Integrand size = 108, antiderivative size = 25 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=\log \left (\frac {(4-x)^2 \log \left (1+x-x^2\right )}{-1+x+\log (4)}\right ) \] Output:
ln((4-x)^2/(x+2*ln(2)-1)*ln(-x^2+x+1))
Time = 5.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=\frac {(6+\log (16)) \log (4-x)}{3+\log (4)}-\log (1-x-\log (4))+\log \left (\log \left (1+x-x^2\right )\right ) \] Input:
Integrate[(-4 + 13*x - 11*x^2 + 2*x^3 + (4 - 9*x + 2*x^2)*Log[4] + (-2 - 3 *x + x^2 + x^3 + (-2 - 2*x + 2*x^2)*Log[4])*Log[1 + x - x^2])/((-4 + x + 8 *x^2 - 6*x^3 + x^4 + (4 + 3*x - 5*x^2 + x^3)*Log[4])*Log[1 + x - x^2]),x]
Output:
((6 + Log[16])*Log[4 - x])/(3 + Log[4]) - Log[1 - x - Log[4]] + Log[Log[1 + x - 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.
| \(\displaystyle \int \frac {2 x^3-11 x^2+\left (2 x^2-9 x+4\right ) \log (4)+\left (x^3+x^2+\left (2 x^2-2 x-2\right ) \log (4)-3 x-2\right ) \log \left (-x^2+x+1\right )+13 x-4}{\left (x^4-6 x^3+8 x^2+\left (x^3-5 x^2+3 x+4\right ) \log (4)+x-4\right ) \log \left (-x^2+x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {(x (4-\log (4))+1-\log (64)) \left (2 x^3-11 x^2+\left (2 x^2-9 x+4\right ) \log (4)+\left (x^3+x^2+\left (2 x^2-2 x-2\right ) \log (4)-3 x-2\right ) \log \left (-x^2+x+1\right )+13 x-4\right )}{11 \left (-x^2+x+1\right ) \left (1-\log ^2(4)+\log (4)\right ) \log \left (-x^2+x+1\right )}+\frac {2 x^3-11 x^2+\left (2 x^2-9 x+4\right ) \log (4)+\left (x^3+x^2+\left (2 x^2-2 x-2\right ) \log (4)-3 x-2\right ) \log \left (-x^2+x+1\right )+13 x-4}{(3+\log (4)) \left (1-\log ^2(4)+\log (4)\right ) (x-1+\log (4)) \log \left (-x^2+x+1\right )}+\frac {2 x^3-11 x^2+\left (2 x^2-9 x+4\right ) \log (4)+\left (x^3+x^2+\left (2 x^2-2 x-2\right ) \log (4)-3 x-2\right ) \log \left (-x^2+x+1\right )+13 x-4}{11 (x-4) (3+\log (4)) \log \left (-x^2+x+1\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^3}{3 (3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}-\frac {(4-\log (4)) x^3}{33 \left (1+\log (4)-\log ^2(4)\right )}+\frac {x^3}{33 (3+\log (4))}-\frac {(9-\log (4) \log (16)+\log (64)) x^2}{22 \left (1+\log (4)-\log ^2(4)\right )}+\frac {(5+\log (16)) x^2}{22 (3+\log (4))}+\frac {(2+\log (4)) x^2}{2 (3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}+\frac {(17+\log (4096)) x}{11 (3+\log (4))}-\frac {\left (1+\log ^2(4)+\log (64)\right ) x}{(3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}-\frac {(2+\log (16)) (1-\log (64)) x}{11 \left (1+\log (4)-\log ^2(4)\right )}+\frac {(6+\log (16)) \log (4-x)}{3+\log (4)}-\log (-x-\log (4)+1)-\frac {\left (15+7 \log ^2(4)-2 \log (2) (5+\log (4096))\right ) \int \frac {1}{\log \left (-x^2+x+1\right )}dx}{11 \left (1+\log (4)-\log ^2(4)\right )}+\frac {4 \int \frac {1}{\log \left (-x^2+x+1\right )}dx}{(3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}+\frac {(1-\log (4)) \int \frac {1}{\log \left (-x^2+x+1\right )}dx}{11 (3+\log (4))}-\frac {\left (5-\sqrt {5}\right ) \left (22+4 \log (4)-\log ^2(4)+6 \log (64)-7 \log (4) \log (64)\right ) \int \frac {1}{\left (-2 x-\sqrt {5}+1\right ) \log \left (-x^2+x+1\right )}dx}{55 \left (1+\log (4)-\log ^2(4)\right )}+\frac {2 \left (11+11 \log (4)+7 \log ^2(4)-6 \log (2) \log (4096)\right ) \int \frac {1}{\left (-2 x+\sqrt {5}+1\right ) \log \left (-x^2+x+1\right )}dx}{11 \sqrt {5} \left (1+\log (4)-\log ^2(4)\right )}-\frac {\left (5+\sqrt {5}\right ) \left (22+4 \log (4)-\log ^2(4)+6 \log (64)-7 \log (4) \log (64)\right ) \int \frac {1}{\left (-2 x+\sqrt {5}+1\right ) \log \left (-x^2+x+1\right )}dx}{55 \left (1+\log (4)-\log ^2(4)\right )}+\frac {\left (34-17 \log (4)+2 \log ^2(4)+\log (4096)\right ) \int \frac {x}{\log \left (-x^2+x+1\right )}dx}{11 \left (1+\log (4)-\log ^2(4)\right )}-\frac {(3-\log (16)) \int \frac {x}{\log \left (-x^2+x+1\right )}dx}{11 (3+\log (4))}-\frac {9 \int \frac {x}{\log \left (-x^2+x+1\right )}dx}{(3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}+\frac {2 \int \frac {x^2}{\log \left (-x^2+x+1\right )}dx}{(3+\log (4)) \left (1+\log (4)-\log ^2(4)\right )}-\frac {2 (4-\log (4)) \int \frac {x^2}{\log \left (-x^2+x+1\right )}dx}{11 \left (1+\log (4)-\log ^2(4)\right )}+\frac {2 \int \frac {x^2}{\log \left (-x^2+x+1\right )}dx}{11 (3+\log (4))}+\frac {2 \left (11+11 \log (4)+7 \log ^2(4)-6 \log (2) \log (4096)\right ) \int \frac {1}{\left (2 x+\sqrt {5}-1\right ) \log \left (-x^2+x+1\right )}dx}{11 \sqrt {5} \left (1+\log (4)-\log ^2(4)\right )}\) |
Input:
Int[(-4 + 13*x - 11*x^2 + 2*x^3 + (4 - 9*x + 2*x^2)*Log[4] + (-2 - 3*x + x ^2 + x^3 + (-2 - 2*x + 2*x^2)*Log[4])*Log[1 + x - x^2])/((-4 + x + 8*x^2 - 6*x^3 + x^4 + (4 + 3*x - 5*x^2 + x^3)*Log[4])*Log[1 + x - x^2]),x]
Output:
$Aborted
Time = 0.85 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
| method | result | size |
| default | \(2 \ln \left (x -4\right )-\ln \left (x +2 \ln \left (2\right )-1\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\) | \(28\) |
| norman | \(2 \ln \left (x -4\right )-\ln \left (x +2 \ln \left (2\right )-1\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\) | \(28\) |
| risch | \(2 \ln \left (x -4\right )-\ln \left (x +2 \ln \left (2\right )-1\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\) | \(28\) |
| parallelrisch | \(2 \ln \left (x -4\right )-\ln \left (x +2 \ln \left (2\right )-1\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\) | \(28\) |
| parts | \(2 \ln \left (x -4\right )-\ln \left (x +2 \ln \left (2\right )-1\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\) | \(28\) |
Input:
int(((2*(2*x^2-2*x-2)*ln(2)+x^3+x^2-3*x-2)*ln(-x^2+x+1)+2*(2*x^2-9*x+4)*ln (2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x^2+3*x+4)*ln(2)+x^4-6*x^3+8*x^2+x-4)/l n(-x^2+x+1),x,method=_RETURNVERBOSE)
Output:
2*ln(x-4)-ln(x+2*ln(2)-1)+ln(ln(-x^2+x+1))
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=-\log \left (x + 2 \, \log \left (2\right ) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \] Input:
integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9 *x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8* x^2+x-4)/log(-x^2+x+1),x, algorithm="fricas")
Output:
-log(x + 2*log(2) - 1) + 2*log(x - 4) + log(log(-x^2 + x + 1))
Time = 1.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=2 \log {\left (x - 4 \right )} - \log {\left (x - 1 + 2 \log {\left (2 \right )} \right )} + \log {\left (\log {\left (- x^{2} + x + 1 \right )} \right )} \] Input:
