Integrand size = 69, antiderivative size = 24 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (20 \log \left (4+\frac {e^x-\frac {e^5}{8 x}}{x^2}\right )\right ) \]
[Out]
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6816} \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (\log \left (-\frac {-32 x^3-8 e^x x+e^5}{8 x^3}\right )\right ) \]
[In]
[Out]
Rule 6816
Rubi steps \begin{align*} \text {integral}& = \log \left (\log \left (-\frac {e^5-8 e^x x-32 x^3}{8 x^3}\right )\right ) \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (\log \left (4-\frac {e^5}{8 x^3}+\frac {e^x}{x^2}\right )\right ) \]
[In]
[Out]
Time = 2.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\ln \left (\ln \left (-\frac {-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}}{8 x^{3}}\right )\right )\) | \(21\) |
norman | \(\ln \left (\ln \left (\frac {8 \,{\mathrm e}^{x} x -{\mathrm e}^{5}+32 x^{3}}{8 x^{3}}\right )\right )\) | \(23\) |
risch | \(\ln \left (\ln \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )+\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{x^{3}}\right ) \operatorname {csgn}\left (i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x^{3}}\right ) {\operatorname {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )}^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-2 \pi {\operatorname {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )}^{3}+6 i \ln \left (2\right )+6 i \ln \left (x \right )+2 \pi \right )}{2}\right )\) | \(314\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (\log \left (\frac {32 \, x^{3} + 8 \, x e^{x} - e^{5}}{8 \, x^{3}}\right )\right ) \]
[In]
[Out]
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log {\left (\log {\left (\frac {4 x^{3} + x e^{x} - \frac {e^{5}}{8}}{x^{3}} \right )} \right )} \]
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (-3 \, \log \left (2\right ) + \log \left (32 \, x^{3} + 8 \, x e^{x} - e^{5}\right ) - 3 \, \log \left (x\right )\right ) \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\log \left (\log \left (\frac {32 \, x^{3} + 8 \, x e^{x} - e^{5}}{8 \, x^{3}}\right )\right ) \]
[In]
[Out]
Time = 11.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {3 e^5+e^x \left (-16 x+8 x^2\right )}{\left (-e^5 x+8 e^x x^2+32 x^4\right ) \log \left (\frac {-e^5+8 e^x x+32 x^3}{8 x^3}\right )} \, dx=\ln \left (\ln \left (\frac {8\,x\,{\mathrm {e}}^x-{\mathrm {e}}^5+32\,x^3}{8\,x^3}\right )\right ) \]
[In]
[Out]