Integrand size = 145, antiderivative size = 28 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=-\left (e^{\left (e^{-1+x}+2 x\right )^2}+x\right )^2+\frac {\log ^2(x)}{x} \]
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Leaf count is larger than twice the leaf count of optimal. \(129\) vs. \(2(28)=56\).
Time = 0.62 (sec) , antiderivative size = 129, normalized size of antiderivative = 4.61, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {14, 6838, 2326, 2341, 2342} \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=-x^2-e^{8 x^2+8 e^{x-1} x+2 e^{2 x-2}}-\frac {2 e^{4 x^2+4 e^{x-1} x+e^{2 x-2}-2} \left (2 e^{x+1} x^2+4 e^2 x^2+e^{2 x} x+2 e^{x+1} x\right )}{2 e^{x-1} x+4 x+2 e^{x-1}+e^{2 x-2}}+\frac {\log ^2(x)}{x} \]
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Rule 14
Rule 2326
Rule 2341
Rule 2342
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \left (-4 e^{-2+2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (2 e+e^x\right ) \left (e^x+2 e x\right )-2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^2+2 e^{2 x} x+4 e^{1+x} x+8 e^2 x^2+4 e^{1+x} x^2\right )+\frac {-2 x^3+2 \log (x)-\log ^2(x)}{x^2}\right ) \, dx \\ & = -\left (2 \int e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^2+2 e^{2 x} x+4 e^{1+x} x+8 e^2 x^2+4 e^{1+x} x^2\right ) \, dx\right )-4 \int e^{-2+2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (2 e+e^x\right ) \left (e^x+2 e x\right ) \, dx+\int \frac {-2 x^3+2 \log (x)-\log ^2(x)}{x^2} \, dx \\ & = -e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+\int \left (-2 x+\frac {2 \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}\right ) \, dx \\ & = -e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+2 \int \frac {\log (x)}{x^2} \, dx-\int \frac {\log ^2(x)}{x^2} \, dx \\ & = -e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-\frac {2}{x}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}-\frac {2 \log (x)}{x}+\frac {\log ^2(x)}{x}-2 \int \frac {\log (x)}{x^2} \, dx \\ & = -e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+\frac {\log ^2(x)}{x} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=\frac {-x \left (e^{\frac {\left (e^x+2 e x\right )^2}{e^2}}+x\right )^2+\log ^2(x)}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(61\) vs. \(2(26)=52\).
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.21
method | result | size |
risch | \(\frac {\ln \left (x \right )^{2}}{x}-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{-2+2 x}+8 x \,{\mathrm e}^{-1+x}+8 x^{2}}-2 x \,{\mathrm e}^{{\mathrm e}^{-2+2 x}+4 x \,{\mathrm e}^{-1+x}+4 x^{2}}\) | \(62\) |
parallelrisch | \(\frac {-2 \,{\mathrm e}^{{\mathrm e}^{-2+2 x}+4 x \,{\mathrm e}^{-1+x}+4 x^{2}} x^{2}-x^{3}+\ln \left (x \right )^{2}-x \,{\mathrm e}^{2 \,{\mathrm e}^{-2+2 x}+8 x \,{\mathrm e}^{-1+x}+8 x^{2}}}{x}\) | \(65\) |
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.29 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=-\frac {x^{3} + 2 \, x^{2} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x - 1\right )} + e^{\left (2 \, x - 2\right )}\right )} + x e^{\left (8 \, x^{2} + 8 \, x e^{\left (x - 1\right )} + 2 \, e^{\left (2 \, x - 2\right )}\right )} - \log \left (x\right )^{2}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=- x^{2} - 2 x e^{4 x^{2} + 4 x e^{x - 1} + e^{2 x - 2}} - e^{8 x^{2} + 8 x e^{x - 1} + 2 e^{2 x - 2}} + \frac {\log {\left (x \right )}^{2}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (26) = 52\).
Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=-x^{2} - \frac {2 \, x^{2} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x - 1\right )} + e^{\left (2 \, x - 2\right )}\right )} + x e^{\left (8 \, x^{2} + 8 \, x e^{\left (x - 1\right )} + 2 \, e^{\left (2 \, x - 2\right )}\right )} - \log \left (x\right )^{2}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (26) = 52\).
Time = 0.34 (sec) , antiderivative size = 220, normalized size of antiderivative = 7.86 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=-\frac {{\left (x - 1\right )}^{3} + 2 \, {\left (x - 1\right )}^{2} e^{\left (4 \, {\left (x - 1\right )}^{2} + 4 \, {\left (x - 1\right )} e^{\left (x - 1\right )} + 8 \, x + e^{\left (2 \, x - 2\right )} + 4 \, e^{\left (x - 1\right )} - 4\right )} + 3 \, {\left (x - 1\right )}^{2} + {\left (x - 1\right )} e^{\left (8 \, {\left (x - 1\right )}^{2} + 8 \, {\left (x - 1\right )} e^{\left (x - 1\right )} + 16 \, x + 2 \, e^{\left (2 \, x - 2\right )} + 8 \, e^{\left (x - 1\right )} - 8\right )} + 4 \, {\left (x - 1\right )} e^{\left (4 \, {\left (x - 1\right )}^{2} + 4 \, {\left (x - 1\right )} e^{\left (x - 1\right )} + 8 \, x + e^{\left (2 \, x - 2\right )} + 4 \, e^{\left (x - 1\right )} - 4\right )} - \log \left (x\right )^{2} + 2 \, x + e^{\left (8 \, {\left (x - 1\right )}^{2} + 8 \, {\left (x - 1\right )} e^{\left (x - 1\right )} + 16 \, x + 2 \, e^{\left (2 \, x - 2\right )} + 8 \, e^{\left (x - 1\right )} - 8\right )} + 2 \, e^{\left (4 \, {\left (x - 1\right )}^{2} + 4 \, {\left (x - 1\right )} e^{\left (x - 1\right )} + 8 \, x + e^{\left (2 \, x - 2\right )} + 4 \, e^{\left (x - 1\right )} - 4\right )} - 2}{x} \]
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Time = 9.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.75 \[ \int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} \left (-8 x^2-8 x^3\right )\right )+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} \left (-8 x^3-8 x^4\right )\right )+2 \log (x)-\log ^2(x)}{x^2} \, dx=\frac {{\ln \left (x\right )}^2+2\,\ln \left (x\right )+2}{x}-{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}+8\,x^2+8\,x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}-\frac {2\,\left (\ln \left (x\right )+1\right )}{x}-x^2-2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}+4\,x^2+4\,x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x} \]
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