Integrand size = 85, antiderivative size = 23 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\frac {2 x}{3 \left (e^{16}+x\right )}+\log \left (\frac {4 e^x}{5}+x\right ) \]
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\[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{3 \left (e^{16}+x\right )^2 \left (4 e^x+5 x\right )} \, dx \\ & = \frac {1}{3} \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{\left (e^{16}+x\right )^2 \left (4 e^x+5 x\right )} \, dx \\ & = \frac {1}{3} \int \left (-\frac {15 (-1+x)}{4 e^x+5 x}+\frac {e^{16} \left (2+3 e^{16}\right )+6 e^{16} x+3 x^2}{\left (e^{16}+x\right )^2}\right ) \, dx \\ & = \frac {1}{3} \int \frac {e^{16} \left (2+3 e^{16}\right )+6 e^{16} x+3 x^2}{\left (e^{16}+x\right )^2} \, dx-5 \int \frac {-1+x}{4 e^x+5 x} \, dx \\ & = \frac {1}{3} \int \left (3+\frac {2 e^{16}}{\left (e^{16}+x\right )^2}\right ) \, dx-5 \int \left (-\frac {1}{4 e^x+5 x}+\frac {x}{4 e^x+5 x}\right ) \, dx \\ & = x-\frac {2 e^{16}}{3 \left (e^{16}+x\right )}+5 \int \frac {1}{4 e^x+5 x} \, dx-5 \int \frac {x}{4 e^x+5 x} \, dx \\ \end{align*}
Time = 1.62 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\frac {1}{3} \left (-\frac {2 e^{16}}{e^{16}+x}+3 \log \left (4 e^x+5 x\right )\right ) \]
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Time = 0.66 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {2 \,{\mathrm e}^{16}}{3 \left (x +{\mathrm e}^{16}\right )}+\ln \left (\frac {5 x}{4}+{\mathrm e}^{x}\right )\) | \(19\) |
norman | \(-\frac {2 \,{\mathrm e}^{16}}{3 \left (x +{\mathrm e}^{16}\right )}+\ln \left (5 x +4 \,{\mathrm e}^{x}\right )\) | \(21\) |
parallelrisch | \(\frac {3 \ln \left (x +\frac {4 \,{\mathrm e}^{x}}{5}\right ) {\mathrm e}^{16}+3 \ln \left (x +\frac {4 \,{\mathrm e}^{x}}{5}\right ) x -2 \,{\mathrm e}^{16}}{3 x +3 \,{\mathrm e}^{16}}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\frac {3 \, {\left (x + e^{16}\right )} \log \left (5 \, x + 4 \, e^{x}\right ) - 2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\log {\left (\frac {5 x}{4} + e^{x} \right )} - \frac {2 e^{16}}{3 x + 3 e^{16}} \]
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=-\frac {2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} + \log \left (\frac {5}{4} \, x + e^{x}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).
Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\frac {3 \, x \log \left (5 \, x + 4 \, e^{x}\right ) + 3 \, e^{16} \log \left (5 \, x + 4 \, e^{x}\right ) - 2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx=\ln \left (x+\frac {4\,{\mathrm {e}}^x}{5}\right )+\frac {2\,x}{3\,x+3\,{\mathrm {e}}^{16}} \]
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