Integrand size = 42, antiderivative size = 28 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=5 \left (3-\frac {-4+\frac {3}{x}}{5 x^2}\right ) \left (-4+e^x+\frac {4}{x}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {14, 2230, 2225, 2208, 2209} \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=-\frac {12}{x^4}-\frac {3 e^x}{x^3}+\frac {28}{x^3}+\frac {4 e^x}{x^2}-\frac {16}{x^2}+15 e^x+\frac {60}{x} \]
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Rule 14
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4 \left (-12+21 x-8 x^2+15 x^3\right )}{x^5}+\frac {e^x \left (9-11 x+4 x^2+15 x^4\right )}{x^4}\right ) \, dx \\ & = -\left (4 \int \frac {-12+21 x-8 x^2+15 x^3}{x^5} \, dx\right )+\int \frac {e^x \left (9-11 x+4 x^2+15 x^4\right )}{x^4} \, dx \\ & = -\left (4 \int \left (-\frac {12}{x^5}+\frac {21}{x^4}-\frac {8}{x^3}+\frac {15}{x^2}\right ) \, dx\right )+\int \left (15 e^x+\frac {9 e^x}{x^4}-\frac {11 e^x}{x^3}+\frac {4 e^x}{x^2}\right ) \, dx \\ & = -\frac {12}{x^4}+\frac {28}{x^3}-\frac {16}{x^2}+\frac {60}{x}+4 \int \frac {e^x}{x^2} \, dx+9 \int \frac {e^x}{x^4} \, dx-11 \int \frac {e^x}{x^3} \, dx+15 \int e^x \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {11 e^x}{2 x^2}+\frac {60}{x}-\frac {4 e^x}{x}+3 \int \frac {e^x}{x^3} \, dx+4 \int \frac {e^x}{x} \, dx-\frac {11}{2} \int \frac {e^x}{x^2} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x}+\frac {3 e^x}{2 x}+4 \operatorname {ExpIntegralEi}(x)+\frac {3}{2} \int \frac {e^x}{x^2} \, dx-\frac {11}{2} \int \frac {e^x}{x} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x}-\frac {3 \operatorname {ExpIntegralEi}(x)}{2}+\frac {3}{2} \int \frac {e^x}{x} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x} \\ \end{align*}
Time = 0.82 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=e^x \left (15-\frac {3}{x^3}+\frac {4}{x^2}\right )-\frac {12}{x^4}+\frac {28}{x^3}-\frac {16}{x^2}+\frac {60}{x} \]
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Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {60 x^{3}-16 x^{2}+28 x -12}{x^{4}}+\frac {\left (15 x^{3}+4 x -3\right ) {\mathrm e}^{x}}{x^{3}}\) | \(37\) |
norman | \(\frac {-12+28 x -16 x^{2}+60 x^{3}-3 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{x} x^{4}}{x^{4}}\) | \(39\) |
parallelrisch | \(\frac {-12+28 x -16 x^{2}+60 x^{3}-3 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{x} x^{4}}{x^{4}}\) | \(39\) |
default | \(-\frac {12}{x^{4}}+\frac {28}{x^{3}}-\frac {16}{x^{2}}+\frac {60}{x}-\frac {3 \,{\mathrm e}^{x}}{x^{3}}+\frac {4 \,{\mathrm e}^{x}}{x^{2}}+15 \,{\mathrm e}^{x}\) | \(40\) |
parts | \(-\frac {12}{x^{4}}+\frac {28}{x^{3}}-\frac {16}{x^{2}}+\frac {60}{x}-\frac {3 \,{\mathrm e}^{x}}{x^{3}}+\frac {4 \,{\mathrm e}^{x}}{x^{2}}+15 \,{\mathrm e}^{x}\) | \(40\) |
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Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {60 \, x^{3} - 16 \, x^{2} + {\left (15 \, x^{4} + 4 \, x^{2} - 3 \, x\right )} e^{x} + 28 \, x - 12}{x^{4}} \]
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Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {\left (15 x^{3} + 4 x - 3\right ) e^{x}}{x^{3}} - \frac {- 60 x^{3} + 16 x^{2} - 28 x + 12}{x^{4}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {60}{x} - \frac {16}{x^{2}} + \frac {28}{x^{3}} - \frac {12}{x^{4}} + 15 \, e^{x} + 4 \, \Gamma \left (-1, -x\right ) + 11 \, \Gamma \left (-2, -x\right ) + 9 \, \Gamma \left (-3, -x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {15 \, x^{4} e^{x} + 60 \, x^{3} + 4 \, x^{2} e^{x} - 16 \, x^{2} - 3 \, x e^{x} + 28 \, x - 12}{x^{4}} \]
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Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=15\,{\mathrm {e}}^x-\frac {x\,\left (3\,{\mathrm {e}}^x-28\right )-x^2\,\left (4\,{\mathrm {e}}^x-16\right )-60\,x^3+12}{x^4} \]
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