Integrand size = 264, antiderivative size = 27 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\left (1-\frac {e^2}{2}+e^x+x\right )^2\right )\right ) \]
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\[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 e^x x^2+\left (4-2 e^2\right ) x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx \\ & = \int \left (\frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{2 \left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x}\right ) \, dx \\ & = \frac {1}{2} \int \frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx \\ & = \frac {1}{2} \int \left (\frac {3 \log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )}+\frac {2 \left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx \\ & = \frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \frac {\left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx \\ & = \frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \left (\frac {2 x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {e^2 \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx \\ & = \frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+2 \int \frac {x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+e^2 \int \frac {\log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 25.99 (sec) , antiderivative size = 249, normalized size of antiderivative = 9.22
method | result | size |
risch | \(\ln \left (x \right ) \ln \left (-2 \ln \left (2\right )+2 \ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )\right )}^{2}}{2}+x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right ) \ln \left (\ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )+\frac {i \left (-\pi {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )}^{2} \operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right ) {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )}^{2}-\pi {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )}^{3}+4 i \ln \left (2\right )-2 i x \right )}{4}\right )}{4}\) | \(249\) |
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Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right ) \]
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Timed out. \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\text {Timed out} \]
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Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\log \left (x - 2 \, \log \left (2\right ) + 2 \, \log \left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )\right ) \log \left (x\right ) \]
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\[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\int { \frac {{\left (2 \, x^{2} - x e^{2} + 6 \, x e^{x} + 6 \, x\right )} \log \left (x^{2}\right ) + 2 \, {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + {\left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right ) + 2 \, x\right )} \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}{2 \, {\left (2 \, x^{3} - x^{2} e^{2} + 2 \, x^{2} e^{x} + 2 \, x^{2} + {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + 2 \, x\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}} \,d x } \]
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Time = 13.95 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {\ln \left (x^2\right )\,\ln \left (x+\ln \left (2\,x+{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^4}{4}+\frac {{\mathrm {e}}^x\,\left (8\,x-4\,{\mathrm {e}}^2+8\right )}{4}+x^2-\frac {{\mathrm {e}}^2\,\left (4\,x+4\right )}{4}+1\right )\right )}{2} \]
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