Integrand size = 44, antiderivative size = 29 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {6 e^{\frac {4 (16-x)}{x}}}{-1+e^3}+5 \left (-5+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6, 12, 14, 2240} \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=5 x^2-\frac {6 e^{\frac {64}{x}-4}}{1-e^3} \]
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Rule 6
Rule 12
Rule 14
Rule 2240
Rubi steps \begin{align*} \text {integral}& = \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{\left (-1+e^3\right ) x^2} \, dx \\ & = \int \frac {-384 e^{\frac {4 (16-x)}{x}}+\left (-10+10 e^3\right ) x^3}{\left (-1+e^3\right ) x^2} \, dx \\ & = \frac {\int \frac {-384 e^{\frac {4 (16-x)}{x}}+\left (-10+10 e^3\right ) x^3}{x^2} \, dx}{-1+e^3} \\ & = \frac {\int \left (-\frac {384 e^{-4+\frac {64}{x}}}{x^2}+10 \left (-1+e^3\right ) x\right ) \, dx}{-1+e^3} \\ & = 5 x^2+\frac {384 \int \frac {e^{-4+\frac {64}{x}}}{x^2} \, dx}{1-e^3} \\ & = -\frac {6 e^{-4+\frac {64}{x}}}{1-e^3}+5 x^2 \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {2 \left (3 e^{64/x}-\frac {5}{2} e^4 \left (1-e^3\right ) x^2\right )}{e^4 \left (-1+e^3\right )} \]
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Time = 0.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(5 x^{2}+\frac {6 \,{\mathrm e}^{-\frac {4 \left (x -16\right )}{x}}}{{\mathrm e}^{3}-1}\) | \(24\) |
parts | \(5 x^{2}+\frac {6 \,{\mathrm e}^{\frac {-4 x +64}{x}}}{{\mathrm e}^{3}-1}\) | \(27\) |
norman | \(\frac {5 x^{3}+\frac {6 x \,{\mathrm e}^{\frac {-4 x +64}{x}}}{{\mathrm e}^{3}-1}}{x}\) | \(32\) |
derivativedivides | \(-\frac {1280 x^{2}}{256 \,{\mathrm e}^{3}-256}+\frac {6 \,{\mathrm e}^{-4+\frac {64}{x}}}{{\mathrm e}^{3}-1}+\frac {1280 \,{\mathrm e}^{3} x^{2}}{256 \,{\mathrm e}^{3}-256}\) | \(48\) |
default | \(-\frac {1280 x^{2}}{256 \,{\mathrm e}^{3}-256}+\frac {6 \,{\mathrm e}^{-4+\frac {64}{x}}}{{\mathrm e}^{3}-1}+\frac {1280 \,{\mathrm e}^{3} x^{2}}{256 \,{\mathrm e}^{3}-256}\) | \(48\) |
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {5 \, x^{2} e^{3} - 5 \, x^{2} + 6 \, e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{e^{3} - 1} \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=5 x^{2} + \frac {6 e^{\frac {4 \cdot \left (16 - x\right )}{x}}}{-1 + e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {5 \, x^{2} e^{3}}{e^{3} - 1} - \frac {5 \, x^{2}}{e^{3} - 1} + \frac {6 \, e^{\frac {64}{x}}}{e^{7} - e^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.45 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (x - 16\right )}^{2} e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{x^{2}} - \frac {6 \, {\left (x - 16\right )} e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )}}{x} + 640 \, e^{3} + 3 \, e^{\left (-\frac {4 \, {\left (x - 16\right )}}{x}\right )} - 640\right )}}{\frac {{\left (x - 16\right )}^{2} e^{3}}{x^{2}} - \frac {2 \, {\left (x - 16\right )} e^{3}}{x} - \frac {{\left (x - 16\right )}^{2}}{x^{2}} + \frac {2 \, {\left (x - 16\right )}}{x} + e^{3} - 1} \]
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Time = 11.72 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {-384 e^{\frac {4 (16-x)}{x}}-10 x^3+10 e^3 x^3}{-x^2+e^3 x^2} \, dx=\frac {6\,{\mathrm {e}}^{\frac {64}{x}-4}}{{\mathrm {e}}^3-1}+\frac {x^2\,\left (5\,{\mathrm {e}}^3-5\right )}{{\mathrm {e}}^3-1} \]
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