2-270*x-50 )*ln(1/4+5/4*x)+5*x**2)/(5*x**4+x**3)/ln(1/4+5/4*x),x)
Output:
log(log(5*x/4 + 1/4))/x - (-20*x - 25)/x**2
Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x \log \left (-2 \, \log \left (2\right ) + \log \left (5 \, x + 1\right )\right ) + 20 \, x + 25}{x^{2}} \] Input:
integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-5 0)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)/log(1/4+5/4*x),x, algorithm="maxima")
Output:
(x*log(-2*log(2) + log(5*x + 1)) + 20*x + 25)/x^2
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {\log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right )}{x} + \frac {5 \, {\left (4 \, x + 5\right )}}{x^{2}} \] Input:
integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-5 0)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)/log(1/4+5/4*x),x, algorithm="giac")
Output:
log(log(5/4*x + 1/4))/x + 5*(4*x + 5)/x^2
Time = 3.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x\,\left (\ln \left (\ln \left (\frac {5\,x}{4}+\frac {1}{4}\right )\right )+20\right )+25}{x^2} \] Input:
int(-(log((5*x)/4 + 1/4)*(270*x + 100*x^2 + 50) - 5*x^2 + log(log((5*x)/4 + 1/4))*log((5*x)/4 + 1/4)*(x + 5*x^2))/(log((5*x)/4 + 1/4)*(x^3 + 5*x^4)) ,x)
Output:
(x*(log(log((5*x)/4 + 1/4)) + 20) + 25)/x^2
Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {5 x}{4}+\frac {1}{4}\right )\right ) x +20 x +25}{x^{2}} \] Input:
int(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-50)*log (1/4+5/4*x)+5*x^2)/(5*x^4+x^3)/log(1/4+5/4*x),x)
Output:
(log(log((5*x + 1)/4))*x + 20*x + 25)/x**2