Integrand size = 62, antiderivative size = 25 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=9+x+\left (e^{2 x}-\frac {4}{1+x}\right ) \left (x+\frac {2}{\log (4)}\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 27, 6874, 2207, 2225, 697} \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=x+\frac {e^{2 x} (x \log (16)+4+\log (4))}{2 \log (4)}-\frac {e^{2 x} \log (16)}{4 \log (4)}-\frac {2 (4-\log (16))}{(x+1) \log (4)} \]
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Rule 12
Rule 27
Rule 697
Rule 2207
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{1+2 x+x^2} \, dx}{\log (4)} \\ & = \frac {\int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{(1+x)^2} \, dx}{\log (4)} \\ & = \frac {\int \left (e^{2 x} (4+\log (4)+x \log (16))+\frac {8-3 \log (4)+x^2 \log (4)+x \log (16)}{(1+x)^2}\right ) \, dx}{\log (4)} \\ & = \frac {\int e^{2 x} (4+\log (4)+x \log (16)) \, dx}{\log (4)}+\frac {\int \frac {8-3 \log (4)+x^2 \log (4)+x \log (16)}{(1+x)^2} \, dx}{\log (4)} \\ & = \frac {e^{2 x} (4+\log (4)+x \log (16))}{2 \log (4)}+\frac {\int \left (\log (4)-\frac {2 (-4+\log (16))}{(1+x)^2}\right ) \, dx}{\log (4)}-\frac {\log (16) \int e^{2 x} \, dx}{2 \log (4)} \\ & = x-\frac {2 (4-\log (16))}{(1+x) \log (4)}-\frac {e^{2 x} \log (16)}{4 \log (4)}+\frac {e^{2 x} (4+\log (4)+x \log (16))}{2 \log (4)} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {-16+x \log (16)+x^2 \log (16)+e^{2 x} (1+x) (4+x \log (16))+\log (65536)}{2 (1+x) \log (4)} \]
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Time = 7.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
parts | \(x \,{\mathrm e}^{2 x}+\frac {{\mathrm e}^{2 x}}{\ln \left (2\right )}+x +\frac {4}{1+x}-\frac {4}{\ln \left (2\right ) \left (1+x \right )}\) | \(36\) |
risch | \(x -\frac {4}{\ln \left (2\right ) \left (1+x \right )}+\frac {4}{1+x}+\frac {\left (2+2 x \ln \left (2\right )\right ) {\mathrm e}^{2 x}}{2 \ln \left (2\right )}\) | \(38\) |
norman | \(\frac {x^{2}+{\mathrm e}^{2 x} x^{2}+\frac {{\mathrm e}^{2 x}}{\ln \left (2\right )}+\frac {\left (1+\ln \left (2\right )\right ) x \,{\mathrm e}^{2 x}}{\ln \left (2\right )}+\frac {3 \ln \left (2\right )-4}{\ln \left (2\right )}}{1+x}\) | \(53\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{2 x} \ln \left (2\right ) x^{2}+2 \,{\mathrm e}^{2 x} \ln \left (2\right ) x +2 x^{2} \ln \left (2\right )+2 x \,{\mathrm e}^{2 x}-8+2 \,{\mathrm e}^{2 x}+6 \ln \left (2\right )}{2 \ln \left (2\right ) \left (1+x \right )}\) | \(58\) |
default | \(\frac {-\frac {8}{1+x}+\frac {6 \ln \left (2\right )}{1+x}+4 \ln \left (2\right ) \left (\frac {1}{1+x}+\ln \left (1+x \right )\right )+2 \ln \left (2\right ) \left (x -\frac {1}{1+x}-2 \ln \left (1+x \right )\right )+2 \,{\mathrm e}^{2 x}+2 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{2 x}}{1+x}-2 \,{\mathrm e}^{-2} \operatorname {Ei}_{1}\left (-2-2 x \right )\right )+8 \ln \left (2\right ) \left (\frac {{\mathrm e}^{2 x}}{1+x}+{\mathrm e}^{-2} \operatorname {Ei}_{1}\left (-2-2 x \right )\right )+5 \ln \left (2\right ) {\mathrm e}^{2 x}-\frac {10 \ln \left (2\right ) {\mathrm e}^{2 x}}{1+x}+4 \ln \left (2\right ) \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {5 \,{\mathrm e}^{2 x}}{4}+\frac {{\mathrm e}^{2 x}}{1+x}-{\mathrm e}^{-2} \operatorname {Ei}_{1}\left (-2-2 x \right )\right )}{2 \ln \left (2\right )}\) | \(175\) |
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Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {{\left ({\left (x^{2} + x\right )} \log \left (2\right ) + x + 1\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + x + 4\right )} \log \left (2\right ) - 4}{{\left (x + 1\right )} \log \left (2\right )} \]
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Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=x + \frac {\left (x \log {\left (2 \right )} + 1\right ) e^{2 x}}{\log {\left (2 \right )}} + \frac {-4 + 4 \log {\left (2 \right )}}{x \log {\left (2 \right )} + \log {\left (2 \right )}} \]
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\[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\int { \frac {{\left (2 \, x^{2} + {\left (2 \, x^{3} + 5 \, x^{2} + 4 \, x + 1\right )} \log \left (2\right ) + 4 \, x + 2\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 2 \, x - 3\right )} \log \left (2\right ) + 4}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (2\right )} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (2\right ) + x^{2} \log \left (2\right ) + x e^{\left (2 \, x\right )} \log \left (2\right ) + x e^{\left (2 \, x\right )} + x \log \left (2\right ) + e^{\left (2 \, x\right )} + 4 \, \log \left (2\right ) - 4}{{\left (x + 1\right )} \log \left (2\right )} \]
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Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {{\mathrm {e}}^{2\,x}+x\,\ln \left (2\right )+x\,{\mathrm {e}}^{2\,x}\,\ln \left (2\right )}{\ln \left (2\right )}+\frac {\ln \left (2\right )+\ln \left (8\right )-4}{\ln \left (2\right )\,\left (x+1\right )} \]
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