Integrand size = 87, antiderivative size = 23 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\frac {9 (3+x)^2}{4 \left (5+e^x-x\right )^2 \log ^4(5)} \]
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\[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9 (3+x) \left (8-e^x (2+x)\right )}{2 \left (5+e^x-x\right )^3 \log ^4(5)} \, dx \\ & = \frac {9 \int \frac {(3+x) \left (8-e^x (2+x)\right )}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)} \\ & = \frac {9 \int \left (-\frac {(-6+x) (3+x)^2}{\left (5+e^x-x\right )^3}-\frac {6+5 x+x^2}{\left (5+e^x-x\right )^2}\right ) \, dx}{2 \log ^4(5)} \\ & = -\frac {9 \int \frac {(-6+x) (3+x)^2}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}-\frac {9 \int \frac {6+5 x+x^2}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)} \\ & = -\frac {9 \int \left (\frac {6}{\left (5+e^x-x\right )^2}+\frac {5 x}{\left (5+e^x-x\right )^2}+\frac {x^2}{\left (5+e^x-x\right )^2}\right ) \, dx}{2 \log ^4(5)}-\frac {9 \int \left (-\frac {54}{\left (5+e^x-x\right )^3}-\frac {27 x}{\left (5+e^x-x\right )^3}+\frac {x^3}{\left (5+e^x-x\right )^3}\right ) \, dx}{2 \log ^4(5)} \\ & = -\frac {9 \int \frac {x^2}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)}-\frac {9 \int \frac {x^3}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}-\frac {45 \int \frac {x}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)}-\frac {27 \int \frac {1}{\left (5+e^x-x\right )^2} \, dx}{\log ^4(5)}+\frac {243 \int \frac {x}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}+\frac {243 \int \frac {1}{\left (5+e^x-x\right )^3} \, dx}{\log ^4(5)} \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\frac {9 (3+x)^2}{4 \left (5+e^x-x\right )^2 \log ^4(5)} \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {\frac {9}{4} x^{2}+\frac {27}{2} x +\frac {81}{4}}{\ln \left (5\right )^{4} \left (-{\mathrm e}^{x}+x -5\right )^{2}}\) | \(24\) |
parallelrisch | \(\frac {9 x^{2}+54 x +81}{4 \ln \left (5\right )^{4} \left (x^{2}-2 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}-10 x +10 \,{\mathrm e}^{x}+25\right )}\) | \(40\) |
norman | \(\frac {\frac {36 x}{\ln \left (5\right )}-\frac {9 \,{\mathrm e}^{2 x}}{4 \ln \left (5\right )}-\frac {45 \,{\mathrm e}^{x}}{2 \ln \left (5\right )}+\frac {9 \,{\mathrm e}^{x} x}{2 \ln \left (5\right )}-\frac {36}{\ln \left (5\right )}}{\ln \left (5\right )^{3} \left (-{\mathrm e}^{x}+x -5\right )^{2}}\) | \(56\) |
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=-\frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (2 \, {\left (x - 5\right )} e^{x} \log \left (5\right )^{4} - {\left (x^{2} - 10 \, x + 25\right )} \log \left (5\right )^{4} - e^{\left (2 \, x\right )} \log \left (5\right )^{4}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.83 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\frac {9 x^{2} + 54 x + 81}{4 x^{2} \log {\left (5 \right )}^{4} - 40 x \log {\left (5 \right )}^{4} + \left (- 8 x \log {\left (5 \right )}^{4} + 40 \log {\left (5 \right )}^{4}\right ) e^{x} + 4 e^{2 x} \log {\left (5 \right )}^{4} + 100 \log {\left (5 \right )}^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (x^{2} \log \left (5\right )^{4} - 10 \, x \log \left (5\right )^{4} + e^{\left (2 \, x\right )} \log \left (5\right )^{4} + 25 \, \log \left (5\right )^{4} - 2 \, {\left (x \log \left (5\right )^{4} - 5 \, \log \left (5\right )^{4}\right )} e^{x}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (20) = 40\).
Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (x^{2} \log \left (5\right )^{4} - 2 \, x e^{x} \log \left (5\right )^{4} - 10 \, x \log \left (5\right )^{4} + e^{\left (2 \, x\right )} \log \left (5\right )^{4} + 10 \, e^{x} \log \left (5\right )^{4} + 25 \, \log \left (5\right )^{4}\right )}} \]
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Timed out. \[ \int \frac {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx=\int \frac {72\,x-{\mathrm {e}}^x\,\left (9\,x^2+45\,x+54\right )+216}{2\,{\mathrm {e}}^{3\,x}\,{\ln \left (5\right )}^4-{\ln \left (5\right )}^4\,\left (2\,x^3-30\,x^2+150\,x-250\right )+{\mathrm {e}}^x\,{\ln \left (5\right )}^4\,\left (6\,x^2-60\,x+150\right )-{\mathrm {e}}^{2\,x}\,{\ln \left (5\right )}^4\,\left (6\,x-30\right )} \,d x \]
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