Integrand size = 63, antiderivative size = 22 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{3+\frac {3}{5-\frac {e^4}{3}+2 e^x}} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {12, 2320, 2262, 2240} \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{\frac {9}{6 e^x+15-e^4}+3} \]
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Rule 12
Rule 2240
Rule 2262
Rule 2320
Rubi steps \begin{align*} \text {integral}& = -\left (54 \int \frac {e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx\right ) \\ & = -\left (54 \text {Subst}\left (\int \frac {e^{\frac {3 \left (-18+e^4-6 x\right )}{-15+e^4-6 x}}}{\left (15-e^4+6 x\right )^2} \, dx,x,e^x\right )\right ) \\ & = -\left (54 \text {Subst}\left (\int \frac {\exp \left (3+\frac {6 \left (-18+e^4\right )-6 \left (-15+e^4\right )}{2 \left (-15+e^4-6 x\right )}\right )}{\left (15-e^4+6 x\right )^2} \, dx,x,e^x\right )\right ) \\ & = e^{3+\frac {9}{15-e^4+6 e^x}} \\ \end{align*}
Time = 0.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{3+\frac {9}{15-e^4+6 e^x}} \]
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Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \({\mathrm e}^{\frac {-18 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{4}-54}{-6 \,{\mathrm e}^{x}+{\mathrm e}^{4}-15}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {-18 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{4}-54}{-6 \,{\mathrm e}^{x}+{\mathrm e}^{4}-15}}\) | \(26\) |
norman | \(\frac {\left ({\mathrm e}^{4}-15\right ) {\mathrm e}^{\frac {18 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{4}+54}{6 \,{\mathrm e}^{x}-{\mathrm e}^{4}+15}}-6 \,{\mathrm e}^{x} {\mathrm e}^{\frac {18 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{4}+54}{6 \,{\mathrm e}^{x}-{\mathrm e}^{4}+15}}}{-6 \,{\mathrm e}^{x}+{\mathrm e}^{4}-15}\) | \(82\) |
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{\left (-x + \frac {{\left (x + 3\right )} e^{4} - 6 \, {\left (x + 3\right )} e^{x} - 15 \, x - 54}{e^{4} - 6 \, e^{x} - 15}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{\frac {18 e^{x} - 3 e^{4} + 54}{6 e^{x} - e^{4} + 15}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (15) = 30\).
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=e^{\left (\frac {3 \, e^{4}}{e^{4} - 6 \, e^{x} - 15} - \frac {18 \, e^{x}}{e^{4} - 6 \, e^{x} - 15} - \frac {54}{e^{4} - 6 \, e^{x} - 15}\right )} \]
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\[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx=\int { \frac {54 \, e^{\left (x + \frac {3 \, {\left (e^{4} - 6 \, e^{x} - 18\right )}}{e^{4} - 6 \, e^{x} - 15}\right )}}{12 \, {\left (e^{4} - 15\right )} e^{x} - e^{8} + 30 \, e^{4} - 36 \, e^{\left (2 \, x\right )} - 225} \,d x } \]
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Time = 11.97 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.27 \[ \int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx={\mathrm {e}}^{\frac {18\,{\mathrm {e}}^x}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}}\,{\mathrm {e}}^{\frac {54}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^4}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}} \]
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