Integrand size = 65, antiderivative size = 25 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\frac {-e^x-e^{\log ^2(x)}+2 x}{5+x^4} \]
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Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(25)=50\).
Time = 0.58 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1608, 28, 6874, 2326, 391} \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\frac {2 x}{x^4+5}-\frac {e^x}{x^4+5}-\frac {e^{\log ^2(x)} \left (x^4 \log (x)+5 \log (x)\right )}{\left (x^4+5\right )^2 \log (x)} \]
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Rule 28
Rule 391
Rule 1608
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{x \left (25+10 x^4+x^8\right )} \, dx \\ & = \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{x \left (5+x^4\right )^2} \, dx \\ & = \int \left (-\frac {-10+5 e^x-4 e^x x^3+6 x^4+e^x x^4}{\left (5+x^4\right )^2}-\frac {2 e^{\log ^2(x)} \left (-2 x^4+5 \log (x)+x^4 \log (x)\right )}{x \left (5+x^4\right )^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{\log ^2(x)} \left (-2 x^4+5 \log (x)+x^4 \log (x)\right )}{x \left (5+x^4\right )^2} \, dx\right )-\int \frac {-10+5 e^x-4 e^x x^3+6 x^4+e^x x^4}{\left (5+x^4\right )^2} \, dx \\ & = -\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)}-\int \left (\frac {e^x \left (5-4 x^3+x^4\right )}{\left (5+x^4\right )^2}+\frac {2 \left (-5+3 x^4\right )}{\left (5+x^4\right )^2}\right ) \, dx \\ & = -\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)}-2 \int \frac {-5+3 x^4}{\left (5+x^4\right )^2} \, dx-\int \frac {e^x \left (5-4 x^3+x^4\right )}{\left (5+x^4\right )^2} \, dx \\ & = -\frac {e^x}{5+x^4}+\frac {2 x}{5+x^4}-\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)} \\ \end{align*}
Time = 0.63 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=-\frac {e^x+e^{\log ^2(x)}-2 x}{5+x^4} \]
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Time = 3.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(-\frac {-10 x +5 \,{\mathrm e}^{x}+5 \,{\mathrm e}^{\ln \left (x \right )^{2}}}{5 \left (x^{4}+5\right )}\) | \(25\) |
risch | \(\frac {2 x -{\mathrm e}^{x}}{x^{4}+5}-\frac {{\mathrm e}^{\ln \left (x \right )^{2}}}{x^{4}+5}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\frac {2 \, x - e^{\left (\log \left (x\right )^{2}\right )} - e^{x}}{x^{4} + 5} \]
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Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\frac {2 x}{x^{4} + 5} - \frac {e^{x}}{x^{4} + 5} - \frac {e^{\log {\left (x \right )}^{2}}}{x^{4} + 5} \]
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\frac {2 \, x}{x^{4} + 5} - \frac {e^{\left (\log \left (x\right )^{2}\right )} + e^{x}}{x^{4} + 5} \]
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\[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=\int { -\frac {6 \, x^{5} - 2 \, {\left (2 \, x^{4} - {\left (x^{4} + 5\right )} \log \left (x\right )\right )} e^{\left (\log \left (x\right )^{2}\right )} + {\left (x^{5} - 4 \, x^{4} + 5 \, x\right )} e^{x} - 10 \, x}{x^{9} + 10 \, x^{5} + 25 \, x} \,d x } \]
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Time = 12.46 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{25 x+10 x^5+x^9} \, dx=-\frac {{\mathrm {e}}^{{\ln \left (x\right )}^2}-2\,x+{\mathrm {e}}^x}{x^4+5} \]
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