Integrand size = 25, antiderivative size = 25 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=-e^3+x+\frac {(2-x) \left (5-e^x+\log (3)\right )}{x} \]
[Out]
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {14, 2230, 2225, 2208, 2209} \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=x+e^x-\frac {2 e^x}{x}+\frac {2 (5+\log (3))}{x} \]
[In]
[Out]
Rule 14
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^x \left (2-2 x+x^2\right )}{x^2}+\frac {-10+x^2-2 \log (3)}{x^2}\right ) \, dx \\ & = \int \frac {e^x \left (2-2 x+x^2\right )}{x^2} \, dx+\int \frac {-10+x^2-2 \log (3)}{x^2} \, dx \\ & = \int \left (e^x+\frac {2 e^x}{x^2}-\frac {2 e^x}{x}\right ) \, dx+\int \left (1-\frac {2 (5+\log (3))}{x^2}\right ) \, dx \\ & = x+\frac {2 (5+\log (3))}{x}+2 \int \frac {e^x}{x^2} \, dx-2 \int \frac {e^x}{x} \, dx+\int e^x \, dx \\ & = e^x-\frac {2 e^x}{x}+x-2 \operatorname {ExpIntegralEi}(x)+\frac {2 (5+\log (3))}{x}+2 \int \frac {e^x}{x} \, dx \\ & = e^x-\frac {2 e^x}{x}+x+\frac {2 (5+\log (3))}{x} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=\frac {10+e^x (-2+x)+x^2+\log (9)}{x} \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {x^{2}+{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+2 \ln \left (3\right )+10}{x}\) | \(22\) |
parallelrisch | \(\frac {x^{2}+{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+2 \ln \left (3\right )+10}{x}\) | \(22\) |
parts | \(-\frac {2 \,{\mathrm e}^{x}}{x}+{\mathrm e}^{x}+x +\frac {2 \ln \left (3\right )+10}{x}\) | \(22\) |
default | \(x +\frac {10}{x}+\frac {2 \ln \left (3\right )}{x}-\frac {2 \,{\mathrm e}^{x}}{x}+{\mathrm e}^{x}\) | \(24\) |
risch | \(x +\frac {2 \ln \left (3\right )}{x}+\frac {10}{x}+\frac {\left (-2+x \right ) {\mathrm e}^{x}}{x}\) | \(24\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=\frac {x^{2} + {\left (x - 2\right )} e^{x} + 2 \, \log \left (3\right ) + 10}{x} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=x + \frac {\left (x - 2\right ) e^{x}}{x} + \frac {2 \log {\left (3 \right )} + 10}{x} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=x + \frac {2 \, \log \left (3\right )}{x} + \frac {10}{x} - 2 \, {\rm Ei}\left (x\right ) + e^{x} + 2 \, \Gamma \left (-1, -x\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=\frac {x^{2} + x e^{x} - 2 \, e^{x} + 2 \, \log \left (3\right ) + 10}{x} \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {-10+x^2+e^x \left (2-2 x+x^2\right )-2 \log (3)}{x^2} \, dx=x+{\mathrm {e}}^x+\frac {\ln \left (9\right )-2\,{\mathrm {e}}^x+10}{x} \]
[In]
[Out]