Integrand size = 132, antiderivative size = 26 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{x^2 \log \left (\frac {24}{5}-\frac {x}{5}+x^2+\log (4-x)\right )} \] Output:
3/ln(ln(4-x)+x^2-1/5*x+24/5)/x^2
Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{x^2 \log \left (\frac {24}{5}-\frac {x}{5}+x^2+\log (4-x)\right )} \] Input:
Integrate[(-27*x + 123*x^2 - 30*x^3 + (576 - 168*x + 126*x^2 - 30*x^3 + (1 20 - 30*x)*Log[4 - x])*Log[(24 - x + 5*x^2 + 5*Log[4 - x])/5])/((-96*x^3 + 28*x^4 - 21*x^5 + 5*x^6 + (-20*x^3 + 5*x^4)*Log[4 - x])*Log[(24 - x + 5*x ^2 + 5*Log[4 - x])/5]^2),x]
Output:
3/(x^2*Log[24/5 - x/5 + x^2 + Log[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.
\(\displaystyle \int \frac {-30 x^3+123 x^2+\left (-30 x^3+126 x^2-168 x+(120-30 x) \log (4-x)+576\right ) \log \left (\frac {1}{5} \left (5 x^2-x+5 \log (4-x)+24\right )\right )-27 x}{\left (5 x^6-21 x^5+28 x^4-96 x^3+\left (5 x^4-20 x^3\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (5 x^2-x+5 \log (4-x)+24\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {30 x^3-123 x^2-\left (-30 x^3+126 x^2-168 x+(120-30 x) \log (4-x)+576\right ) \log \left (\frac {1}{5} \left (5 x^2-x+5 \log (4-x)+24\right )\right )+27 x}{(4-x) x^3 \left (5 x^2-x+5 \log (4-x)+24\right ) \log ^2\left (\frac {1}{5} \left (5 x^2-x+5 \log (4-x)+24\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {3 \left (10 x^2-41 x+9\right )}{(x-4) x^2 \left (5 x^2-x+5 \log (4-x)+24\right ) \log ^2\left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}-\frac {6}{x^3 \log \left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {15}{16} \int \frac {1}{(x-4) \left (5 x^2-x+5 \log (4-x)+24\right ) \log ^2\left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}dx+\frac {27}{4} \int \frac {1}{x^2 \left (5 x^2-x+5 \log (4-x)+24\right ) \log ^2\left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}dx-\frac {465}{16} \int \frac {1}{x \left (5 x^2-x+5 \log (4-x)+24\right ) \log ^2\left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}dx-6 \int \frac {1}{x^3 \log \left (x^2-\frac {x}{5}+\log (4-x)+\frac {24}{5}\right )}dx\) |
Input:
Int[(-27*x + 123*x^2 - 30*x^3 + (576 - 168*x + 126*x^2 - 30*x^3 + (120 - 3 0*x)*Log[4 - x])*Log[(24 - x + 5*x^2 + 5*Log[4 - x])/5])/((-96*x^3 + 28*x^ 4 - 21*x^5 + 5*x^6 + (-20*x^3 + 5*x^4)*Log[4 - x])*Log[(24 - x + 5*x^2 + 5 *Log[4 - x])/5]^2),x]
Output:
$Aborted
Time = 1.81 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {3}{\ln \left (\ln \left (-x +4\right )+x^{2}-\frac {x}{5}+\frac {24}{5}\right ) x^{2}}\) | \(23\) |
parallelrisch | \(\frac {3}{\ln \left (\ln \left (-x +4\right )+x^{2}-\frac {x}{5}+\frac {24}{5}\right ) x^{2}}\) | \(23\) |
default | \(-\frac {3}{x^{2} \left (\ln \left (5\right )-\ln \left (5 \left (-x +4\right )^{2}+39 x -56+5 \ln \left (-x +4\right )\right )\right )}\) | \(36\) |
Input:
int((((-30*x+120)*ln(-x+4)-30*x^3+126*x^2-168*x+576)*ln(ln(-x+4)+x^2-1/5*x +24/5)-30*x^3+123*x^2-27*x)/((5*x^4-20*x^3)*ln(-x+4)+5*x^6-21*x^5+28*x^4-9 6*x^3)/ln(ln(-x+4)+x^2-1/5*x+24/5)^2,x,method=_RETURNVERBOSE)
