Integrand size = 48, antiderivative size = 21 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=\left (-\frac {4}{x^4}+x+x^2\right ) \left (x+4 x \left (e^x+x\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(21)=42\).
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29, number of steps used = 22, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2230, 2208, 2209, 2207, 2225} \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=4 x^4+4 e^x x^3+5 x^3-\frac {16 e^x}{x^3}-\frac {4}{x^3}+4 e^x x^2+x^2-\frac {16}{x^2} \]
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Rule 14
Rule 2207
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4 e^x \left (12-4 x+2 x^5+4 x^6+x^7\right )}{x^4}+\frac {12+32 x+2 x^5+15 x^6+16 x^7}{x^4}\right ) \, dx \\ & = 4 \int \frac {e^x \left (12-4 x+2 x^5+4 x^6+x^7\right )}{x^4} \, dx+\int \frac {12+32 x+2 x^5+15 x^6+16 x^7}{x^4} \, dx \\ & = 4 \int \left (\frac {12 e^x}{x^4}-\frac {4 e^x}{x^3}+2 e^x x+4 e^x x^2+e^x x^3\right ) \, dx+\int \left (\frac {12}{x^4}+\frac {32}{x^3}+2 x+15 x^2+16 x^3\right ) \, dx \\ & = -\frac {4}{x^3}-\frac {16}{x^2}+x^2+5 x^3+4 x^4+4 \int e^x x^3 \, dx+8 \int e^x x \, dx-16 \int \frac {e^x}{x^3} \, dx+16 \int e^x x^2 \, dx+48 \int \frac {e^x}{x^4} \, dx \\ & = -\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+\frac {8 e^x}{x^2}+8 e^x x+x^2+16 e^x x^2+5 x^3+4 e^x x^3+4 x^4-8 \int e^x \, dx-8 \int \frac {e^x}{x^2} \, dx-12 \int e^x x^2 \, dx+16 \int \frac {e^x}{x^3} \, dx-32 \int e^x x \, dx \\ & = -8 e^x-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+\frac {8 e^x}{x}-24 e^x x+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4+8 \int \frac {e^x}{x^2} \, dx-8 \int \frac {e^x}{x} \, dx+24 \int e^x x \, dx+32 \int e^x \, dx \\ & = 24 e^x-\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4-8 \text {Ei}(x)+8 \int \frac {e^x}{x} \, dx-24 \int e^x \, dx \\ & = -\frac {4}{x^3}-\frac {16 e^x}{x^3}-\frac {16}{x^2}+x^2+4 e^x x^2+5 x^3+4 e^x x^3+4 x^4 \\ \end{align*}
Time = 1.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=\frac {\left (1+4 e^x+4 x\right ) \left (-4+x^5+x^6\right )}{x^3} \]
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Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.86
method | result | size |
risch | \(4 x^{4}+5 x^{3}+x^{2}+\frac {-16 x -4}{x^{3}}+\frac {4 \left (x^{6}+x^{5}-4\right ) {\mathrm e}^{x}}{x^{3}}\) | \(39\) |
norman | \(\frac {-4+x^{5}-16 x +5 x^{6}+4 x^{7}+4 x^{5} {\mathrm e}^{x}+4 x^{6} {\mathrm e}^{x}-16 \,{\mathrm e}^{x}}{x^{3}}\) | \(41\) |
parallelrisch | \(\frac {-4+x^{5}-16 x +5 x^{6}+4 x^{7}+4 x^{5} {\mathrm e}^{x}+4 x^{6} {\mathrm e}^{x}-16 \,{\mathrm e}^{x}}{x^{3}}\) | \(41\) |
default | \(x^{2}-\frac {4}{x^{3}}-\frac {16}{x^{2}}+5 x^{3}+4 x^{4}-\frac {16 \,{\mathrm e}^{x}}{x^{3}}+4 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(46\) |
parts | \(x^{2}-\frac {4}{x^{3}}-\frac {16}{x^{2}}+5 x^{3}+4 x^{4}-\frac {16 \,{\mathrm e}^{x}}{x^{3}}+4 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(46\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=\frac {4 \, x^{7} + 5 \, x^{6} + x^{5} + 4 \, {\left (x^{6} + x^{5} - 4\right )} e^{x} - 16 \, x - 4}{x^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=4 x^{4} + 5 x^{3} + x^{2} + \frac {- 16 x - 4}{x^{3}} + \frac {\left (4 x^{6} + 4 x^{5} - 16\right ) e^{x}}{x^{3}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.52 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=4 \, x^{4} + 5 \, x^{3} + x^{2} + 4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} + 16 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 8 \, {\left (x - 1\right )} e^{x} - \frac {16}{x^{2}} - \frac {4}{x^{3}} + 16 \, \Gamma \left (-2, -x\right ) + 48 \, \Gamma \left (-3, -x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=\frac {4 \, x^{7} + 4 \, x^{6} e^{x} + 5 \, x^{6} + 4 \, x^{5} e^{x} + x^{5} - 16 \, x - 16 \, e^{x} - 4}{x^{3}} \]
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Time = 11.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {12+32 x+2 x^5+15 x^6+16 x^7+e^x \left (48-16 x+8 x^5+16 x^6+4 x^7\right )}{x^4} \, dx=\frac {\left (x^6+x^5-4\right )\,\left (4\,x+4\,{\mathrm {e}}^x+1\right )}{x^3} \]
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