Integrand size = 47, antiderivative size = 26 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=e^{25}-2 x+e^x \left (\frac {e^{2 x}}{x}+x\right )^2+\log (x) \]
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Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {14, 2225, 45, 2227, 2207, 2228} \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=e^x x^2+\frac {e^{5 x}}{x^2}-2 x+2 e^{3 x}+\log (x) \]
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Rule 14
Rule 45
Rule 2207
Rule 2225
Rule 2227
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \int \left (6 e^{3 x}+\frac {1-2 x}{x}+e^x x (2+x)+\frac {e^{5 x} (-2+5 x)}{x^3}\right ) \, dx \\ & = 6 \int e^{3 x} \, dx+\int \frac {1-2 x}{x} \, dx+\int e^x x (2+x) \, dx+\int \frac {e^{5 x} (-2+5 x)}{x^3} \, dx \\ & = 2 e^{3 x}+\frac {e^{5 x}}{x^2}+\int \left (-2+\frac {1}{x}\right ) \, dx+\int \left (2 e^x x+e^x x^2\right ) \, dx \\ & = 2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+\log (x)+2 \int e^x x \, dx+\int e^x x^2 \, dx \\ & = 2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+2 e^x x+e^x x^2+\log (x)-2 \int e^x \, dx-2 \int e^x x \, dx \\ & = -2 e^x+2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x)+2 \int e^x \, dx \\ & = 2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x) \]
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Time = 0.81 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
default | \(\ln \left (x \right )-2 x +{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{3 x}+\frac {{\mathrm e}^{5 x}}{x^{2}}\) | \(27\) |
risch | \(\ln \left (x \right )-2 x +{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{3 x}+\frac {{\mathrm e}^{5 x}}{x^{2}}\) | \(27\) |
parts | \(\ln \left (x \right )-2 x +{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{3 x}+\frac {{\mathrm e}^{5 x}}{x^{2}}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{x} x^{4}+x^{2} \ln \left (x \right )+2 x^{2} {\mathrm e}^{3 x}-2 x^{3}+{\mathrm e}^{5 x}}{x^{2}}\) | \(36\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=\frac {x^{4} e^{x} - 2 \, x^{3} + 2 \, x^{2} e^{\left (3 \, x\right )} + x^{2} \log \left (x\right ) + e^{\left (5 \, x\right )}}{x^{2}} \]
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Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=- 2 x + \log {\left (x \right )} + \frac {x^{4} e^{x} + 2 x^{2} e^{3 x} + e^{5 x}}{x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx={\left (x^{2} - 2 \, x + 2\right )} e^{x} + 2 \, {\left (x - 1\right )} e^{x} - 2 \, x + 2 \, e^{\left (3 \, x\right )} + 25 \, \Gamma \left (-1, -5 \, x\right ) + 50 \, \Gamma \left (-2, -5 \, x\right ) + \log \left (x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=\frac {x^{4} e^{x} - 2 \, x^{3} + 2 \, x^{2} e^{\left (3 \, x\right )} + x^{2} \log \left (x\right ) + e^{\left (5 \, x\right )}}{x^{2}} \]
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Time = 8.62 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x \left (2 x^4+x^5\right )}{x^3} \, dx=\ln \left (x\right )+\frac {{\mathrm {e}}^{5\,x}+x^4\,{\mathrm {e}}^x+2\,x^2\,{\mathrm {e}}^{3\,x}-2\,x^3}{x^2} \]
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