Integrand size = 244, antiderivative size = 32 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\left (\frac {x}{8}+e^5 \left (-4+e^{\left (1+x^2\right )^2}-\frac {\log (3)}{1+x}\right )\right )^2 \]
[Out]
\[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x}{32}+8 e^{12+4 x^2+2 x^4} x \left (1+x^2\right )+\frac {1}{4} e^{6+2 x^2+x^4} \left (1+2 x^2\right )^2-\frac {2 e^{10} \log (3) (4+4 x+\log (3))}{(1+x)^3}-\frac {e^5 \left (4+8 x+4 x^2+\log (3)\right )}{4 (1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (16 x^5-\log (3)+4 x (4+\log (3))+4 x^2 (8+\log (3))+4 x^3 (8+\log (3))+4 x^4 (8+\log (3))\right )}{(1+x)^2}\right ) \, dx \\ & = \frac {x^2}{64}+\frac {1}{4} \int e^{6+2 x^2+x^4} \left (1+2 x^2\right )^2 \, dx-2 \int \frac {e^{11+2 x^2+x^4} \left (16 x^5-\log (3)+4 x (4+\log (3))+4 x^2 (8+\log (3))+4 x^3 (8+\log (3))+4 x^4 (8+\log (3))\right )}{(1+x)^2} \, dx+8 \int e^{12+4 x^2+2 x^4} x \left (1+x^2\right ) \, dx-\frac {1}{4} e^5 \int \frac {4+8 x+4 x^2+\log (3)}{(1+x)^2} \, dx-\left (2 e^{10} \log (3)\right ) \int \frac {4+4 x+\log (3)}{(1+x)^3} \, dx \\ & = e^{12+4 x^2+2 x^4}+\frac {x^2}{64}+\frac {e^{10} (4+4 x+\log (3))^2}{(1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (4 x^5+x (4+\log (3))+x^2 (8+\log (3))+x^3 (8+\log (3))+x^4 (8+\log (3))\right )}{(1+x)^2 \left (x+x^3\right )}+\frac {1}{4} \int \left (e^{6+2 x^2+x^4}+4 e^{6+2 x^2+x^4} x^2+4 e^{6+2 x^2+x^4} x^4\right ) \, dx-\frac {1}{4} e^5 \int \left (4+\frac {\log (3)}{(1+x)^2}\right ) \, dx \\ & = e^{12+4 x^2+2 x^4}-e^5 x+\frac {x^2}{64}+\frac {e^5 \log (3)}{4 (1+x)}+\frac {e^{10} (4+4 x+\log (3))^2}{(1+x)^2}-\frac {2 e^{11+2 x^2+x^4} \left (4 x^5+x (4+\log (3))+x^2 (8+\log (3))+x^3 (8+\log (3))+x^4 (8+\log (3))\right )}{(1+x)^2 \left (x+x^3\right )}+\frac {1}{4} \int e^{6+2 x^2+x^4} \, dx+\int e^{6+2 x^2+x^4} x^2 \, dx+\int e^{6+2 x^2+x^4} x^4 \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(109\) vs. \(2(32)=64\).
Time = 0.99 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.41 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=e^{2 \left (6+2 x^2+x^4\right )}+\frac {1}{4} e^{6+2 x^2+x^4} x+\frac {x^2}{64}-\frac {e^5 \left (4+8 x+4 x^2-\log (3)\right )}{4 (1+x)}-\frac {2 e^{11+2 x^2+x^4} (4+4 x+\log (3))}{1+x}+\frac {e^{10} \log (3) (8+8 x+\log (3))}{(1+x)^2} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(28)=56\).
