Integrand size = 96, antiderivative size = 30 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=-3+\frac {3}{x \left (-e^{-4-x+5 x^2} x+x (3+x)\right )} \]
[Out]
\[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=\int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {3 e^{4+x} \left (-3 e^{4+x} (2+x)-e^{5 x^2} \left (-2+x-10 x^2\right )\right )}{x^3 \left (e^{5 x^2}-e^{4+x} (3+x)\right )^2} \, dx \\ & = 3 \int \frac {e^{4+x} \left (-3 e^{4+x} (2+x)-e^{5 x^2} \left (-2+x-10 x^2\right )\right )}{x^3 \left (e^{5 x^2}-e^{4+x} (3+x)\right )^2} \, dx \\ & = 3 \int \left (-\frac {e^{4+x} \left (2-x+10 x^2\right )}{x^3 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )}+\frac {e^{8+2 x} \left (-4+29 x+10 x^2\right )}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2}\right ) \, dx \\ & = -\left (3 \int \frac {e^{4+x} \left (2-x+10 x^2\right )}{x^3 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )} \, dx\right )+3 \int \frac {e^{8+2 x} \left (-4+29 x+10 x^2\right )}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2} \, dx \\ & = 3 \int \left (\frac {10 e^{8+2 x}}{\left (e^{5 x^2}-3 e^{4+x}-e^{4+x} x\right )^2}-\frac {4 e^{8+2 x}}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2}+\frac {29 e^{8+2 x}}{x \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2}\right ) \, dx-3 \int \left (\frac {2 e^{4+x}}{x^3 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )}-\frac {e^{4+x}}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )}+\frac {10 e^{4+x}}{x \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )}\right ) \, dx \\ & = 3 \int \frac {e^{4+x}}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )} \, dx-6 \int \frac {e^{4+x}}{x^3 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )} \, dx-12 \int \frac {e^{8+2 x}}{x^2 \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2} \, dx+30 \int \frac {e^{8+2 x}}{\left (e^{5 x^2}-3 e^{4+x}-e^{4+x} x\right )^2} \, dx-30 \int \frac {e^{4+x}}{x \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )} \, dx+87 \int \frac {e^{8+2 x}}{x \left (-e^{5 x^2}+3 e^{4+x}+e^{4+x} x\right )^2} \, dx \\ \end{align*}
Time = 5.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=\frac {3 e^{4+x}}{x^2 \left (-e^{5 x^2}+e^{4+x} (3+x)\right )} \]
[In]
[Out]
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37
method | result | size |
norman | \(\frac {3 \,{\mathrm e}^{-5 x^{2}+x +4}}{x^{2} \left ({\mathrm e}^{-5 x^{2}+x +4} x +3 \,{\mathrm e}^{-5 x^{2}+x +4}-1\right )}\) | \(41\) |
parallelrisch | \(\frac {3 \,{\mathrm e}^{-5 x^{2}+x +4}}{x^{2} \left ({\mathrm e}^{-5 x^{2}+x +4} x +3 \,{\mathrm e}^{-5 x^{2}+x +4}-1\right )}\) | \(41\) |
risch | \(\frac {3}{x^{2} \left (3+x \right )}+\frac {3}{x^{2} \left (3+x \right ) \left ({\mathrm e}^{-\left (4+5 x \right ) \left (-1+x \right )} x +3 \,{\mathrm e}^{-\left (4+5 x \right ) \left (-1+x \right )}-1\right )}\) | \(52\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=-\frac {3 \, e^{\left (-5 \, x^{2} + x + 4\right )}}{x^{2} - {\left (x^{3} + 3 \, x^{2}\right )} e^{\left (-5 \, x^{2} + x + 4\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=\frac {3}{- x^{3} - 3 x^{2} + \left (x^{4} + 6 x^{3} + 9 x^{2}\right ) e^{- 5 x^{2} + x + 4}} + \frac {3}{x^{3} + 3 x^{2}} \]
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=-\frac {3 \, e^{\left (x + 4\right )}}{x^{2} e^{\left (5 \, x^{2}\right )} - {\left (x^{3} e^{4} + 3 \, x^{2} e^{4}\right )} e^{x}} \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=\frac {3 \, e^{\left (-5 \, x^{2} + x + 4\right )}}{x^{3} e^{\left (-5 \, x^{2} + x + 4\right )} + 3 \, x^{2} e^{\left (-5 \, x^{2} + x + 4\right )} - x^{2}} \]
[In]
[Out]
Time = 0.62 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {e^{8+2 x-10 x^2} (-18-9 x)+e^{4+x-5 x^2} \left (6-3 x+30 x^2\right )}{x^3+e^{4+x-5 x^2} \left (-6 x^3-2 x^4\right )+e^{8+2 x-10 x^2} \left (9 x^3+6 x^4+x^5\right )} \, dx=\frac {3\,{\mathrm {e}}^4\,{\mathrm {e}}^{-5\,x^2}\,{\mathrm {e}}^x}{3\,x^2\,{\mathrm {e}}^4\,{\mathrm {e}}^{-5\,x^2}\,{\mathrm {e}}^x-x^2+x^3\,{\mathrm {e}}^4\,{\mathrm {e}}^{-5\,x^2}\,{\mathrm {e}}^x} \]
[In]
[Out]