Integrand size = 158, antiderivative size = 24 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (2-\frac {x \left (x+\log \left (-4+\left (2+e^2\right )^2 x\right )\right )}{\log (5)}\right ) \]
[Out]
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6, 6820, 6816} \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x^2+x \log \left (\left (2+e^2\right )^2 x-4\right )-2 \log (5)\right ) \]
[In]
[Out]
Rule 6
Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+e^4 x^3+\left (4+4 e^2\right ) x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx \\ & = \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+\left (4+4 e^2+e^4\right ) x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx \\ & = \int \frac {-x \left (-4+8 x+e^4 (1+2 x)+e^2 (4+8 x)\right )-\left (-4+\left (2+e^2\right )^2 x\right ) \log \left (-4+\left (2+e^2\right )^2 x\right )}{\left (4-\left (2+e^2\right )^2 x\right ) \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right )} \, dx \\ & = \log \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right ) \]
[In]
[Out]
Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) x +x^{2}-2 \ln \left (5\right )\right )\) | \(29\) |
norman | \(\ln \left (-\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) x -x^{2}+2 \ln \left (5\right )\right )\) | \(32\) |
risch | \(\ln \left (x \right )+\ln \left (\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-\frac {-x^{2}+2 \ln \left (5\right )}{x}\right )\) | \(36\) |
default | \(\ln \left (-2 \,{\mathrm e}^{8} \ln \left (5\right )+\left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-16 \,{\mathrm e}^{6} \ln \left (5\right )+4 \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-48 \,{\mathrm e}^{4} \ln \left (5\right )+4 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{4}+\left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )^{2}+4 \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-64 \,{\mathrm e}^{2} \ln \left (5\right )+16 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{2}+8 x \,{\mathrm e}^{4}+32 \,{\mathrm e}^{2} x +32 x -16-32 \ln \left (5\right )+16 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )\right )\) | \(244\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x^{2} + x \log \left (x e^{4} + 4 \, x e^{2} + 4 \, x - 4\right ) - 2 \, \log \left (5\right )}{x}\right ) \]
[In]
[Out]
Time = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (4 x + 4 x e^{2} + x e^{4} - 4 \right )} + \frac {x^{2} - 2 \log {\left (5 \right )}}{x} \right )} \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x^{2} + x \log \left (x {\left (e^{4} + 4 \, e^{2} + 4\right )} - 4\right ) - 2 \, \log \left (5\right )}{x}\right ) \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (-x^{2} - x \log \left (x e^{4} + 4 \, x e^{2} + 4 \, x - 4\right ) + 2 \, \log \left (5\right )\right ) \]
[In]
[Out]
Time = 12.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\ln \left (\frac {x^2-\ln \left (25\right )+x\,\ln \left (4\,x+4\,x\,{\mathrm {e}}^2+x\,{\mathrm {e}}^4-4\right )}{x}\right )+\ln \left (x\right ) \]
[In]
[Out]