Integrand size = 167, antiderivative size = 31 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2 e^{-x+\frac {x+\log \left (e^x \left (4+\log \left (x^2\right )\right )\right )}{-\frac {4}{x^2}+x}} \]
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\[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=\int \frac {\exp \left (\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}\right ) \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx \]
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Rubi steps Aborted
Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(31)=62\).
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.06 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2 e^{\frac {x \left (4+x^2-x^3-x \log \left (4+\log \left (x^2\right )\right )+x \log \left (e^x \left (4+\log \left (x^2\right )\right )\right )\right )}{-4+x^3}} \left (4+\log \left (x^2\right )\right )^{\frac {x^2}{-4+x^3}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.20 (sec) , antiderivative size = 569, normalized size of antiderivative = 18.35
\[2 \left (\frac {1}{2}\right )^{\frac {x^{2}}{x^{3}-4}} \left ({\mathrm e}^{x}\right )^{\frac {x^{2}}{x^{3}-4}} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )^{\frac {x^{2}}{x^{3}-4}} {\mathrm e}^{-\frac {x \left (i x \pi \,\operatorname {csgn}\left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 \ln \left (x \right )+8\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )\right )^{2}-i x \pi \,\operatorname {csgn}\left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 \ln \left (x \right )+8\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-i x \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )\right )^{3}-i x \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+2 i x \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (8 i+4 i \ln \left (x \right )+\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^{2}\right )^{3}\right )\right )^{2}-2 i \pi x +2 x^{3}-2 x^{2}-8\right )}{2 \left (x^{3}-4\right )}}\]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2 \, e^{\left (-\frac {x^{4} - x^{3} - x^{2} \log \left (e^{x} \log \left (x^{2}\right ) + 4 \, e^{x}\right ) - 4 \, x}{x^{3} - 4}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=\text {Timed out} \]
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Time = 0.46 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2 \, e^{\left (\frac {x^{2} \log \left (2\right )}{x^{3} - 4} + \frac {x^{2} \log \left (\log \left (x\right ) + 2\right )}{x^{3} - 4} - x + \frac {8}{x^{3} - 4} + 2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (29) = 58\).
Time = 0.66 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2 \, e^{\left (-\frac {x^{4}}{x^{3} - 4} + \frac {x^{3}}{x^{3} - 4} + \frac {x^{2} \log \left (e^{x} \log \left (x^{2}\right ) + 4 \, e^{x}\right )}{x^{3} - 4} + \frac {4 \, x}{x^{3} - 4}\right )} \]
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Time = 14.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {4 x+x^3-x^4+x^2 \log \left (4 e^x+e^x \log \left (x^2\right )\right )}{-4+x^3}} \left (-128-16 x-128 x^2+64 x^3+4 x^4+8 x^5-8 x^6+\left (-32-32 x^2+16 x^3+2 x^5-2 x^6\right ) \log \left (x^2\right )+\left (-64 x-8 x^4+\left (-16 x-2 x^4\right ) \log \left (x^2\right )\right ) \log \left (4 e^x+e^x \log \left (x^2\right )\right )\right )}{64-32 x^3+4 x^6+\left (16-8 x^3+x^6\right ) \log \left (x^2\right )} \, dx=2\,{\mathrm {e}}^{\frac {x^3}{x^3-4}}\,{\mathrm {e}}^{-\frac {x^4}{x^3-4}}\,{\mathrm {e}}^{\frac {4\,x}{x^3-4}}\,{\left (4\,{\mathrm {e}}^x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )}^{\frac {x^2}{x^3-4}} \]
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