Integrand size = 68, antiderivative size = 32 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=4-e^x-\frac {\log \left (\frac {4}{3}\right )}{5 \left (-e^{e^4}+2 e^x\right )}+\log (2) \]
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Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {2320, 12, 697} \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=\frac {\log \left (\frac {16}{9}\right )}{10 \left (e^{e^4}-2 e^x\right )}-e^x \]
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Rule 12
Rule 697
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-5 e^{2 e^4}+20 e^{e^4} x-20 x^2+\log \left (\frac {16}{9}\right )}{5 \left (e^{e^4}-2 x\right )^2} \, dx,x,e^x\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \frac {-5 e^{2 e^4}+20 e^{e^4} x-20 x^2+\log \left (\frac {16}{9}\right )}{\left (e^{e^4}-2 x\right )^2} \, dx,x,e^x\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \left (-5+\frac {\log \left (\frac {16}{9}\right )}{\left (e^{e^4}-2 x\right )^2}\right ) \, dx,x,e^x\right ) \\ & = -e^x+\frac {\log \left (\frac {16}{9}\right )}{10 \left (e^{e^4}-2 e^x\right )} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=\frac {1}{5} \left (-5 e^x+\frac {\log \left (\frac {16}{9}\right )}{2 \left (e^{e^4}-2 e^x\right )}\right ) \]
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Time = 2.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {5 \,{\mathrm e}^{2 \,{\mathrm e}^{4}}-20 \,{\mathrm e}^{2 x}-2 \ln \left (\frac {4}{3}\right )}{10 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}\) | \(31\) |
norman | \(\frac {2 \,{\mathrm e}^{2 x}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2}-\frac {\ln \left (3\right )}{5}+\frac {2 \ln \left (2\right )}{5}}{{\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}}\) | \(34\) |
risch | \(-{\mathrm e}^{x}-\frac {\ln \left (3\right )}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}+\frac {2 \ln \left (2\right )}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}\) | \(34\) |
default | \(\frac {10 \,{\mathrm e}^{2 x}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{4}}}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-10 \,{\mathrm e}^{x}}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2 \,{\mathrm e}^{{\mathrm e}^{4}}-4 \,{\mathrm e}^{x}}+\frac {2 \ln \left (2\right )}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}-\frac {\ln \left (3\right )}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-2 \,{\mathrm e}^{x}\right )}\) | \(73\) |
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Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=-\frac {e^{\left (4 \, e^{4}\right )} \log \left (\frac {4}{3}\right ) + 10 \, e^{\left (2 \, x + 4 \, e^{4}\right )} - 5 \, e^{\left (x + 5 \, e^{4}\right )}}{5 \, {\left (2 \, e^{\left (x + 4 \, e^{4}\right )} - e^{\left (5 \, e^{4}\right )}\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=- e^{x} + \frac {- 2 \log {\left (2 \right )} + \log {\left (3 \right )}}{10 e^{x} - 5 e^{e^{4}}} \]
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none
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=-\frac {\log \left (\frac {4}{3}\right )}{5 \, {\left (2 \, e^{x} - e^{\left (e^{4}\right )}\right )}} - e^{x} \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=\frac {\log \left (3\right ) - 2 \, \log \left (2\right )}{5 \, {\left (2 \, e^{x} - e^{\left (e^{4}\right )}\right )}} - e^{x} \]
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Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {-20 e^{3 x}-5 e^{2 e^4+x}+20 e^{e^4+2 x}+2 e^x \log \left (\frac {4}{3}\right )}{5 e^{2 e^4}+20 e^{2 x}-20 e^{e^4+x}} \, dx=\frac {10\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{x-{\mathrm {e}}^4}\,\left (5\,{\mathrm {e}}^{2\,{\mathrm {e}}^4}-\ln \left (\frac {16}{9}\right )\right )}{5\,{\mathrm {e}}^{{\mathrm {e}}^4}-10\,{\mathrm {e}}^x} \]
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