Integrand size = 169, antiderivative size = 26 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{-4+x-\log (3)-\log \left (-3+x-x^2-\log (4)\right )} \]
[Out]
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6820, 12, 6818} \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=-\frac {5}{\log \left (-3 \left (x^2-x+3+\log (4)\right )\right )-x+4} \]
[In]
[Out]
Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (-4+3 x-x^2-\log (4)\right )}{\left (3-x+x^2+\log (4)\right ) \left (4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )\right )^2} \, dx \\ & = 5 \int \frac {-4+3 x-x^2-\log (4)}{\left (3-x+x^2+\log (4)\right ) \left (4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )\right )^2} \, dx \\ & = -\frac {5}{4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=-\frac {5}{4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )} \]
[In]
[Out]
Time = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
norman | \(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\) | \(25\) |
risch | \(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\) | \(25\) |
parallelrisch | \(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\) | \(25\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]
[In]
[Out]
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=- \frac {5}{- x + \log {\left (- x^{2} + x - 3 - 2 \log {\left (2 \right )} \right )} + \log {\left (3 \right )} + 4} \]
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]
[In]
[Out]
none
Time = 0.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]
[In]
[Out]
Timed out. \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\int -\frac {5\,x^2-15\,x+10\,\ln \left (2\right )+20}{{\ln \left (3\right )}^2\,\left (x^2-x+3\right )-40\,x+\ln \left (-x^2+x-2\,\ln \left (2\right )-3\right )\,\left (\ln \left (3\right )\,\left (2\,x^2-2\,x+6\right )-14\,x+2\,\ln \left (2\right )\,\left (2\,\ln \left (3\right )-2\,x+8\right )+10\,x^2-2\,x^3+24\right )-\ln \left (3\right )\,\left (2\,x^3-10\,x^2+14\,x-24\right )+2\,\ln \left (2\right )\,\left ({\ln \left (3\right )}^2-\ln \left (3\right )\,\left (2\,x-8\right )-8\,x+x^2+16\right )+27\,x^2-9\,x^3+x^4+{\ln \left (-x^2+x-2\,\ln \left (2\right )-3\right )}^2\,\left (x^2-x+2\,\ln \left (2\right )+3\right )+48} \,d x \]
[In]
[Out]