Integrand size = 74, antiderivative size = 29 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=5 x+\frac {4 \left (4+\frac {5}{x^2}\right )^2+x}{\frac {2}{e^{15}}-x} \]
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Timed out. \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=\text {\$Aborted} \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{x^5 \left (4-4 e^{15} x+e^{30} x^2\right )} \, dx \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=\frac {-2 x^4 (1+5 x)+e^{15} \left (-100-160 x^2-64 x^4+5 x^6\right )}{x^4 \left (-2+e^{15} x\right )} \]
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Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34
method | result | size |
risch | \(5 x +\frac {\left (-64 \,{\mathrm e}^{15}-2\right ) x^{4}-160 x^{2} {\mathrm e}^{15}-100 \,{\mathrm e}^{15}}{x^{4} \left (x \,{\mathrm e}^{15}-2\right )}\) | \(39\) |
norman | \(\frac {5 \,{\mathrm e}^{15} x^{6}+\left (-32 \,{\mathrm e}^{30}-{\mathrm e}^{15}-10\right ) x^{5}-160 x^{2} {\mathrm e}^{15}-100 \,{\mathrm e}^{15}}{x^{4} \left (x \,{\mathrm e}^{15}-2\right )}\) | \(58\) |
gosper | \(-\frac {32 x^{5} {\mathrm e}^{30}-5 \,{\mathrm e}^{15} x^{6}+x^{5} {\mathrm e}^{15}+160 x^{2} {\mathrm e}^{15}+10 x^{5}+100 \,{\mathrm e}^{15}}{x^{4} \left (x \,{\mathrm e}^{15}-2\right )}\) | \(63\) |
parallelrisch | \(-\frac {64 x^{5} {\mathrm e}^{30}-10 \,{\mathrm e}^{15} x^{6}+2 x^{5} {\mathrm e}^{15}+320 x^{2} {\mathrm e}^{15}+20 x^{5}+200 \,{\mathrm e}^{15}}{2 x^{4} \left (x \,{\mathrm e}^{15}-2\right )}\) | \(64\) |
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Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=-\frac {10 \, x^{5} + 2 \, x^{4} - {\left (5 \, x^{6} - 64 \, x^{4} - 160 \, x^{2} - 100\right )} e^{15}}{x^{5} e^{15} - 2 \, x^{4}} \]
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Time = 1.73 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=5 x + \frac {x^{4} \left (- 64 e^{15} - 2\right ) - 160 x^{2} e^{15} - 100 e^{15}}{x^{5} e^{15} - 2 x^{4}} \]
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Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=5 \, x - \frac {2 \, {\left (x^{4} {\left (32 \, e^{15} + 1\right )} + 80 \, x^{2} e^{15} + 50 \, e^{15}\right )}}{x^{5} e^{15} - 2 \, x^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=5 \, x - \frac {25 \, e^{75} + 160 \, e^{45} + 256 \, e^{15} + 8}{4 \, {\left (x e^{15} - 2\right )}} + \frac {5 \, {\left (5 \, x^{3} e^{60} + 32 \, x^{3} e^{30} + 10 \, x^{2} e^{45} + 64 \, x^{2} e^{15} + 20 \, x e^{30} + 40 \, e^{15}\right )}}{4 \, x^{4}} \]
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Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {20 x^5+e^{15} \left (-800-640 x^2+2 x^5-20 x^6\right )+e^{30} \left (500 x+480 x^3+64 x^5+5 x^7\right )}{4 x^5-4 e^{15} x^6+e^{30} x^7} \, dx=5\,x-\frac {\left (64\,{\mathrm {e}}^{15}+2\right )\,x^4+160\,{\mathrm {e}}^{15}\,x^2+100\,{\mathrm {e}}^{15}}{x^4\,\left (x\,{\mathrm {e}}^{15}-2\right )} \]
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