Integrand size = 34, antiderivative size = 25 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-x-2 e^5 x^2 \left (1+e^x x^3\right )+\log (3)+\log (x) \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 17, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2227, 2207, 2225} \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-2 e^{x+5} x^5-2 e^5 x^2-x+\log (x) \]
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Rule 14
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \int \left (-2 e^{5+x} x^4 (5+x)+\frac {1-x-4 e^5 x^2}{x}\right ) \, dx \\ & = -\left (2 \int e^{5+x} x^4 (5+x) \, dx\right )+\int \frac {1-x-4 e^5 x^2}{x} \, dx \\ & = -\left (2 \int \left (5 e^{5+x} x^4+e^{5+x} x^5\right ) \, dx\right )+\int \left (-1+\frac {1}{x}-4 e^5 x\right ) \, dx \\ & = -x-2 e^5 x^2+\log (x)-2 \int e^{5+x} x^5 \, dx-10 \int e^{5+x} x^4 \, dx \\ & = -x-2 e^5 x^2-10 e^{5+x} x^4-2 e^{5+x} x^5+\log (x)+10 \int e^{5+x} x^4 \, dx+40 \int e^{5+x} x^3 \, dx \\ & = -x-2 e^5 x^2+40 e^{5+x} x^3-2 e^{5+x} x^5+\log (x)-40 \int e^{5+x} x^3 \, dx-120 \int e^{5+x} x^2 \, dx \\ & = -x-2 e^5 x^2-120 e^{5+x} x^2-2 e^{5+x} x^5+\log (x)+120 \int e^{5+x} x^2 \, dx+240 \int e^{5+x} x \, dx \\ & = -x+240 e^{5+x} x-2 e^5 x^2-2 e^{5+x} x^5+\log (x)-240 \int e^{5+x} \, dx-240 \int e^{5+x} x \, dx \\ & = -240 e^{5+x}-x-2 e^5 x^2-2 e^{5+x} x^5+\log (x)+240 \int e^{5+x} \, dx \\ & = -x-2 e^5 x^2-2 e^{5+x} x^5+\log (x) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-x \left (1+2 e^5 x+2 e^{5+x} x^4\right )+\log (x) \]
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Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
norman | \(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5} {\mathrm e}^{x}+\ln \left (x \right )\) | \(23\) |
risch | \(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5+x}+\ln \left (x \right )\) | \(23\) |
parallelrisch | \(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5} {\mathrm e}^{x}+\ln \left (x \right )\) | \(23\) |
parts | \(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5} {\mathrm e}^{x}+\ln \left (x \right )\) | \(23\) |
default | \(\ln \left (x \right )-x -2 x^{2} {\mathrm e}^{5}-10 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{3}+12 \,{\mathrm e}^{x} x^{2}-24 \,{\mathrm e}^{x} x +24 \,{\mathrm e}^{x}\right )-2 \,{\mathrm e}^{5} \left (x^{5} {\mathrm e}^{x}-5 \,{\mathrm e}^{x} x^{4}+20 \,{\mathrm e}^{x} x^{3}-60 \,{\mathrm e}^{x} x^{2}+120 \,{\mathrm e}^{x} x -120 \,{\mathrm e}^{x}\right )\) | \(89\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-2 \, x^{5} e^{\left (x + 5\right )} - 2 \, x^{2} e^{5} - x + \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=- 2 x^{5} e^{5} e^{x} - 2 x^{2} e^{5} - x + \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (23) = 46\).
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.52 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-2 \, x^{2} e^{5} - 2 \, {\left (x^{5} e^{5} - 5 \, x^{4} e^{5} + 20 \, x^{3} e^{5} - 60 \, x^{2} e^{5} + 120 \, x e^{5} - 120 \, e^{5}\right )} e^{x} - 10 \, {\left (x^{4} e^{5} - 4 \, x^{3} e^{5} + 12 \, x^{2} e^{5} - 24 \, x e^{5} + 24 \, e^{5}\right )} e^{x} - x + \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=-2 \, x^{5} e^{\left (x + 5\right )} - 2 \, x^{2} e^{5} - x + \log \left (x\right ) \]
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Time = 8.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1-x-4 e^5 x^2+e^{5+x} \left (-10 x^5-2 x^6\right )}{x} \, dx=\ln \left (x\right )-x-2\,x^5\,{\mathrm {e}}^{x+5}-2\,x^2\,{\mathrm {e}}^5 \]
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