Integrand size = 126, antiderivative size = 30 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=4^{\frac {4}{\frac {e^4 x}{4}+\frac {e^x+(5-x) x}{x}}} \]
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Time = 0.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6, 6820, 12, 6838} \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=4^{\frac {16 x}{e^4 x^2+4 (5-x) x+4 e^x}} \]
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Rule 6
Rule 12
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+\left (16+e^8\right ) x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx \\ & = \int \frac {4^{2+\frac {16 x}{4 e^x-4 (-5+x) x+e^4 x^2}} \left (-4 e^x (-1+x)+4 \left (1-\frac {e^4}{4}\right ) x^2\right ) \log (4)}{\left (4 e^x-4 (-5+x) x+e^4 x^2\right )^2} \, dx \\ & = \log (4) \int \frac {4^{2+\frac {16 x}{4 e^x-4 (-5+x) x+e^4 x^2}} \left (-4 e^x (-1+x)+4 \left (1-\frac {e^4}{4}\right ) x^2\right )}{\left (4 e^x-4 (-5+x) x+e^4 x^2\right )^2} \, dx \\ & = 4^{\frac {16 x}{4 e^x+4 (5-x) x+e^4 x^2}} \\ \end{align*}
Time = 1.61 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \]
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Time = 2.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
risch | \({\mathrm e}^{\frac {32 x \ln \left (2\right )}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}\) | \(28\) |
parallelrisch | \({\mathrm e}^{\frac {32 x \ln \left (2\right )}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}\) | \(28\) |
norman | \(\frac {\left ({\mathrm e}^{4}-4\right ) x^{2} {\mathrm e}^{\frac {32 x \ln \left (2\right )}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}+20 x \,{\mathrm e}^{\frac {32 x \ln \left (2\right )}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}+4 \,{\mathrm e}^{x} {\mathrm e}^{\frac {32 x \ln \left (2\right )}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}\) | \(120\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=2^{\frac {32 \, x}{x^{2} e^{4} - 4 \, x^{2} + 20 \, x + 4 \, e^{x}}} \]
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Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=e^{\frac {32 x \log {\left (2 \right )}}{- 4 x^{2} + x^{2} e^{4} + 20 x + 4 e^{x}}} \]
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Time = 0.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=2^{\frac {32 \, x}{x^{2} {\left (e^{4} - 4\right )} + 20 \, x + 4 \, e^{x}}} \]
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Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx=2^{\frac {32 \, x}{x^{2} e^{4} - 4 \, x^{2} + 20 \, x + 4 \, e^{x}}} \]
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Time = 13.99 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx={\mathrm {e}}^{\frac {32\,x\,\ln \left (2\right )}{20\,x+4\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^4-4\,x^2}} \]
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