Integrand size = 108, antiderivative size = 23 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=6-x-\log \left (\log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )\right ) \]
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=-x-\log \left (\log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )\right ) \]
Integrate[(-2 + 6*x - 2*x^2 + (3*x^2 - x^3 + (3 - x)*Log[(9 - 6*x + x^2)/4 ])*Log[x^2 + Log[(9 - 6*x + x^2)/4]])/((-3*x^2 + x^3 + (-3 + x)*Log[(9 - 6 *x + x^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]]),x]
Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {7239, 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.
\(\displaystyle \int \frac {-2 x^2+\left (-x^3+3 x^2+(3-x) \log \left (\frac {1}{4} \left (x^2-6 x+9\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (x^2-6 x+9\right )\right )\right )+6 x-2}{\left (x^3-3 x^2+(x-3) \log \left (\frac {1}{4} \left (x^2-6 x+9\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (x^2-6 x+9\right )\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {2 \left (x^2-3 x+1\right )}{(x-3) \left (x^2+\log \left (\frac {1}{4} (x-3)^2\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} (x-3)^2\right )\right )}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log \left (\log \left (x^2+\log \left (\frac {1}{4} (3-x)^2\right )\right )\right )-x\) |
Int[(-2 + 6*x - 2*x^2 + (3*x^2 - x^3 + (3 - x)*Log[(9 - 6*x + x^2)/4])*Log [x^2 + Log[(9 - 6*x + x^2)/4]])/((-3*x^2 + x^3 + (-3 + x)*Log[(9 - 6*x + x ^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]]),x]
3.19.26.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
norman | \(-x -\ln \left (\ln \left (\ln \left (\frac {1}{4} x^{2}-\frac {3}{2} x +\frac {9}{4}\right )+x^{2}\right )\right )\) | \(24\) |
parallelrisch | \(-\frac {3}{2}-\ln \left (\ln \left (\ln \left (\frac {1}{4} x^{2}-\frac {3}{2} x +\frac {9}{4}\right )+x^{2}\right )\right )-x\) | \(25\) |
int((((-x+3)*ln(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*ln(ln(1/4*x^2-3/2*x+9/4)+x^2 )-2*x^2+6*x-2)/((-3+x)*ln(1/4*x^2-3/2*x+9/4)+x^3-3*x^2)/ln(ln(1/4*x^2-3/2* x+9/4)+x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=-x - \log \left (\log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )\right ) \]
integrate((((-x+3)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x +9/4)+x^2)-2*x^2+6*x-2)/((-3+x)*log(1/4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log( 1/4*x^2-3/2*x+9/4)+x^2),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=- x - \log {\left (\log {\left (x^{2} + \log {\left (\frac {x^{2}}{4} - \frac {3 x}{2} + \frac {9}{4} \right )} \right )} \right )} \]
integrate((((-x+3)*ln(1/4*x**2-3/2*x+9/4)-x**3+3*x**2)*ln(ln(1/4*x**2-3/2* x+9/4)+x**2)-2*x**2+6*x-2)/((-3+x)*ln(1/4*x**2-3/2*x+9/4)+x**3-3*x**2)/ln( ln(1/4*x**2-3/2*x+9/4)+x**2),x)
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=-x - \log \left (\log \left (x^{2} - 2 \, \log \left (2\right ) + 2 \, \log \left (x - 3\right )\right )\right ) \]
integrate((((-x+3)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x +9/4)+x^2)-2*x^2+6*x-2)/((-3+x)*log(1/4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log( 1/4*x^2-3/2*x+9/4)+x^2),x, algorithm=\
\[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=\int { -\frac {2 \, x^{2} + {\left (x^{3} - 3 \, x^{2} + {\left (x - 3\right )} \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )} \log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right ) - 6 \, x + 2}{{\left (x^{3} - 3 \, x^{2} + {\left (x - 3\right )} \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )} \log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )} \,d x } \]
integrate((((-x+3)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x +9/4)+x^2)-2*x^2+6*x-2)/((-3+x)*log(1/4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log( 1/4*x^2-3/2*x+9/4)+x^2),x, algorithm=\
integrate(-(2*x^2 + (x^3 - 3*x^2 + (x - 3)*log(1/4*x^2 - 3/2*x + 9/4))*log (x^2 + log(1/4*x^2 - 3/2*x + 9/4)) - 6*x + 2)/((x^3 - 3*x^2 + (x - 3)*log( 1/4*x^2 - 3/2*x + 9/4))*log(x^2 + log(1/4*x^2 - 3/2*x + 9/4))), x)
Time = 10.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-2+6 x-2 x^2+\left (3 x^2-x^3+(3-x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )}{\left (-3 x^2+x^3+(-3+x) \log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} \left (9-6 x+x^2\right )\right )\right )} \, dx=-x-\ln \left (\ln \left (\ln \left (\frac {x^2}{4}-\frac {3\,x}{2}+\frac {9}{4}\right )+x^2\right )\right ) \]