Integrand size = 44, antiderivative size = 21 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=e^5 \log \left (\frac {e^{\frac {2}{1+x^2}}}{\log ^2(x)}\right ) \]
[Out]
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1608, 28, 6820, 267, 2339, 29} \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=\frac {2 e^5}{x^2+1}-2 e^5 \log (\log (x)) \]
[In]
[Out]
Rule 28
Rule 29
Rule 267
Rule 1608
Rule 2339
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{x \left (1+2 x^2+x^4\right ) \log (x)} \, dx \\ & = \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{x \left (1+x^2\right )^2 \log (x)} \, dx \\ & = \int \left (-\frac {4 e^5 x}{\left (1+x^2\right )^2}-\frac {2 e^5}{x \log (x)}\right ) \, dx \\ & = -\left (\left (2 e^5\right ) \int \frac {1}{x \log (x)} \, dx\right )-\left (4 e^5\right ) \int \frac {x}{\left (1+x^2\right )^2} \, dx \\ & = \frac {2 e^5}{1+x^2}-\left (2 e^5\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \frac {2 e^5}{1+x^2}-2 e^5 \log (\log (x)) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=\frac {2 e^5}{1+x^2}-2 e^5 \log (\log (x)) \]
[In]
[Out]
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
default | \(-2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\) | \(20\) |
norman | \(-2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\) | \(20\) |
risch | \(-2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\) | \(20\) |
parts | \(-2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\) | \(20\) |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right ) x^{2}+2 \,{\mathrm e}^{5} \ln \left (\ln \left (x \right )\right )-2 \,{\mathrm e}^{5}}{x^{2}+1}\) | \(32\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=-\frac {2 \, {\left ({\left (x^{2} + 1\right )} e^{5} \log \left (\log \left (x\right )\right ) - e^{5}\right )}}{x^{2} + 1} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=- 2 e^{5} \log {\left (\log {\left (x \right )} \right )} + \frac {4 e^{5}}{2 x^{2} + 2} \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=-2 \, e^{5} \log \left (\log \left (x\right )\right ) + \frac {2 \, e^{5}}{x^{2} + 1} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=-\frac {2 \, {\left (x^{2} e^{5} \log \left (\log \left (x\right )\right ) + e^{5} \log \left (\log \left (x\right )\right ) - e^{5}\right )}}{x^{2} + 1} \]
[In]
[Out]
Time = 13.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^5 \left (-2-4 x^2-2 x^4\right )-4 e^5 x^2 \log (x)}{\left (x+2 x^3+x^5\right ) \log (x)} \, dx=\frac {2\,{\mathrm {e}}^5}{x^2+1}-2\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^5 \]
[In]
[Out]