Integrand size = 26, antiderivative size = 27 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\left (4+\frac {-x+\frac {1}{3} \left (\frac {1}{3} \left (-5+e^5\right )+x\right )}{x}\right )^2 \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 192, 37} \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {\left (\left (e^5-5\right )^2-30 \left (5-e^5\right ) x\right )^2}{81 \left (5-e^5\right )^2 x^2} \]
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Rule 12
Rule 37
Rule 192
Rubi steps \begin{align*} \text {integral}& = \frac {1}{81} \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{x^3} \, dx \\ & = \frac {1}{81} \int \frac {-2 \left (5-e^5\right )^2+60 \left (5-e^5\right ) x}{x^3} \, dx \\ & = \frac {\left (\left (-5+e^5\right )^2-30 \left (5-e^5\right ) x\right )^2}{81 \left (5-e^5\right )^2 x^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {\left (-5+e^5\right ) \left (-5+e^5+60 x\right )}{81 x^2} \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {\left ({\mathrm e}^{5}-5\right ) \left ({\mathrm e}^{5}+60 x -5\right )}{81 x^{2}}\) | \(17\) |
risch | \(\frac {\left (60 \,{\mathrm e}^{5}-300\right ) x +{\mathrm e}^{10}-10 \,{\mathrm e}^{5}+25}{81 x^{2}}\) | \(22\) |
parallelrisch | \(\frac {60 x \,{\mathrm e}^{5}+{\mathrm e}^{10}-10 \,{\mathrm e}^{5}-300 x +25}{81 x^{2}}\) | \(24\) |
norman | \(\frac {\left (\frac {20 \,{\mathrm e}^{5}}{27}-\frac {100}{27}\right ) x +\frac {{\mathrm e}^{10}}{81}-\frac {10 \,{\mathrm e}^{5}}{81}+\frac {25}{81}}{x^{2}}\) | \(25\) |
default | \(\frac {\left (2 \,{\mathrm e}^{5}-10\right ) \left (\frac {30}{x}-\frac {5-{\mathrm e}^{5}}{2 x^{2}}\right )}{81}\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {10 \, {\left (6 \, x - 1\right )} e^{5} - 300 \, x + e^{10} + 25}{81 \, x^{2}} \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {x \left (-300 + 60 e^{5}\right ) - 10 e^{5} + 25 + e^{10}}{81 x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {60 \, x {\left (e^{5} - 5\right )} + e^{10} - 10 \, e^{5} + 25}{81 \, x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {60 \, x e^{5} - 300 \, x + e^{10} - 10 \, e^{5} + 25}{81 \, x^{2}} \]
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Time = 13.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {{\mathrm {e}}^{10}-10\,{\mathrm {e}}^5+x\,\left (60\,{\mathrm {e}}^5-300\right )+25}{81\,x^2} \]
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