Integrand size = 35, antiderivative size = 23 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-4-e^{4 \left (-2+3 e^5+x^2\right )} (-4+x)+\log (x) \]
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Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2258, 2235, 2240, 2243} \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-e^{4 x^2-4 \left (2-3 e^5\right )} x+4 e^{4 x^2-4 \left (2-3 e^5\right )}+\log (x) \]
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Rule 14
Rule 2235
Rule 2240
Rule 2243
Rule 2258
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x}-e^{-8+12 e^5+4 x^2} \left (1-32 x+8 x^2\right )\right ) \, dx \\ & = \log (x)-\int e^{-8+12 e^5+4 x^2} \left (1-32 x+8 x^2\right ) \, dx \\ & = \log (x)-\int \left (e^{-8+12 e^5+4 x^2}-32 e^{-8+12 e^5+4 x^2} x+8 e^{-8+12 e^5+4 x^2} x^2\right ) \, dx \\ & = \log (x)-8 \int e^{-8+12 e^5+4 x^2} x^2 \, dx+32 \int e^{-8+12 e^5+4 x^2} x \, dx-\int e^{-8+12 e^5+4 x^2} \, dx \\ & = 4 e^{-4 \left (2-3 e^5\right )+4 x^2}-e^{-4 \left (2-3 e^5\right )+4 x^2} x-\frac {1}{4} e^{-8+12 e^5} \sqrt {\pi } \text {erfi}(2 x)+\log (x)+\int e^{-8+12 e^5+4 x^2} \, dx \\ & = 4 e^{-4 \left (2-3 e^5\right )+4 x^2}-e^{-4 \left (2-3 e^5\right )+4 x^2} x+\log (x) \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\frac {e^{4 \left (-2+3 e^5+x^2\right )} \left (32 x-8 x^2\right )}{8 x}+\log (x) \]
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Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\ln \left (x \right )+\left (-x +4\right ) {\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}\) | \(22\) |
norman | \(-x \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+\ln \left (x \right )\) | \(33\) |
parallelrisch | \(-x \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+\ln \left (x \right )\) | \(33\) |
default | \(\ln \left (x \right )-\frac {{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{4}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{4 x^{2}} {\mathrm e}^{-8}-8 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \left (\frac {x \,{\mathrm e}^{4 x^{2}}}{8}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{32}\right )\) | \(63\) |
parts | \(\ln \left (x \right )-\frac {{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{4}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{4 x^{2}} {\mathrm e}^{-8}-8 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \left (\frac {x \,{\mathrm e}^{4 x^{2}}}{8}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{32}\right )\) | \(63\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-{\left (x - 4\right )} e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\left (4 - x\right ) e^{4 x^{2} - 8 + 12 e^{5}} + \log {\left (x \right )} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-x e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + 4 \, e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-x e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + 4 \, e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]
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Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\ln \left (x\right )+4\,{\mathrm {e}}^{12\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-8}\,{\mathrm {e}}^{4\,x^2}-x\,{\mathrm {e}}^{12\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-8}\,{\mathrm {e}}^{4\,x^2} \]
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