Integrand size = 107, antiderivative size = 33 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^3-e^x \left (-4+\log \left (e^x+\frac {2-x}{2 x}+2 x\right )\right ) \]
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Time = 2.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6820, 6874, 2225, 2634} \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=4 e^x-e^x \log \left (2 x+e^x+\frac {1}{x}-\frac {1}{2}\right ) \]
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Rule 2225
Rule 2634
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (2+8 x+\left (-8+6 e^x\right ) x^2+16 x^3-x \left (2+\left (-1+2 e^x\right ) x+4 x^2\right ) \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right )}{x \left (2+\left (-1+2 e^x\right ) x+4 x^2\right )} \, dx \\ & = \int \left (3 e^x+\frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )}-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right ) \, dx \\ & = 3 \int e^x \, dx+\int \frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx-\int e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right ) \, dx \\ & = 3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+\int \frac {e^x \left (2+e^x-\frac {1}{x^2}\right )}{-\frac {1}{2}+e^x+\frac {1}{x}+2 x} \, dx+\int \left (\frac {2 e^x}{2-x+2 e^x x+4 x^2}+\frac {2 e^x}{x \left (2-x+2 e^x x+4 x^2\right )}-\frac {5 e^x x}{2-x+2 e^x x+4 x^2}+\frac {4 e^x x^2}{2-x+2 e^x x+4 x^2}\right ) \, dx \\ & = 3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx+\int \left (e^x-\frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )}\right ) \, dx \\ & = 3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx+\int e^x \, dx-\int \frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx \\ & = 4 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx-\int \left (\frac {2 e^x}{2-x+2 e^x x+4 x^2}+\frac {2 e^x}{x \left (2-x+2 e^x x+4 x^2\right )}-\frac {5 e^x x}{2-x+2 e^x x+4 x^2}+\frac {4 e^x x^2}{2-x+2 e^x x+4 x^2}\right ) \, dx \\ & = 4 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right ) \\ \end{align*}
Time = 5.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^x \left (-4+\log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(28)=56\).
Time = 0.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.27
method | result | size |
parallelrisch | \(2-{\mathrm e}^{x} \ln \left (\frac {2 \,{\mathrm e}^{x} x +4 x^{2}-x +2}{2 x}\right )-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (\frac {1}{2}+x^{2}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}\right )}{2}+4 \,{\mathrm e}^{x}-\frac {\ln \left (\frac {2 \,{\mathrm e}^{x} x +4 x^{2}-x +2}{2 x}\right )}{2}\) | \(75\) |
risch | \(-{\mathrm e}^{x} \ln \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )+{\mathrm e}^{x} \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{2} {\mathrm e}^{x}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}+\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{3} {\mathrm e}^{x}}{2}-{\mathrm e}^{x} \ln \left (2\right )+4 \,{\mathrm e}^{x}\) | \(193\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \]
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Timed out. \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=\text {Timed out} \]
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Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx={\left (\log \left (2\right ) + \log \left (x\right ) + 4\right )} e^{x} - e^{x} \log \left (4 \, x^{2} + 2 \, x e^{x} - x + 2\right ) \]
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Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \]
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Time = 12.81 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-{\mathrm {e}}^x\,\left (\ln \left (\frac {x\,{\mathrm {e}}^x-\frac {x}{2}+2\,x^2+1}{x}\right )-4\right ) \]
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