Integrand size = 96, antiderivative size = 22 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\log (-1+x) \log \left (4+4 \left (x-x^2\right )\right )} \] Output:
1/ln(-1+x)*x/ln(-4*x^2+4*x+4)
Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\log (-1+x) \log \left (4+4 x-4 x^2\right )} \] Input:
Integrate[((x + x^2 - x^3)*Log[4 + 4*x - 4*x^2] + Log[-1 + x]*(-x + 3*x^2 - 2*x^3 + (1 - 2*x^2 + x^3)*Log[4 + 4*x - 4*x^2]))/((1 - 2*x^2 + x^3)*Log[ -1 + x]^2*Log[4 + 4*x - 4*x^2]^2),x]
Output:
x/(Log[-1 + x]*Log[4 + 4*x - 4*x^2])
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.
\(\displaystyle \int \frac {\left (-x^3+x^2+x\right ) \log \left (-4 x^2+4 x+4\right )+\log (x-1) \left (-2 x^3+3 x^2+\left (x^3-2 x^2+1\right ) \log \left (-4 x^2+4 x+4\right )-x\right )}{\left (x^3-2 x^2+1\right ) \log ^2(x-1) \log ^2\left (-4 x^2+4 x+4\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-x^3+x^2+x\right ) \log \left (-4 x^2+4 x+4\right )+\log (x-1) \left (-2 x^3+3 x^2+\left (x^3-2 x^2+1\right ) \log \left (-4 x^2+4 x+4\right )-x\right )}{(1-x) \log ^2(x-1) \log ^2\left (-4 x^2+4 x+4\right )}+\frac {x \left (\left (-x^3+x^2+x\right ) \log \left (-4 x^2+4 x+4\right )+\log (x-1) \left (-2 x^3+3 x^2+\left (x^3-2 x^2+1\right ) \log \left (-4 x^2+4 x+4\right )-x\right )\right )}{\left (x^2-x-1\right ) \log ^2(x-1) \log ^2\left (-4 x^2+4 x+4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{\log (x-1) \log ^2\left (-4 x^2+4 x+4\right )}dx+\frac {4 \int \frac {1}{\left (-2 x+\sqrt {5}+1\right ) \log (x-1) \log ^2\left (-4 x^2+4 x+4\right )}dx}{\sqrt {5}}-\frac {1}{5} \left (5+\sqrt {5}\right ) \int \frac {1}{\left (2 x-\sqrt {5}-1\right ) \log (x-1) \log ^2\left (-4 x^2+4 x+4\right )}dx-\frac {1}{5} \left (5-\sqrt {5}\right ) \int \frac {1}{\left (2 x+\sqrt {5}-1\right ) \log (x-1) \log ^2\left (-4 x^2+4 x+4\right )}dx+\frac {4 \int \frac {1}{\left (2 x+\sqrt {5}-1\right ) \log (x-1) \log ^2\left (-4 x^2+4 x+4\right )}dx}{\sqrt {5}}-\int \frac {1}{\log ^2(x-1) \log \left (-4 x^2+4 x+4\right )}dx-\int \frac {1}{(x-1) \log ^2(x-1) \log \left (-4 x^2+4 x+4\right )}dx+\int \frac {1}{\log (x-1) \log \left (-4 x^2+4 x+4\right )}dx\) |
Input:
Int[((x + x^2 - x^3)*Log[4 + 4*x - 4*x^2] + Log[-1 + x]*(-x + 3*x^2 - 2*x^ 3 + (1 - 2*x^2 + x^3)*Log[4 + 4*x - 4*x^2]))/((1 - 2*x^2 + x^3)*Log[-1 + x ]^2*Log[4 + 4*x - 4*x^2]^2),x]
Output:
$Aborted
