Integrand size = 68, antiderivative size = 30 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2 e^{-2-2 x^2}}{x^2 \left (x+5 e^{6-x^2} x\right )} \]
[Out]
\[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )}{e^2 \left (5 e^6+e^{x^2}\right )^2 x^4} \, dx \\ & = \frac {\int \frac {-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )}{\left (5 e^6+e^{x^2}\right )^2 x^4} \, dx}{e^2} \\ & = \frac {\int \left (-\frac {4}{\left (5 e^6+e^{x^2}\right )^2 x^2}-\frac {2 e^{-6-x^2} \left (3+2 x^2\right )}{5 x^4}+\frac {2 \left (3+2 x^2\right )}{5 e^6 \left (5 e^6+e^{x^2}\right ) x^4}\right ) \, dx}{e^2} \\ & = \frac {2 \int \frac {3+2 x^2}{\left (5 e^6+e^{x^2}\right ) x^4} \, dx}{5 e^8}-\frac {2 \int \frac {e^{-6-x^2} \left (3+2 x^2\right )}{x^4} \, dx}{5 e^2}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2} \\ & = \frac {2 e^{-8-x^2}}{5 x^3}+\frac {2 \int \left (\frac {3}{\left (5 e^6+e^{x^2}\right ) x^4}+\frac {2}{\left (5 e^6+e^{x^2}\right ) x^2}\right ) \, dx}{5 e^8}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2} \\ & = \frac {2 e^{-8-x^2}}{5 x^3}+\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right ) x^2} \, dx}{5 e^8}+\frac {6 \int \frac {1}{\left (5 e^6+e^{x^2}\right ) x^4} \, dx}{5 e^8}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2} \\ \end{align*}
Time = 1.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2 e^{-2-x^2}}{\left (5 e^6+e^{x^2}\right ) x^3} \]
[In]
[Out]
Time = 2.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{-2 x^{2}-2}}{x^{3} \left (1+5 \,{\mathrm e}^{-x^{2}+6}\right )}\) | \(28\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{-2 x^{2}-2}}{x^{3} \left (1+5 \,{\mathrm e}^{-x^{2}+6}\right )}\) | \(28\) |
norman | \(\frac {2 \,{\mathrm e}^{-14} {\mathrm e}^{-2 x^{2}+12}}{x^{3} \left (1+5 \,{\mathrm e}^{-x^{2}+6}\right )}\) | \(34\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2 \, e^{\left (-2 \, x^{2} + 12\right )}}{x^{3} e^{14} + 5 \, x^{3} e^{\left (-x^{2} + 20\right )}} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2}{125 x^{3} e^{14} e^{6 - x^{2}} + 25 x^{3} e^{14}} + \frac {2 e^{6 - x^{2}}}{5 x^{3} e^{14}} - \frac {2}{25 x^{3} e^{14}} \]
[In]
[Out]
\[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\int { -\frac {2 \, {\left (4 \, x^{2} + 5 \, {\left (2 \, x^{2} + 3\right )} e^{\left (-x^{2} + 6\right )} + 3\right )} e^{\left (-2 \, x^{2} - 2\right )}}{10 \, x^{4} e^{\left (-x^{2} + 6\right )} + 25 \, x^{4} e^{\left (-2 \, x^{2} + 12\right )} + x^{4}} \,d x } \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2 \, {\left (x^{2} e^{\left (-x^{2} + 6\right )} + 5 \, x^{2} e^{\left (-2 \, x^{2} + 12\right )}\right )}}{10 \, x^{5} e^{14} + x^{5} e^{\left (x^{2} + 8\right )} + 25 \, x^{5} e^{\left (-x^{2} + 20\right )}} \]
[In]
[Out]
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx=\frac {2\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{-2\,x^2}}{x^3\,{\mathrm {e}}^{14}+5\,x^3\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-x^2}} \]
[In]
[Out]