Integrand size = 45, antiderivative size = 21 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=1+2 x+x \left (-1+5 \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6820, 6874, 45, 2436, 2332} \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+\frac {5}{2} \left (2 x+e^4\right ) \log \left (\frac {1}{\left (2 x+e^4\right )^2}\right )+5 e^4 \log \left (2 x+e^4\right ) \]
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Rule 45
Rule 2332
Rule 2436
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4-18 x+5 \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )}{e^4+2 x} \, dx \\ & = \int \left (\frac {e^4-18 x}{e^4+2 x}+5 \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )\right ) \, dx \\ & = 5 \int \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right ) \, dx+\int \frac {e^4-18 x}{e^4+2 x} \, dx \\ & = \frac {5}{2} \text {Subst}\left (\int \log \left (\frac {1}{x^2}\right ) \, dx,x,e^4+2 x\right )+\int \left (-9+\frac {10 e^4}{e^4+2 x}\right ) \, dx \\ & = x+\frac {5}{2} \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )+5 e^4 \log \left (e^4+2 x\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+\frac {5}{2} \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )+5 e^4 \log \left (e^4+2 x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05
method | result | size |
risch | \(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\) | \(22\) |
default | \(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\) | \(24\) |
norman | \(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\) | \(24\) |
parts | \(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\) | \(24\) |
parallelrisch | \(5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x -{\mathrm e}^{4}+x\) | \(28\) |
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=5 \, x \log \left (\frac {1}{4 \, x^{2} + 4 \, x e^{4} + e^{8}}\right ) + x \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=5 x \log {\left (\frac {1}{4 x^{2} + 4 x e^{4} + e^{8}} \right )} + x \]
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Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (20) = 40\).
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 5.14 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=-\frac {5}{2} \, e^{4} \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) \log \left (2 \, x + e^{4}\right ) - \frac {5}{2} \, e^{4} \log \left (2 \, x + e^{4}\right )^{2} + \frac {5}{2} \, {\left (\log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) \log \left (2 \, x + e^{4}\right ) - \log \left (2 \, x + e^{4}\right )^{2}\right )} e^{4} + \frac {5}{2} \, {\left (e^{4} \log \left (2 \, x + e^{4}\right ) - 2 \, x\right )} \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) + x \]
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Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=-5 \, x \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) + x \]
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Time = 9.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+5\,x\,\ln \left (\frac {1}{4\,x^2+4\,{\mathrm {e}}^4\,x+{\mathrm {e}}^8}\right ) \]
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