Time = 262.52 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x}{\ln \left (-1+x \right ) \ln \left (-4 x^{2}+4 x +4\right )}\) | \(22\) |
parallelrisch | \(\frac {x}{\ln \left (-1+x \right ) \ln \left (-4 x^{2}+4 x +4\right )}\) | \(22\) |
Input:
int((((x^3-2*x^2+1)*ln(-4*x^2+4*x+4)-2*x^3+3*x^2-x)*ln(-1+x)+(-x^3+x^2+x)* ln(-4*x^2+4*x+4))/(x^3-2*x^2+1)/ln(-4*x^2+4*x+4)^2/ln(-1+x)^2,x,method=_RE TURNVERBOSE)
Output:
1/ln(-1+x)*x/ln(-4*x^2+4*x+4)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\log \left (-4 \, x^{2} + 4 \, x + 4\right ) \log \left (x - 1\right )} \] Input:
integrate((((x^3-2*x^2+1)*log(-4*x^2+4*x+4)-2*x^3+3*x^2-x)*log(-1+x)+(-x^3 +x^2+x)*log(-4*x^2+4*x+4))/(x^3-2*x^2+1)/log(-4*x^2+4*x+4)^2/log(-1+x)^2,x , algorithm="fricas")
Output:
x/(log(-4*x^2 + 4*x + 4)*log(x - 1))
Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\log {\left (x - 1 \right )} \log {\left (- 4 x^{2} + 4 x + 4 \right )}} \] Input:
integrate((((x**3-2*x**2+1)*ln(-4*x**2+4*x+4)-2*x**3+3*x**2-x)*ln(-1+x)+(- x**3+x**2+x)*ln(-4*x**2+4*x+4))/(x**3-2*x**2+1)/ln(-4*x**2+4*x+4)**2/ln(-1 +x)**2,x)
Output:
x/(log(x - 1)*log(-4*x**2 + 4*x + 4))
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{{\left (i \, \pi + 2 \, \log \left (2\right )\right )} \log \left (x - 1\right ) + \log \left (x^{2} - x - 1\right ) \log \left (x - 1\right )} \] Input:
integrate((((x^3-2*x^2+1)*log(-4*x^2+4*x+4)-2*x^3+3*x^2-x)*log(-1+x)+(-x^3 +x^2+x)*log(-4*x^2+4*x+4))/(x^3-2*x^2+1)/log(-4*x^2+4*x+4)^2/log(-1+x)^2,x , algorithm="maxima")
Output:
x/((I*pi + 2*log(2))*log(x - 1) + log(x^2 - x - 1)*log(x - 1))
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\log \left (-4 \, x^{2} + 4 \, x + 4\right ) \log \left (x - 1\right )} \] Input:
integrate((((x^3-2*x^2+1)*log(-4*x^2+4*x+4)-2*x^3+3*x^2-x)*log(-1+x)+(-x^3 +x^2+x)*log(-4*x^2+4*x+4))/(x^3-2*x^2+1)/log(-4*x^2+4*x+4)^2/log(-1+x)^2,x , algorithm="giac")
Output:
x/(log(-4*x^2 + 4*x + 4)*log(x - 1))
Time = 4.08 (sec) , antiderivative size = 349, normalized size of antiderivative = 15.86 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{4}-\frac {5\,\ln \left (x-1\right )}{8}+\frac {\frac {x}{\ln \left (x-1\right )}-\frac {\ln \left (-4\,x^2+4\,x+4\right )\,\left (-x^2+x+1\right )\,\left (x+\ln \left (x-1\right )-x\,\ln \left (x-1\right )\right )}{{\ln \left (x-1\right )}^2\,\left (2\,x-1\right )\,\left (x-1\right )}}{\ln \left (-4\,x^2+4\,x+4\right )}+\frac {\frac {\ln \left (x-1\right )\,\left (4\,x^4-11\,x^3+10\,x^2-2\right )}{2\,{\left (2\,x-1\right )}^2\,\left (x-1\right )}+\frac {x\,\left (-x^2+x+1\right )}{\left (2\,x-1\right )\,\left (x-1\right )}-\frac {{\ln \left (x-1\right )}^2\,\left (x-1\right )\,\left (2\,x^2-2\,x+3\right )}{2\,{\left (2\,x-1\right )}^2}}{{\ln \left (x-1\right )}^2}+\frac {\frac {13\,x^3}{16}-\frac {39\,x^2}{32}+\frac {3\,x}{8}+\frac {3}{32}}{x^4-\frac {5\,x^3}{2}+\frac {9\,x^2}{4}-\frac {7\,x}{8}+\frac {1}{8}}+\frac {\frac {x^3-6\,x^2+4\,x}{2\,{\left (2\,x-1\right )}^2\,\left (x-1\right )}+\frac {{\ln \left (x-1\right )}^2\,\left (x-1\right )\,\left (-4\,x^3+6\,x^2+2\,x-7\right )}{2\,{\left (2\,x-1\right )}^3}+\frac {\ln \left (x-1\right )\,\left (-8\,x^3+7\,x^2+4\,x-4\right )}{2\,{\left (2\,x-1\right )}^3\,\left (x-1\right )}}{\ln \left (x-1\right )}+\frac {\ln \left (x-1\right )\,\left (\frac {x^4}{4}-\frac {11\,x^2}{16}+\frac {33\,x}{32}-\frac {33}{64}\right )}{x^3-\frac {3\,x^2}{2}+\frac {3\,x}{4}-\frac {1}{8}} \] Input:
int((log(4*x - 4*x^2 + 4)*(x + x^2 - x^3) - log(x - 1)*(x - log(4*x - 4*x^ 2 + 4)*(x^3 - 2*x^2 + 1) - 3*x^2 + 2*x^3))/(log(x - 1)^2*log(4*x - 4*x^2 + 4)^2*(x^3 - 2*x^2 + 1)),x)
Output:
x/4 - (5*log(x - 1))/8 + (x/log(x - 1) - (log(4*x - 4*x^2 + 4)*(x - x^2 + 1)*(x + log(x - 1) - x*log(x - 1)))/(log(x - 1)^2*(2*x - 1)*(x - 1)))/log( 4*x - 4*x^2 + 4) + ((log(x - 1)*(10*x^2 - 11*x^3 + 4*x^4 - 2))/(2*(2*x - 1 )^2*(x - 1)) + (x*(x - x^2 + 1))/((2*x - 1)*(x - 1)) - (log(x - 1)^2*(x - 1)*(2*x^2 - 2*x + 3))/(2*(2*x - 1)^2))/log(x - 1)^2 + ((3*x)/8 - (39*x^2)/ 32 + (13*x^3)/16 + 3/32)/((9*x^2)/4 - (7*x)/8 - (5*x^3)/2 + x^4 + 1/8) + ( (4*x - 6*x^2 + x^3)/(2*(2*x - 1)^2*(x - 1)) + (log(x - 1)^2*(x - 1)*(2*x + 6*x^2 - 4*x^3 - 7))/(2*(2*x - 1)^3) + (log(x - 1)*(4*x + 7*x^2 - 8*x^3 - 4))/(2*(2*x - 1)^3*(x - 1)))/log(x - 1) + (log(x - 1)*((33*x)/32 - (11*x^2 )/16 + x^4/4 - 33/64))/((3*x)/4 - (3*x^2)/2 + x^3 - 1/8)
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {\left (x+x^2-x^3\right ) \log \left (4+4 x-4 x^2\right )+\log (-1+x) \left (-x+3 x^2-2 x^3+\left (1-2 x^2+x^3\right ) \log \left (4+4 x-4 x^2\right )\right )}{\left (1-2 x^2+x^3\right ) \log ^2(-1+x) \log ^2\left (4+4 x-4 x^2\right )} \, dx=\frac {x}{\mathrm {log}\left (-4 x^{2}+4 x +4\right ) \mathrm {log}\left (x -1\right )} \] Input:
int((((x^3-2*x^2+1)*log(-4*x^2+4*x+4)-2*x^3+3*x^2-x)*log(-1+x)+(-x^3+x^2+x )*log(-4*x^2+4*x+4))/(x^3-2*x^2+1)/log(-4*x^2+4*x+4)^2/log(-1+x)^2,x)
Output:
x/(log( - 4*x**2 + 4*x + 4)*log(x - 1))