Integrand size = 92, antiderivative size = 28 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=e^{-\frac {8}{x}-x+\left (\frac {4}{x}+x^2+\log \left (\frac {16 x}{3}\right )\right )^2} \]
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\[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=\int \frac {\exp \left (\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}\right ) \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx \\ & = \int \left (\frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \left (-32+16 x+7 x^3+2 x^4+4 x^6\right )}{x^3}+\frac {2 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \left (-4+x+2 x^3\right ) \log \left (\frac {16 x}{3}\right )}{x^2}\right ) \, dx \\ & = 2 \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \left (-4+x+2 x^3\right ) \log \left (\frac {16 x}{3}\right )}{x^2} \, dx+\int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \left (-32+16 x+7 x^3+2 x^4+4 x^6\right )}{x^3} \, dx \\ & = 2 \int \left (-\frac {4 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \log \left (\frac {16 x}{3}\right )}{x}+2 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x \log \left (\frac {16 x}{3}\right )\right ) \, dx+\int \left (7 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right )-\frac {32 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right )}{x^3}+\frac {16 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right )}{x^2}+2 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x+4 \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x^3\right ) \, dx \\ & = 2 \int \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x \, dx+2 \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \log \left (\frac {16 x}{3}\right )}{x} \, dx+4 \int \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x^3 \, dx+4 \int \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) x \log \left (\frac {16 x}{3}\right ) \, dx+7 \int \exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \, dx-8 \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right ) \log \left (\frac {16 x}{3}\right )}{x^2} \, dx+16 \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right )}{x^2} \, dx-32 \int \frac {\exp \left (\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )}{x^2}+\log ^2\left (\frac {16 x}{3}\right )\right )}{x^3} \, dx \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=e^{\frac {16}{x^2}-\frac {8}{x}+7 x+x^4+\frac {2 \left (4+x^3\right ) \log \left (\frac {16 x}{3}\right )}{x}+\log ^2\left (\frac {16 x}{3}\right )} \]
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Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x^{2} \ln \left (\frac {16 x}{3}\right )^{2}+\left (2 x^{4}+8 x \right ) \ln \left (\frac {16 x}{3}\right )+x^{6}+7 x^{3}-8 x +16}{x^{2}}}\) | \(43\) |
risch | \(\left (\frac {16 x}{3}\right )^{2 x^{2}} \left (\frac {16 x}{3}\right )^{\frac {8}{x}} {\mathrm e}^{\frac {x^{6}+x^{2} \ln \left (\frac {16 x}{3}\right )^{2}+7 x^{3}-8 x +16}{x^{2}}}\) | \(48\) |
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Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=e^{\left (\frac {x^{6} + x^{2} \log \left (\frac {16}{3} \, x\right )^{2} + 7 \, x^{3} + 2 \, {\left (x^{4} + 4 \, x\right )} \log \left (\frac {16}{3} \, x\right ) - 8 \, x + 16}{x^{2}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=e^{\frac {x^{6} + 7 x^{3} + x^{2} \log {\left (\frac {16 x}{3} \right )}^{2} - 8 x + \left (2 x^{4} + 8 x\right ) \log {\left (\frac {16 x}{3} \right )} + 16}{x^{2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (25) = 50\).
Time = 0.48 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.39 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=\frac {e^{\left (x^{4} - 2 \, x^{2} \log \left (3\right ) + 8 \, x^{2} \log \left (2\right ) + 2 \, x^{2} \log \left (x\right ) + \log \left (3\right )^{2} + 16 \, \log \left (2\right )^{2} - 2 \, \log \left (3\right ) \log \left (x\right ) + 8 \, \log \left (2\right ) \log \left (x\right ) + \log \left (x\right )^{2} + 7 \, x - \frac {8 \, \log \left (3\right )}{x} + \frac {32 \, \log \left (2\right )}{x} + \frac {8 \, \log \left (x\right )}{x} - \frac {8}{x} + \frac {16}{x^{2}}\right )}}{2^{8 \, \log \left (3\right )}} \]
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Time = 0.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=e^{\left (x^{4} + 2 \, x^{2} \log \left (\frac {16}{3} \, x\right ) + \log \left (\frac {16}{3} \, x\right )^{2} + 7 \, x + \frac {8 \, \log \left (\frac {16}{3} \, x\right )}{x} - \frac {8}{x} + \frac {16}{x^{2}}\right )} \]
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Time = 8.79 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.79 \[ \int \frac {e^{\frac {16-8 x+7 x^3+x^6+\left (8 x+2 x^4\right ) \log \left (\frac {16 x}{3}\right )+x^2 \log ^2\left (\frac {16 x}{3}\right )}{x^2}} \left (-32+16 x+7 x^3+2 x^4+4 x^6+\left (-8 x+2 x^2+4 x^4\right ) \log \left (\frac {16 x}{3}\right )\right )}{x^3} \, dx=\frac {2^{8\,x^2}\,2^{32/x}\,x^{2\,x^2}\,x^{8/x}\,x^{8\,\ln \left (2\right )}\,{\mathrm {e}}^{{\ln \left (3\right )}^2}\,{\mathrm {e}}^{7\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{16\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{-\frac {8}{x}}\,{\mathrm {e}}^{\frac {16}{x^2}}\,{\mathrm {e}}^{{\ln \left (x\right )}^2}}{2^{8\,\ln \left (3\right )}\,3^{2\,x^2}\,3^{8/x}\,x^{2\,\ln \left (3\right )}} \]
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