Integrand size = 40, antiderivative size = 30 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\log \left (\frac {e^{4+e^{-3+x+\left (4-x+x^2\right ) \log (4)}-4 (4+x)}}{x}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 45, 2325, 12, 2268} \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\frac {2^{2 x^2+7} \log (16) e^{x (1-\log (4))-3}}{\log (4)}-4 x-\log (x) \]
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Rule 12
Rule 14
Rule 45
Rule 2268
Rule 2325
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-1-4 x}{x}-4^{4-x+x^2} e^{-3+x} (-1+\log (4)-x \log (16))\right ) \, dx \\ & = \int \frac {-1-4 x}{x} \, dx-\int 4^{4-x+x^2} e^{-3+x} (-1+\log (4)-x \log (16)) \, dx \\ & = \int \left (-4-\frac {1}{x}\right ) \, dx-\int 256 e^{-3+x (1-\log (4))+x^2 \log (4)} (-1+\log (4)-x \log (16)) \, dx \\ & = -4 x-\log (x)-256 \int e^{-3+x (1-\log (4))+x^2 \log (4)} (-1+\log (4)-x \log (16)) \, dx \\ & = -4 x+\frac {2^{7+2 x^2} e^{-3+x (1-\log (4))} \log (16)}{\log (4)}-\log (x) \\ \end{align*}
\[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx \]
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Time = 1.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83
method | result | size |
norman | \(-4 x +{\mathrm e}^{2 \left (x^{2}-x +4\right ) \ln \left (2\right )+x -3}-\ln \left (x \right )\) | \(25\) |
parallelrisch | \(-4 x +{\mathrm e}^{2 \left (x^{2}-x +4\right ) \ln \left (2\right )+x -3}-\ln \left (x \right )\) | \(25\) |
risch | \(-4 x +2^{2 x^{2}-2 x +8} {\mathrm e}^{-3+x}-\ln \left (x \right )\) | \(26\) |
parts | \({\mathrm e}^{2 x^{2} \ln \left (2\right )+\left (1-2 \ln \left (2\right )\right ) x +8 \ln \left (2\right )-3}-\ln \left (x \right )-4 x\) | \(31\) |
default | \({\mathrm e}^{2 x^{2} \ln \left (2\right )+\left (1-2 \ln \left (2\right )\right ) x +8 \ln \left (2\right )-3}+\frac {i \left (1-2 \ln \left (2\right )\right ) \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (1-2 \ln \left (2\right )\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (1-2 \ln \left (2\right )\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{4 \sqrt {\ln \left (2\right )}}+\frac {i \sqrt {\ln \left (2\right )}\, \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (1-2 \ln \left (2\right )\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (1-2 \ln \left (2\right )\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{2}-\frac {i \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (1-2 \ln \left (2\right )\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (1-2 \ln \left (2\right )\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{4 \sqrt {\ln \left (2\right )}}-\ln \left (x \right )-4 x\) | \(226\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-4 \, x + e^{\left (2 \, {\left (x^{2} - x + 4\right )} \log \left (2\right ) + x - 3\right )} - \log \left (x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=- 4 x + e^{x + \left (2 x^{2} - 2 x + 8\right ) \log {\left (2 \right )} - 3} - \log {\left (x \right )} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.34 (sec) , antiderivative size = 269, normalized size of antiderivative = 8.97 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \left (2\right )} + \frac {\sqrt {2} {\left (2 \, \log \left (2\right ) - 1\right )}}{4 \, \sqrt {-\log \left (2\right )}}\right ) e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )} \log \left (2\right )}{\sqrt {-\log \left (2\right )}} + 64 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\pi } {\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{\log \left (2\right )}}\right ) - 1\right )} {\left (2 \, \log \left (2\right ) - 1\right )}}{\sqrt {-\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{\log \left (2\right )}} \log \left (2\right )^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} e^{\left (\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{8 \, \log \left (2\right )}\right )}}{\sqrt {\log \left (2\right )}}\right )} e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )} \sqrt {\log \left (2\right )} + \frac {64 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \left (2\right )} + \frac {\sqrt {2} {\left (2 \, \log \left (2\right ) - 1\right )}}{4 \, \sqrt {-\log \left (2\right )}}\right ) e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )}}{\sqrt {-\log \left (2\right )}} - 4 \, x - \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-{\left (4 \, x e^{3} + e^{3} \log \left (x\right ) - 256 \, e^{\left (2 \, x^{2} \log \left (2\right ) - 2 \, x \log \left (2\right ) + x\right )}\right )} e^{\left (-3\right )} \]
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Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\frac {256\,2^{2\,x^2}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{2^{2\,x}}-\ln \left (x\right )-4\,x \]
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