Output:
3/ln(ln(-x+4)+x^2-1/5*x+24/5)/x^2
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{x^{2} \log \left (x^{2} - \frac {1}{5} \, x + \log \left (-x + 4\right ) + \frac {24}{5}\right )} \] Input:
integrate((((-30*x+120)*log(-x+4)-30*x^3+126*x^2-168*x+576)*log(log(-x+4)+ x^2-1/5*x+24/5)-30*x^3+123*x^2-27*x)/((5*x^4-20*x^3)*log(-x+4)+5*x^6-21*x^ 5+28*x^4-96*x^3)/log(log(-x+4)+x^2-1/5*x+24/5)^2,x, algorithm="fricas")
Output:
3/(x^2*log(x^2 - 1/5*x + log(-x + 4) + 24/5))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{x^{2} \log {\left (x^{2} - \frac {x}{5} + \log {\left (4 - x \right )} + \frac {24}{5} \right )}} \] Input:
integrate((((-30*x+120)*ln(-x+4)-30*x**3+126*x**2-168*x+576)*ln(ln(-x+4)+x **2-1/5*x+24/5)-30*x**3+123*x**2-27*x)/((5*x**4-20*x**3)*ln(-x+4)+5*x**6-2 1*x**5+28*x**4-96*x**3)/ln(ln(-x+4)+x**2-1/5*x+24/5)**2,x)
Output:
3/(x**2*log(x**2 - x/5 + log(4 - x) + 24/5))
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=-\frac {3}{x^{2} \log \left (5\right ) - x^{2} \log \left (5 \, x^{2} - x + 5 \, \log \left (-x + 4\right ) + 24\right )} \] Input:
integrate((((-30*x+120)*log(-x+4)-30*x^3+126*x^2-168*x+576)*log(log(-x+4)+ x^2-1/5*x+24/5)-30*x^3+123*x^2-27*x)/((5*x^4-20*x^3)*log(-x+4)+5*x^6-21*x^ 5+28*x^4-96*x^3)/log(log(-x+4)+x^2-1/5*x+24/5)^2,x, algorithm="maxima")
Output:
-3/(x^2*log(5) - x^2*log(5*x^2 - x + 5*log(-x + 4) + 24))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=-\frac {3}{x^{2} \log \left (5\right ) - x^{2} \log \left (5 \, x^{2} - x + 5 \, \log \left (-x + 4\right ) + 24\right )} \] Input:
integrate((((-30*x+120)*log(-x+4)-30*x^3+126*x^2-168*x+576)*log(log(-x+4)+ x^2-1/5*x+24/5)-30*x^3+123*x^2-27*x)/((5*x^4-20*x^3)*log(-x+4)+5*x^6-21*x^ 5+28*x^4-96*x^3)/log(log(-x+4)+x^2-1/5*x+24/5)^2,x, algorithm="giac")
Output:
-3/(x^2*log(5) - x^2*log(5*x^2 - x + 5*log(-x + 4) + 24))
Time = 3.48 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{x^2\,\ln \left (\ln \left (4-x\right )-\frac {x}{5}+x^2+\frac {24}{5}\right )} \] Input:
int((27*x + log(log(4 - x) - x/5 + x^2 + 24/5)*(168*x + log(4 - x)*(30*x - 120) - 126*x^2 + 30*x^3 - 576) - 123*x^2 + 30*x^3)/(log(log(4 - x) - x/5 + x^2 + 24/5)^2*(log(4 - x)*(20*x^3 - 5*x^4) + 96*x^3 - 28*x^4 + 21*x^5 - 5*x^6)),x)
Output:
3/(x^2*log(log(4 - x) - x/5 + x^2 + 24/5))
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-27 x+123 x^2-30 x^3+\left (576-168 x+126 x^2-30 x^3+(120-30 x) \log (4-x)\right ) \log \left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )}{\left (-96 x^3+28 x^4-21 x^5+5 x^6+\left (-20 x^3+5 x^4\right ) \log (4-x)\right ) \log ^2\left (\frac {1}{5} \left (24-x+5 x^2+5 \log (4-x)\right )\right )} \, dx=\frac {3}{\mathrm {log}\left (\mathrm {log}\left (-x +4\right )+x^{2}-\frac {x}{5}+\frac {24}{5}\right ) x^{2}} \] Input:
int((((-30*x+120)*log(-x+4)-30*x^3+126*x^2-168*x+576)*log(log(-x+4)+x^2-1/ 5*x+24/5)-30*x^3+123*x^2-27*x)/((5*x^4-20*x^3)*log(-x+4)+5*x^6-21*x^5+28*x ^4-96*x^3)/log(log(-x+4)+x^2-1/5*x+24/5)^2,x)
Output:
3/(log((5*log( - x + 4) + 5*x**2 - x + 24)/5)*x**2)