Integrand size = 80, antiderivative size = 21 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \]
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Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6820, 12, 2326} \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{e^x-4} \log \left (\log \left (-x^4\right )-4 x+4\right ) \]
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Rule 12
Rule 2326
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int 4 e^{-4+e^x} \left (\frac {4-4 x}{4 x-4 x^2+x \log \left (-x^4\right )}+e^x \log \left (4-4 x+\log \left (-x^4\right )\right )\right ) \, dx \\ & = 4 \int e^{-4+e^x} \left (\frac {4-4 x}{4 x-4 x^2+x \log \left (-x^4\right )}+e^x \log \left (4-4 x+\log \left (-x^4\right )\right )\right ) \, dx \\ & = 4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \]
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Time = 3.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (\ln \left (-x^{4}\right )-4 x +4\right )\) | \(20\) |
risch | \(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (i \pi +4 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{3}\right )\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}+i \pi \operatorname {csgn}\left (i x^{4}\right )^{2} \left (\operatorname {csgn}\left (i x^{4}\right )-1\right )-4 x +4\right )\) | \(158\) |
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \]
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Timed out. \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=\text {Timed out} \]
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Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (i \, \pi - 4 \, x + 4 \, \log \left (x\right ) + 4\right ) \]
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \]
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Time = 9.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4\,\ln \left (\ln \left (-x^4\right )-4\,x+4\right )\,{\mathrm {e}}^{{\mathrm {e}}^x-4} \]
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