Integrand size = 114, antiderivative size = 24 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=\log ^4(3) (x+4 \log (\log (5)+\log ((5-x) (-1+x) x))) \]
Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=\log ^4(3) \left (x+4 \log \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )\right ) \]
Integrate[((20 - 48*x + 12*x^2)*Log[3]^4 + (5*x - 6*x^2 + x^3)*Log[3]^4*Lo g[5] + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[-5*x + 6*x^2 - x^3])/((5*x - 6*x^2 + x^3)*Log[5] + (5*x - 6*x^2 + x^3)*Log[-5*x + 6*x^2 - x^3]),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.
\(\displaystyle \int \frac {\left (12 x^2-48 x+20\right ) \log ^4(3)+\left (x^3-6 x^2+5 x\right ) \log ^4(3) \log \left (-x^3+6 x^2-5 x\right )+\left (x^3-6 x^2+5 x\right ) \log ^4(3) \log (5)}{\left (x^3-6 x^2+5 x\right ) \log \left (-x^3+6 x^2-5 x\right )+\left (x^3-6 x^2+5 x\right ) \log (5)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (12 x^2-48 x+20\right ) \log ^4(3)+\left (x^3-6 x^2+5 x\right ) \log ^4(3) \log \left (-x^3+6 x^2-5 x\right )+\left (x^3-6 x^2+5 x\right ) \log ^4(3) \log (5)}{x \left (x^2-6 x+5\right ) \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {\log ^4(3) \log \left (x \left (-x^2+6 x-5\right )\right )}{\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)}+\frac {\log ^4(3) \left (x^3 \log (5)+6 x^2 (2-\log (5))-x (48-5 \log (5))+20\right )}{(1-x) (5-x) x \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log ^4(3) \log (5) \int \frac {1}{\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)}dx+4 \log ^4(3) \int \frac {1}{(x-5) \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )}dx+4 \log ^4(3) \int \frac {1}{(x-1) \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )}dx+4 \log ^4(3) \int \frac {1}{x \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )}dx+\log ^4(3) \int \frac {\log \left (x \left (-x^2+6 x-5\right )\right )}{\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)}dx\) |
Int[((20 - 48*x + 12*x^2)*Log[3]^4 + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[5] + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[-5*x + 6*x^2 - x^3])/((5*x - 6*x^2 + x^3 )*Log[5] + (5*x - 6*x^2 + x^3)*Log[-5*x + 6*x^2 - x^3]),x]
3.21.82.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38
method | result | size |
norman | \(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\) | \(33\) |
risch | \(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\) | \(33\) |
parallelrisch | \(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\) | \(33\) |
int(((x^3-6*x^2+5*x)*ln(3)^4*ln(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*ln(3)^4*ln (5)+(12*x^2-48*x+20)*ln(3)^4)/((x^3-6*x^2+5*x)*ln(-x^3+6*x^2-5*x)+(x^3-6*x ^2+5*x)*ln(5)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (-x^{3} + 6 \, x^{2} - 5 \, x\right )\right ) \]
integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*lo g(3)^4*log(5)+(12*x^2-48*x+20)*log(3)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5 *x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log {\left (3 \right )}^{4} + 4 \log {\left (3 \right )}^{4} \log {\left (\log {\left (- x^{3} + 6 x^{2} - 5 x \right )} + \log {\left (5 \right )} \right )} \]
integrate(((x**3-6*x**2+5*x)*ln(3)**4*ln(-x**3+6*x**2-5*x)+(x**3-6*x**2+5* x)*ln(3)**4*ln(5)+(12*x**2-48*x+20)*ln(3)**4)/((x**3-6*x**2+5*x)*ln(-x**3+ 6*x**2-5*x)+(x**3-6*x**2+5*x)*ln(5)),x)
Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (x - 1\right ) + \log \left (x\right ) + \log \left (-x + 5\right )\right ) \]
integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*lo g(3)^4*log(5)+(12*x^2-48*x+20)*log(3)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5 *x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (-x^{3} + 6 \, x^{2} - 5 \, x\right )\right ) \]
integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*lo g(3)^4*log(5)+(12*x^2-48*x+20)*log(3)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5 *x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm=\
Time = 15.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x\,{\ln \left (3\right )}^4+4\,{\ln \left (3\right )}^4\,\ln \left (\ln \left (-5\,x^3+30\,x^2-25\,x\right )\right ) \]