Integrand size = 50, antiderivative size = 22 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=\frac {\log (5)}{2+\log \left (e^{\frac {2 e^8}{(2+4 x)^2}}\right )} \]
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Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 1694, 267} \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^8 \log (5)}{2 \left (4 (2 x+1)^2+e^8\right )} \]
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Rule 12
Rule 267
Rule 1694
Rubi steps \begin{align*} \text {integral}& = \left (e^8 \log (5)\right ) \int \frac {8+16 x}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx \\ & = \left (e^8 \log (5)\right ) \text {Subst}\left (\int \frac {16 x}{\left (e^8+16 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right ) \\ & = \left (16 e^8 \log (5)\right ) \text {Subst}\left (\int \frac {x}{\left (e^8+16 x^2\right )^2} \, dx,x,\frac {1}{2}+x\right ) \\ & = -\frac {e^8 \log (5)}{2 \left (e^8+4 (1+2 x)^2\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^8 \log (5)}{2 \left (e^8+4 (1+2 x)^2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\) | \(21\) |
gosper | \(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\) | \(25\) |
norman | \(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\) | \(25\) |
parallelrisch | \(-\frac {{\mathrm e}^{8} \ln \left (5\right )}{2 \left ({\mathrm e}^{8}+16 x^{2}+16 x +4\right )}\) | \(25\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]
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Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=- \frac {e^{8} \log {\left (5 \right )}}{32 x^{2} + 32 x + 8 + 2 e^{8}} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {e^{8} \log \left (5\right )}{2 \, {\left (16 \, x^{2} + 16 \, x + e^{8} + 4\right )}} \]
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Time = 13.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 (8+16 x) \log (5)}{16+e^{16}+128 x+384 x^2+512 x^3+256 x^4+e^8 \left (8+32 x+32 x^2\right )} \, dx=-\frac {{\mathrm {e}}^8\,\ln \left (5\right )}{2\,\left (16\,x^2+16\,x+{\mathrm {e}}^8+4\right )} \]
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