Integrand size = 51, antiderivative size = 26 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=x-\frac {\frac {5}{4}+x-\frac {x}{5+\log (2)}}{16 e^4 x^2} \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6, 12, 14} \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=-\frac {5}{64 e^4 x^2}+x-\frac {8+\log (4)}{32 e^4 x (5+\log (2))} \]
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Rule 6
Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{e^4 x^3 (160+32 \log (2))} \, dx \\ & = \frac {\int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{x^3} \, dx}{32 e^4 (5+\log (2))} \\ & = \frac {\int \left (32 e^4 (5+\log (2))+\frac {5 (5+\log (2))}{x^3}+\frac {8+\log (4)}{x^2}\right ) \, dx}{32 e^4 (5+\log (2))} \\ & = -\frac {5}{64 e^4 x^2}+x-\frac {8+\log (4)}{32 e^4 x (5+\log (2))} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=-\frac {25-64 e^4 x^3 (5+\log (2))+2 x (8+\log (4))+\log (32)}{64 e^4 x^2 (5+\log (2))} \]
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Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x +\frac {\left (\left (-\frac {\ln \left (2\right )}{16}-\frac {1}{4}\right ) x -\frac {5 \ln \left (2\right )}{64}-\frac {25}{64}\right ) {\mathrm e}^{-4}}{x^{2} \left (\ln \left (2\right )+5\right )}\) | \(29\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{2}-\frac {5 \,{\mathrm e}^{-2}}{64}-\frac {\left (4+\ln \left (2\right )\right ) {\mathrm e}^{-2} x}{16 \left (\ln \left (2\right )+5\right )}\right ) {\mathrm e}^{-2}}{x^{2}}\) | \(39\) |
default | \(\frac {{\mathrm e}^{-4} \left (32 x \,{\mathrm e}^{4} \ln \left (2\right )+160 x \,{\mathrm e}^{4}-\frac {8+2 \ln \left (2\right )}{x}-\frac {5 \ln \left (2\right )+25}{2 x^{2}}\right )}{32 \ln \left (2\right )+160}\) | \(48\) |
gosper | \(\frac {\left (64 x^{3} {\mathrm e}^{4} \ln \left (2\right )+320 x^{3} {\mathrm e}^{4}-4 x \ln \left (2\right )-5 \ln \left (2\right )-16 x -25\right ) {\mathrm e}^{-4}}{64 x^{2} \left (\ln \left (2\right )+5\right )}\) | \(50\) |
parallelrisch | \(\frac {\left (64 x^{3} {\mathrm e}^{4} \ln \left (2\right )+320 x^{3} {\mathrm e}^{4}-4 x \ln \left (2\right )-5 \ln \left (2\right )-16 x -25\right ) {\mathrm e}^{-4}}{64 x^{2} \left (\ln \left (2\right )+5\right )}\) | \(50\) |
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=\frac {320 \, x^{3} e^{4} + {\left (64 \, x^{3} e^{4} - 4 \, x - 5\right )} \log \left (2\right ) - 16 \, x - 25}{64 \, {\left (x^{2} e^{4} \log \left (2\right ) + 5 \, x^{2} e^{4}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=\frac {x \left (32 e^{4} \log {\left (2 \right )} + 160 e^{4}\right ) + \frac {x \left (-16 - 4 \log {\left (2 \right )}\right ) - 25 - 5 \log {\left (2 \right )}}{2 x^{2}}}{32 e^{4} \log {\left (2 \right )} + 160 e^{4}} \]
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none
Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=x - \frac {4 \, x {\left (\log \left (2\right ) + 4\right )} + 5 \, \log \left (2\right ) + 25}{64 \, {\left (e^{4} \log \left (2\right ) + 5 \, e^{4}\right )} x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=\frac {x e^{4} \log \left (2\right ) + 5 \, x e^{4}}{e^{4} \log \left (2\right ) + 5 \, e^{4}} - \frac {4 \, x \log \left (2\right ) + 16 \, x + 5 \, \log \left (2\right ) + 25}{64 \, {\left (e^{4} \log \left (2\right ) + 5 \, e^{4}\right )} x^{2}} \]
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Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx=x-\frac {\frac {\ln \left (32\right )}{2}+x\,\left (\ln \left (4\right )+8\right )+\frac {25}{2}}{x^2\,\left (160\,{\mathrm {e}}^4+32\,{\mathrm {e}}^4\,\ln \left (2\right )\right )} \]
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