Integrand size = 68, antiderivative size = 21 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {\left (-17+e^4 x\right ) \log ^2\left (e^5 x\right )}{(1+x)^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(208\) vs. \(2(21)=42\).
Time = 0.45 (sec) , antiderivative size = 208, normalized size of antiderivative = 9.90, number of steps used = 24, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6820, 6874, 46, 37, 2404, 2338, 2356, 2351, 31, 2354, 2438, 2398} \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=-\frac {15 e^4 x^2}{2 (x+1)^2}+\frac {10 \left (17+e^4\right )}{x+1}-\frac {170}{x+1}-\frac {5 \left (136+7 e^4\right )}{2 (x+1)^2}-\frac {85}{(x+1)^2}-\frac {\left (-e^4 x+e^4+34\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (x+1)^2}+\frac {\left (34+e^4\right )^2 \log ^2(x)}{4 \left (17+e^4\right )}-17 \log ^2(x)-\frac {2 \left (17+e^4\right ) x \log (x)}{x+1}+\frac {2 \left (17-4 e^4\right ) x \log (x)}{x+1}-\frac {10 \left (17+e^4\right ) \log (x)}{(x+1)^2}+10 \left (17+e^4\right ) \log (x)-170 \log (x)-8 \left (17+e^4\right ) \log (x+1)-2 \left (17-4 e^4\right ) \log (x+1)+170 \log (x+1) \]
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Rule 31
Rule 37
Rule 46
Rule 2338
Rule 2351
Rule 2354
Rule 2356
Rule 2398
Rule 2404
Rule 2438
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {(5+\log (x)) \left (-34+\left (136+7 e^4\right ) x-3 e^4 x^2+\left (34-e^4 (-1+x)\right ) x \log (x)\right )}{x (1+x)^3} \, dx \\ & = \int \left (\frac {5 \left (136+7 e^4\right )}{(1+x)^3}-\frac {170}{x (1+x)^3}-\frac {15 e^4 x}{(1+x)^3}+\frac {2 \left (-17+3 \left (51+2 e^4\right ) x-4 e^4 x^2\right ) \log (x)}{x (1+x)^3}-\frac {\left (-34-e^4+e^4 x\right ) \log ^2(x)}{(1+x)^3}\right ) \, dx \\ & = -\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}+2 \int \frac {\left (-17+3 \left (51+2 e^4\right ) x-4 e^4 x^2\right ) \log (x)}{x (1+x)^3} \, dx-170 \int \frac {1}{x (1+x)^3} \, dx-\left (15 e^4\right ) \int \frac {x}{(1+x)^3} \, dx-\int \frac {\left (-34-e^4+e^4 x\right ) \log ^2(x)}{(1+x)^3} \, dx \\ & = -\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}-\frac {15 e^4 x^2}{2 (1+x)^2}-\frac {\left (34+e^4-e^4 x\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (1+x)^2}+2 \int \left (-\frac {17 \log (x)}{x}+\frac {10 \left (17+e^4\right ) \log (x)}{(1+x)^3}+\frac {\left (17-4 e^4\right ) \log (x)}{(1+x)^2}+\frac {17 \log (x)}{1+x}\right ) \, dx-170 \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^3}-\frac {1}{(1+x)^2}\right ) \, dx+\frac {\int \frac {\left (-34-e^4+e^4 x\right )^2 \log (x)}{x (1+x)^2} \, dx}{2 \left (17+e^4\right )} \\ & = -\frac {85}{(1+x)^2}-\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}-\frac {15 e^4 x^2}{2 (1+x)^2}-\frac {170}{1+x}-170 \log (x)-\frac {\left (34+e^4-e^4 x\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (1+x)^2}+170 \log (1+x)-34 \int \frac {\log (x)}{x} \, dx+34 \int \frac {\log (x)}{1+x} \, dx+\left (2 \left (17-4 e^4\right )\right ) \int \frac {\log (x)}{(1+x)^2} \, dx+\frac {\int \left (\frac {\left (34+e^4\right )^2 \log (x)}{x}-\frac {4 \left (17+e^4\right )^2 \log (x)}{(1+x)^2}-\frac {68 \left (17+e^4\right ) \log (x)}{1+x}\right ) \, dx}{2 \left (17+e^4\right )}+\left (20 \left (17+e^4\right )\right ) \int \frac {\log (x)}{(1+x)^3} \, dx \\ & = -\frac {85}{(1+x)^2}-\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}-\frac {15 e^4 x^2}{2 (1+x)^2}-\frac {170}{1+x}-170 \log (x)-\frac {10 \left (17+e^4\right ) \log (x)}{(1+x)^2}+\frac {2 \left (17-4 e^4\right ) x \log (x)}{1+x}-17 \log ^2(x)-\frac {\left (34+e^4-e^4 x\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (1+x)^2}+170 \log (1+x)+34 \log (x) \log (1+x)-34 \int \frac {\log (x)}{1+x} \, dx-34 \int \frac {\log (1+x)}{x} \, dx-\left (2 \left (17-4 e^4\right )\right ) \int \frac {1}{1+x} \, dx-\left (2 \left (17+e^4\right )\right ) \int \frac {\log (x)}{(1+x)^2} \, dx+\left (10 \left (17+e^4\right )\right ) \int \frac {1}{x (1+x)^2} \, dx+\frac {\left (34+e^4\right )^2 \int \frac {\log (x)}{x} \, dx}{2 \left (17+e^4\right )} \\ & = -\frac {85}{(1+x)^2}-\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}-\frac {15 e^4 x^2}{2 (1+x)^2}-\frac {170}{1+x}-170 \log (x)-\frac {10 \left (17+e^4\right ) \log (x)}{(1+x)^2}+\frac {2 \left (17-4 e^4\right ) x \log (x)}{1+x}-\frac {2 \left (17+e^4\right ) x \log (x)}{1+x}-17 \log ^2(x)+\frac {\left (34+e^4\right )^2 \log ^2(x)}{4 \left (17+e^4\right )}-\frac {\left (34+e^4-e^4 x\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (1+x)^2}+170 \log (1+x)-2 \left (17-4 e^4\right ) \log (1+x)+34 \operatorname {PolyLog}(2,-x)+34 \int \frac {\log (1+x)}{x} \, dx+\left (2 \left (17+e^4\right )\right ) \int \frac {1}{1+x} \, dx+\left (10 \left (17+e^4\right )\right ) \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^2}\right ) \, dx \\ & = -\frac {85}{(1+x)^2}-\frac {5 \left (136+7 e^4\right )}{2 (1+x)^2}-\frac {15 e^4 x^2}{2 (1+x)^2}-\frac {170}{1+x}+\frac {10 \left (17+e^4\right )}{1+x}-170 \log (x)+10 \left (17+e^4\right ) \log (x)-\frac {10 \left (17+e^4\right ) \log (x)}{(1+x)^2}+\frac {2 \left (17-4 e^4\right ) x \log (x)}{1+x}-\frac {2 \left (17+e^4\right ) x \log (x)}{1+x}-17 \log ^2(x)+\frac {\left (34+e^4\right )^2 \log ^2(x)}{4 \left (17+e^4\right )}-\frac {\left (34+e^4-e^4 x\right )^2 \log ^2(x)}{4 \left (17+e^4\right ) (1+x)^2}+170 \log (1+x)-2 \left (17-4 e^4\right ) \log (1+x)-8 \left (17+e^4\right ) \log (1+x) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {\left (-17+e^4 x\right ) (5+\log (x))^2}{(1+x)^2} \]
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Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19
method | result | size |
risch | \(\frac {\left (x \,{\mathrm e}^{4}-17\right ) \ln \left (x \,{\mathrm e}^{5}\right )^{2}}{x^{2}+2 x +1}\) | \(25\) |
norman | \(\frac {x \,{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{5}\right )^{2}-17 \ln \left (x \,{\mathrm e}^{5}\right )^{2}}{\left (1+x \right )^{2}}\) | \(28\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{5}\right )^{2}-17 \ln \left (x \,{\mathrm e}^{5}\right )^{2}}{x^{2}+2 x +1}\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {{\left (x e^{4} - 17\right )} \log \left (x e^{5}\right )^{2}}{x^{2} + 2 \, x + 1} \]
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Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {\left (x e^{4} - 17\right ) \log {\left (x e^{5} \right )}^{2}}{x^{2} + 2 x + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.86 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {{\left (x e^{4} - 17\right )} \log \left (x\right )^{2} + 25 \, x e^{4} + 10 \, {\left (x e^{4} - 17\right )} \log \left (x\right ) - 425}{x^{2} + 2 \, x + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.05 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {x e^{4} \log \left (x\right )^{2} + 10 \, x e^{4} \log \left (x\right ) + 25 \, x e^{4} - 17 \, \log \left (x\right )^{2} - 170 \, \log \left (x\right ) - 425}{x^{2} + 2 \, x + 1} \]
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Time = 13.58 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-34-34 x+e^4 \left (2 x+2 x^2\right )\right ) \log \left (e^5 x\right )+\left (34 x+e^4 \left (x-x^2\right )\right ) \log ^2\left (e^5 x\right )}{x+3 x^2+3 x^3+x^4} \, dx=\frac {{\ln \left (x\,{\mathrm {e}}^5\right )}^2\,\left (x\,{\mathrm {e}}^4-17\right )}{{\left (x+1\right )}^2} \]
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