Integrand size = 85, antiderivative size = 21 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=x-4 \left (x+\log ^2\left (3+2 \left (-1+e^5+x\right )\right )\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(21)=42\).
Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.29, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6874, 2458, 2388, 2338, 2332, 2436, 2333, 2437, 2339, 30} \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=-4 x^2+x-4 \log ^4\left (2 x+2 e^5+1\right )-4 \left (2 x+2 e^5+1\right ) \log ^2\left (2 x+2 e^5+1\right )+4 \left (1+2 e^5\right ) \log ^2\left (2 x+2 e^5+1\right ) \]
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Rule 30
Rule 2332
Rule 2333
Rule 2338
Rule 2339
Rule 2388
Rule 2436
Rule 2437
Rule 2458
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (1-8 x-\frac {32 x \log \left (1+2 e^5+2 x\right )}{1+2 e^5+2 x}-8 \log ^2\left (1+2 e^5+2 x\right )-\frac {32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x}\right ) \, dx \\ & = x-4 x^2-8 \int \log ^2\left (1+2 e^5+2 x\right ) \, dx-32 \int \frac {x \log \left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx-32 \int \frac {\log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx \\ & = x-4 x^2-4 \text {Subst}\left (\int \log ^2(x) \, dx,x,1+2 e^5+2 x\right )-16 \text {Subst}\left (\int \frac {\left (\frac {1}{2} \left (-1-2 e^5\right )+\frac {x}{2}\right ) \log (x)}{x} \, dx,x,1+2 e^5+2 x\right )-16 \text {Subst}\left (\int \frac {\log ^3(x)}{x} \, dx,x,1+2 e^5+2 x\right ) \\ & = x-4 x^2-4 \left (1+2 e^5+2 x\right ) \log ^2\left (1+2 e^5+2 x\right )-16 \text {Subst}\left (\int x^3 \, dx,x,\log \left (1+2 e^5+2 x\right )\right )+\left (8 \left (1+2 e^5\right )\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+2 e^5+2 x\right ) \\ & = x-4 x^2+4 \left (1+2 e^5\right ) \log ^2\left (1+2 e^5+2 x\right )-4 \left (1+2 e^5+2 x\right ) \log ^2\left (1+2 e^5+2 x\right )-4 \log ^4\left (1+2 e^5+2 x\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=x-4 x^2-8 x \log ^2\left (1+2 e^5+2 x\right )-4 \log ^4\left (1+2 e^5+2 x\right ) \]
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Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76
method | result | size |
norman | \(x -4 x^{2}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} x\) | \(37\) |
risch | \(x -4 x^{2}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} x\) | \(37\) |
parts | \(-4 x^{2}+x -4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} \left (2 \,{\mathrm e}^{5}+2 x +1\right )-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}+8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} {\mathrm e}^{5}+4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}\) | \(75\) |
derivativedivides | \(-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} \left (2 \,{\mathrm e}^{5}+2 x +1\right )+5 \,{\mathrm e}^{5}+5 x +\frac {5}{2}+8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} {\mathrm e}^{5}+4 \,{\mathrm e}^{5} \left (2 \,{\mathrm e}^{5}+2 x +1\right )-\left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}+4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}\) | \(103\) |
default | \(-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} \left (2 \,{\mathrm e}^{5}+2 x +1\right )+5 \,{\mathrm e}^{5}+5 x +\frac {5}{2}+8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} {\mathrm e}^{5}+4 \,{\mathrm e}^{5} \left (2 \,{\mathrm e}^{5}+2 x +1\right )-\left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}+4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}\) | \(103\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=-4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 8 \, x \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 4 \, x^{2} + x \]
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Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=- 4 x^{2} - 8 x \log {\left (2 x + 1 + 2 e^{5} \right )}^{2} + x - 4 \log {\left (2 x + 1 + 2 e^{5} \right )}^{4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (20) = 40\).
Time = 0.19 (sec) , antiderivative size = 262, normalized size of antiderivative = 12.48 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=\frac {4}{3} \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - \frac {8}{3} \, e^{5} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - 4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 4 \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - \frac {4}{3} \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - 4 \, {\left (\log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 2 \, \log \left (2 \, x + 2 \, e^{5} + 1\right ) + 2\right )} {\left (2 \, x + 2 \, e^{5} + 1\right )} - 4 \, x^{2} + 4 \, x {\left (2 \, e^{5} + 1\right )} + 4 \, {\left ({\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, x\right )} e^{5} + 8 \, {\left ({\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, x\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, {\left (4 \, e^{10} + 4 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - \frac {13}{2} \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) + e^{5} \log \left (2 \, x + 2 \, e^{5} + 1\right ) + 13 \, x + \frac {1}{2} \, \log \left (2 \, x + 2 \, e^{5} + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=-4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 8 \, x \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 4 \, x^{2} + x \]
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Time = 8.77 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log \left (1+2 e^5+2 x\right )+\left (-8-16 e^5-16 x\right ) \log ^2\left (1+2 e^5+2 x\right )-32 \log ^3\left (1+2 e^5+2 x\right )}{1+2 e^5+2 x} \, dx=-4\,x^2-8\,x\,{\ln \left (2\,x+2\,{\mathrm {e}}^5+1\right )}^2+x-4\,{\ln \left (2\,x+2\,{\mathrm {e}}^5+1\right )}^4 \]
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