integrate(((2*(2*x**2-2*x-2)*ln(2)+x**3+x**2-3*x-2)*ln(-x**2+x+1)+2*(2*x** 2-9*x+4)*ln(2)+2*x**3-11*x**2+13*x-4)/(2*(x**3-5*x**2+3*x+4)*ln(2)+x**4-6* x**3+8*x**2+x-4)/ln(-x**2+x+1),x)
Output:
2*log(x - 4) - log(x - 1 + 2*log(2)) + log(log(-x**2 + x + 1))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=-\log \left (x + 2 \, \log \left (2\right ) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \] Input:
integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9 *x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8* x^2+x-4)/log(-x^2+x+1),x, algorithm="maxima")
Output:
-log(x + 2*log(2) - 1) + 2*log(x - 4) + log(log(-x^2 + x + 1))
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=-\log \left (x + 2 \, \log \left (2\right ) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \] Input:
integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9 *x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8* x^2+x-4)/log(-x^2+x+1),x, algorithm="giac")
Output:
-log(x + 2*log(2) - 1) + 2*log(x - 4) + log(log(-x^2 + x + 1))
Time = 3.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.28 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=\ln \left (\ln \left (-x^2+x+1\right )\right )+\frac {\ln \left (x-4\right )\,\left (2\,\ln \left (16\right )-\ln \left (4\right )+\sqrt {4\,\ln \left (256\right )-10\,\ln \left (4\right )+{\ln \left (4\right )}^2+9}+9\right )}{2\,\sqrt {4\,\ln \left (256\right )-10\,\ln \left (4\right )+{\ln \left (4\right )}^2+9}}+\frac {\ln \left (x+\ln \left (4\right )-1\right )\,\left (\ln \left (4\right )-2\,\ln \left (16\right )+\sqrt {4\,\ln \left (256\right )-10\,\ln \left (4\right )+{\ln \left (4\right )}^2+9}-9\right )}{2\,\sqrt {4\,\ln \left (256\right )-10\,\ln \left (4\right )+{\ln \left (4\right )}^2+9}} \] Input:
int((13*x + 2*log(2)*(2*x^2 - 9*x + 4) - log(x - x^2 + 1)*(3*x + 2*log(2)* (2*x - 2*x^2 + 2) - x^2 - x^3 + 2) - 11*x^2 + 2*x^3 - 4)/(log(x - x^2 + 1) *(x + 2*log(2)*(3*x - 5*x^2 + x^3 + 4) + 8*x^2 - 6*x^3 + x^4 - 4)),x)
Output:
log(log(x - x^2 + 1)) + (log(x - 4)*(2*log(16) - log(4) + (4*log(256) - 10 *log(4) + log(4)^2 + 9)^(1/2) + 9))/(2*(4*log(256) - 10*log(4) + log(4)^2 + 9)^(1/2)) + (log(x + log(4) - 1)*(log(4) - 2*log(16) + (4*log(256) - 10* log(4) + log(4)^2 + 9)^(1/2) - 9))/(2*(4*log(256) - 10*log(4) + log(4)^2 + 9)^(1/2))
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+13 x-11 x^2+2 x^3+\left (4-9 x+2 x^2\right ) \log (4)+\left (-2-3 x+x^2+x^3+\left (-2-2 x+2 x^2\right ) \log (4)\right ) \log \left (1+x-x^2\right )}{\left (-4+x+8 x^2-6 x^3+x^4+\left (4+3 x-5 x^2+x^3\right ) \log (4)\right ) \log \left (1+x-x^2\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (-x^{2}+x +1\right )\right )-\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )+x -1\right )+2 \,\mathrm {log}\left (x -4\right ) \] Input:
int(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9*x+4)* log(2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8*x^2+x- 4)/log(-x^2+x+1),x)
Output:
log(log( - x**2 + x + 1)) - log(2*log(2) + x - 1) + 2*log(x - 4)