Integrand size = 106, antiderivative size = 30 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=x+e^5 \left (\log \left (x^2\right )+\log ^2\left (\frac {1}{3} \left (1+\log \left (\left (-x+x^2\right )^2\right )\right )\right )\right ) \] Output:
x+exp(5)*(ln(x^2)+ln(1/3+1/3*ln((x^2-x)^2))^2)
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=x+2 e^5 \log (x)+e^5 \log ^2\left (\frac {1}{3} \left (1+\log \left ((-1+x)^2 x^2\right )\right )\right ) \] Input:
Integrate[(-x + x^2 + E^5*(-2 + 2*x) + (-x + x^2 + E^5*(-2 + 2*x))*Log[x^2 - 2*x^3 + x^4] + E^5*(-4 + 8*x)*Log[(1 + Log[x^2 - 2*x^3 + x^4])/3])/(-x + x^2 + (-x + x^2)*Log[x^2 - 2*x^3 + x^4]),x]
Output:
x + 2*E^5*Log[x] + E^5*Log[(1 + Log[(-1 + x)^2*x^2])/3]^2
Time = 1.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {x^2+\left (x^2-x+e^5 (2 x-2)\right ) \log \left (x^4-2 x^3+x^2\right )+e^5 (8 x-4) \log \left (\frac {1}{3} \left (\log \left (x^4-2 x^3+x^2\right )+1\right )\right )-x+e^5 (2 x-2)}{x^2+\left (x^2-x\right ) \log \left (x^4-2 x^3+x^2\right )-x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^2-\left (x^2-x+e^5 (2 x-2)\right ) \log \left (x^4-2 x^3+x^2\right )-e^5 (8 x-4) \log \left (\frac {1}{3} \left (\log \left (x^4-2 x^3+x^2\right )+1\right )\right )+x-e^5 (2 x-2)}{(1-x) x \left (\log \left ((x-1)^2 x^2\right )+1\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^5 (2 x-1) \log \left (\frac {1}{3} \left (\log \left ((x-1)^2 x^2\right )+1\right )\right )}{(x-1) x \left (\log \left ((x-1)^2 x^2\right )+1\right )}+\frac {x+2 e^5}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^5 \log ^2\left (\frac {1}{3} \left (\log \left ((1-x)^2 x^2\right )+1\right )\right )+x+2 e^5 \log (x)\) |
Input:
Int[(-x + x^2 + E^5*(-2 + 2*x) + (-x + x^2 + E^5*(-2 + 2*x))*Log[x^2 - 2*x ^3 + x^4] + E^5*(-4 + 8*x)*Log[(1 + Log[x^2 - 2*x^3 + x^4])/3])/(-x + x^2 + (-x + x^2)*Log[x^2 - 2*x^3 + x^4]),x]
Output:
x + 2*E^5*Log[x] + E^5*Log[(1 + Log[(1 - x)^2*x^2])/3]^2
Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
norman | \(x +{\mathrm e}^{5} {\ln \left (\frac {\ln \left (x^{4}-2 x^{3}+x^{2}\right )}{3}+\frac {1}{3}\right )}^{2}+2 \,{\mathrm e}^{5} \ln \left (x \right )\) | \(32\) |
default | \(x +2 \,{\mathrm e}^{5} \ln \left (x \right )-2 \,{\mathrm e}^{5} \ln \left (3\right ) \ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )+{\mathrm e}^{5} {\ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )}^{2}\) | \(52\) |
parts | \(x +2 \,{\mathrm e}^{5} \ln \left (x \right )-2 \,{\mathrm e}^{5} \ln \left (3\right ) \ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )+{\mathrm e}^{5} {\ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )}^{2}\) | \(52\) |
Input:
int(((8*x-4)*exp(5)*ln(1/3*ln(x^4-2*x^3+x^2)+1/3)+((-2+2*x)*exp(5)+x^2-x)* ln(x^4-2*x^3+x^2)+(-2+2*x)*exp(5)+x^2-x)/((x^2-x)*ln(x^4-2*x^3+x^2)+x^2-x) ,x,method=_RETURNVERBOSE)
Output:
x+exp(5)*ln(1/3*ln(x^4-2*x^3+x^2)+1/3)^2+2*exp(5)*ln(x)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=e^{5} \log \left (\frac {1}{3} \, \log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + \frac {1}{3}\right )^{2} + 2 \, e^{5} \log \left (x\right ) + x \] Input:
integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+ x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*exp(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2 )+x^2-x),x, algorithm="fricas")
Output:
e^5*log(1/3*log(x^4 - 2*x^3 + x^2) + 1/3)^2 + 2*e^5*log(x) + x
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=x + 2 e^{5} \log {\left (x \right )} + e^{5} \log {\left (\frac {\log {\left (x^{4} - 2 x^{3} + x^{2} \right )}}{3} + \frac {1}{3} \right )}^{2} \] Input:
integrate(((8*x-4)*exp(5)*ln(1/3*ln(x**4-2*x**3+x**2)+1/3)+((2*x-2)*exp(5) +x**2-x)*ln(x**4-2*x**3+x**2)+(2*x-2)*exp(5)+x**2-x)/((x**2-x)*ln(x**4-2*x **3+x**2)+x**2-x),x)
Output:
x + 2*exp(5)*log(x) + exp(5)*log(log(x**4 - 2*x**3 + x**2)/3 + 1/3)**2
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=-2 \, e^{5} \log \left (3\right ) \log \left (2 \, \log \left (x - 1\right ) + 2 \, \log \left (x\right ) + 1\right ) + e^{5} \log \left (2 \, \log \left (x - 1\right ) + 2 \, \log \left (x\right ) + 1\right )^{2} + 2 \, e^{5} \log \left (x\right ) + x \] Input:
integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+ x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*exp(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2 )+x^2-x),x, algorithm="maxima")
Output:
-2*e^5*log(3)*log(2*log(x - 1) + 2*log(x) + 1) + e^5*log(2*log(x - 1) + 2* log(x) + 1)^2 + 2*e^5*log(x) + x
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx=-2 \, e^{5} \log \left (3\right ) \log \left (\log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + 1\right ) + e^{5} \log \left (\log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + 1\right )^{2} + 2 \, e^{5} \log \left (x\right ) + x \] Input:
integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+ x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*exp(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2 )+x^2-x),x, algorithm="giac")
Output:
-2*e^5*log(3)*log(log(x^4 - 2*x^3 + x^2) + 1) + e^5*log(log(x^4 - 2*x^3 + x^2) + 1)^2 + 2*e^5*log(x) + x
Time = 0.66 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx={\mathrm {e}}^5\,{\ln \left (\frac {\ln \left (x^4-2\,x^3+x^2\right )}{3}+\frac {1}{3}\right )}^2+x+2\,{\mathrm {e}}^5\,\ln \left (x\right ) \] Input:
int(-(log(x^2 - 2*x^3 + x^4)*(x^2 - x + exp(5)*(2*x - 2)) - x + x^2 + exp( 5)*(2*x - 2) + exp(5)*log(log(x^2 - 2*x^3 + x^4)/3 + 1/3)*(8*x - 4))/(x + log(x^2 - 2*x^3 + x^4)*(x - x^2) - x^2),x)
Output:
x + exp(5)*log(log(x^2 - 2*x^3 + x^4)/3 + 1/3)^2 + 2*exp(5)*log(x)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-x+x^2+e^5 (-2+2 x)+\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )+e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{-x+x^2+\left (-x+x^2\right ) \log \left (x^2-2 x^3+x^4\right )} \, dx={\mathrm {log}\left (\frac {\mathrm {log}\left (x^{4}-2 x^{3}+x^{2}\right )}{3}+\frac {1}{3}\right )}^{2} e^{5}+2 \,\mathrm {log}\left (x \right ) e^{5}+x \] Input:
int(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+x^2-x) *log(x^4-2*x^3+x^2)+(2*x-2)*exp(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2)+x^2- x),x)
Output:
log((log(x**4 - 2*x**3 + x**2) + 1)/3)**2*e**5 + 2*log(x)*e**5 + x