Integrand size = 24, antiderivative size = 28 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=\frac {1+e^2-\frac {1}{5} x \left (x+3 \left (-2+2 x^2\right )\right )}{2 x} \]
[Out]
Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14} \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=-\frac {3 x^2}{5}-\frac {x}{10}+\frac {1+e^2}{2 x} \]
[In]
[Out]
Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \frac {1}{10} \int \frac {-5-5 e^2-x^2-12 x^3}{x^2} \, dx \\ & = \frac {1}{10} \int \left (-1-\frac {5 \left (1+e^2\right )}{x^2}-12 x\right ) \, dx \\ & = \frac {1+e^2}{2 x}-\frac {x}{10}-\frac {3 x^2}{5} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=\frac {1}{10} \left (\frac {5 \left (1+e^2\right )}{x}-x-6 x^2\right ) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {3 x^{2}}{5}-\frac {x}{10}-\frac {-5-5 \,{\mathrm e}^{2}}{10 x}\) | \(21\) |
norman | \(\frac {-\frac {x^{2}}{10}-\frac {3 x^{3}}{5}+\frac {{\mathrm e}^{2}}{2}+\frac {1}{2}}{x}\) | \(21\) |
gosper | \(\frac {-6 x^{3}-x^{2}+5 \,{\mathrm e}^{2}+5}{10 x}\) | \(22\) |
risch | \(-\frac {3 x^{2}}{5}-\frac {x}{10}+\frac {{\mathrm e}^{2}}{2 x}+\frac {1}{2 x}\) | \(22\) |
parallelrisch | \(\frac {-6 x^{3}-x^{2}+5 \,{\mathrm e}^{2}+5}{10 x}\) | \(22\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=-\frac {6 \, x^{3} + x^{2} - 5 \, e^{2} - 5}{10 \, x} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=- \frac {3 x^{2}}{5} - \frac {x}{10} - \frac {- 5 e^{2} - 5}{10 x} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=-\frac {3}{5} \, x^{2} - \frac {1}{10} \, x + \frac {e^{2} + 1}{2 \, x} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=-\frac {3}{5} \, x^{2} - \frac {1}{10} \, x + \frac {e^{2} + 1}{2 \, x} \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-5-5 e^2-x^2-12 x^3}{10 x^2} \, dx=\frac {\frac {{\mathrm {e}}^2}{2}+\frac {1}{2}}{x}-\frac {x}{10}-\frac {3\,x^2}{5} \]
[In]
[Out]