Integrand size = 169, antiderivative size = 29 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=e^{-(-16+x)^2 x} \log \left (x+\frac {x}{2 e^x+\frac {2}{x}}\right ) \]
[Out]
Leaf count is larger than twice the leaf count of optimal. \(152\) vs. \(2(29)=58\).
Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 5.24, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {2326} \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=\frac {e^{-x^3+32 x^2-256 x} \left (3 x^4-58 x^3+128 x^2+2 e^{2 x} \left (3 x^5-64 x^4+256 x^3\right )+e^x \left (3 x^5-52 x^4+1024 x^2\right )+512 x\right ) \log \left (\frac {2 e^x x^2+x^2+2 x}{2 \left (e^x x+1\right )}\right )}{\left (3 x^2-64 x+256\right ) \left (2 e^{2 x} x^3+x^2+e^x \left (x^3+4 x^2\right )+2 x\right )} \]
[In]
[Out]
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{-256 x+32 x^2-x^3} \left (512 x+128 x^2-58 x^3+3 x^4+2 e^{2 x} \left (256 x^3-64 x^4+3 x^5\right )+e^x \left (1024 x^2-52 x^4+3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2 \left (1+e^x x\right )}\right )}{\left (256-64 x+3 x^2\right ) \left (2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )\right )} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=e^{-(-16+x)^2 x} \log \left (\frac {x \left (2+x+2 e^x x\right )}{2+2 e^x x}\right ) \]
[In]
[Out]
Time = 103.64 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45
method | result | size |
parallelrisch | \(\ln \left (\frac {2 \,{\mathrm e}^{x} x^{2}+x^{2}+2 x}{2 \,{\mathrm e}^{x} x +2}\right ) {\mathrm e}^{-x^{3}+32 x^{2}-256 x}\) | \(42\) |
risch | \({\mathrm e}^{-\left (x -16\right )^{2} x} \ln \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )+\frac {\left (-i \pi \,\operatorname {csgn}\left (i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x} x +1}\right ) \operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )+i \pi \,\operatorname {csgn}\left (i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x} x +1}\right ) {\operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{2}-i \pi {\operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{3}+i \pi \,\operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right ) {\operatorname {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{2}-i \pi \,\operatorname {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right ) \operatorname {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right ) \operatorname {csgn}\left (i x \right )-i \pi {\operatorname {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{3}+i \pi {\operatorname {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )}^{2} \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x} x +1\right )\right ) {\mathrm e}^{-\left (x -16\right )^{2} x}}{2}\) | \(352\) |
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=e^{\left (-x^{3} + 32 \, x^{2} - 256 \, x\right )} \log \left (\frac {2 \, x^{2} e^{x} + x^{2} + 2 \, x}{2 \, {\left (x e^{x} + 1\right )}}\right ) \]
[In]
[Out]
Timed out. \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=\text {Timed out} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (24) = 48\).
Time = 27.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=-{\left ({\left (\log \left (2\right ) - \log \left (x\right )\right )} e^{\left (-x^{3} + 32 \, x^{2}\right )} - e^{\left (-x^{3} + 32 \, x^{2}\right )} \log \left (2 \, x e^{x} + x + 2\right ) + e^{\left (-x^{3} + 32 \, x^{2}\right )} \log \left (x e^{x} + 1\right )\right )} e^{\left (-256 \, x\right )} \]
[In]
[Out]
none
Time = 0.61 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=e^{\left (-x^{3} + 32 \, x^{2} - 256 \, x\right )} \log \left (\frac {2 \, x^{2} e^{x} + x^{2} + 2 \, x}{2 \, {\left (x e^{x} + 1\right )}}\right ) \]
[In]
[Out]
Time = 14.65 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-256 x+32 x^2-x^3} \left (2+2 x+2 e^{2 x} x^2+e^x \left (4 x+x^2-x^3\right )+\left (-512 x-128 x^2+58 x^3-3 x^4+e^{2 x} \left (-512 x^3+128 x^4-6 x^5\right )+e^x \left (-1024 x^2+52 x^4-3 x^5\right )\right ) \log \left (\frac {2 x+x^2+2 e^x x^2}{2+2 e^x x}\right )\right )}{2 x+x^2+2 e^{2 x} x^3+e^x \left (4 x^2+x^3\right )} \, dx=\ln \left (\frac {2\,x+2\,x^2\,{\mathrm {e}}^x+x^2}{2\,x\,{\mathrm {e}}^x+2}\right )\,{\mathrm {e}}^{-256\,x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{32\,x^2} \]
[In]
[Out]