Time = 0.60 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.56
method | result | size |
risch | \(\frac {x^{2}}{64}-x \,{\mathrm e}^{5}+\frac {\frac {\left (8 \,{\mathrm e}^{5} \ln \left (3\right )+256 \,{\mathrm e}^{10} \ln \left (3\right )\right ) x}{32}+{\mathrm e}^{10} \ln \left (3\right )^{2}+8 \,{\mathrm e}^{10} \ln \left (3\right )+\frac {{\mathrm e}^{5} \ln \left (3\right )}{4}}{x^{2}+2 x +1}+{\mathrm e}^{2 x^{4}+4 x^{2}+12}-\frac {\left (8 \,{\mathrm e}^{5} \ln \left (3\right )+32 x \,{\mathrm e}^{5}-x^{2}+32 \,{\mathrm e}^{5}-x \right ) {\mathrm e}^{x^{4}+2 x^{2}+6}}{4 \left (1+x \right )}\) | \(114\) |
parts | \(\frac {x^{2}}{64}-x \,{\mathrm e}^{5}+\frac {{\mathrm e}^{10} \ln \left (3\right )^{2}}{\left (1+x \right )^{2}}+\frac {\ln \left (3\right ) \left ({\mathrm e}^{5}+32 \,{\mathrm e}^{10}\right )}{4+4 x}+\frac {\left (-2 \,{\mathrm e}^{10} \ln \left (3\right )-8 \,{\mathrm e}^{10}\right ) {\mathrm e}^{x^{4}+2 x^{2}+1}+\left (\frac {{\mathrm e}^{5}}{4}-8 \,{\mathrm e}^{10}\right ) x \,{\mathrm e}^{x^{4}+2 x^{2}+1}+\frac {{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{5} x^{2}}{4}}{1+x}+{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2}\) | \(134\) |
norman | \(\frac {\left (\frac {1}{32}-{\mathrm e}^{5}\right ) x^{3}+\left (-2 \,{\mathrm e}^{10} \ln \left (3\right )-8 \,{\mathrm e}^{10}\right ) {\mathrm e}^{x^{4}+2 x^{2}+1}+\left (-\frac {1}{32}+3 \,{\mathrm e}^{5}+8 \,{\mathrm e}^{10} \ln \left (3\right )+\frac {{\mathrm e}^{5} \ln \left (3\right )}{4}\right ) x +{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2}+\left (\frac {{\mathrm e}^{5}}{2}-8 \,{\mathrm e}^{10}\right ) x^{2} {\mathrm e}^{x^{4}+2 x^{2}+1}+\left (-2 \,{\mathrm e}^{10} \ln \left (3\right )-16 \,{\mathrm e}^{10}+\frac {{\mathrm e}^{5}}{4}\right ) x \,{\mathrm e}^{x^{4}+2 x^{2}+1}+{\mathrm e}^{2 x^{4}+4 x^{2}+2} {\mathrm e}^{10} x^{2}+\frac {x^{4}}{64}+\frac {{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{5} x^{3}}{4}+2 \,{\mathrm e}^{2 x^{4}+4 x^{2}+2} {\mathrm e}^{10} x -\frac {1}{64}+2 \,{\mathrm e}^{5}+8 \,{\mathrm e}^{10} \ln \left (3\right )+\frac {{\mathrm e}^{5} \ln \left (3\right )}{4}+{\mathrm e}^{10} \ln \left (3\right )^{2}}{\left (1+x \right )^{2}}\) | \(235\) |
parallelrisch | \(\frac {-1-2 x +16 x \,{\mathrm e}^{5} \ln \left (3\right )-64 x^{3} {\mathrm e}^{5}-128 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{10} \ln \left (3\right ) x +16 \,{\mathrm e}^{5} \ln \left (3\right )+64 \,{\mathrm e}^{10} \ln \left (3\right )^{2}+192 x \,{\mathrm e}^{5}+128 \,{\mathrm e}^{5}+x^{4}+2 x^{3}+16 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{5} x^{3}-128 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{10} \ln \left (3\right )-1024 \,{\mathrm e}^{x^{4}+2 x^{2}+1} x \,{\mathrm e}^{10}+32 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{5} x^{2}+512 \,{\mathrm e}^{10} \ln \left (3\right ) x +16 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{5} x +64 \,{\mathrm e}^{2 x^{4}+4 x^{2}+2} {\mathrm e}^{10} x^{2}+128 \,{\mathrm e}^{2 x^{4}+4 x^{2}+2} {\mathrm e}^{10} x -512 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{10} x^{2}+64 \,{\mathrm e}^{10} {\mathrm e}^{2 x^{4}+4 x^{2}+2}-512 \,{\mathrm e}^{x^{4}+2 x^{2}+1} {\mathrm e}^{10}+512 \,{\mathrm e}^{10} \ln \left (3\right )}{64 x^{2}+128 x +64}\) | \(289\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (32) = 64\).
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 4.25 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\frac {x^{4} + 2 \, x^{3} + 64 \, e^{10} \log \left (3\right )^{2} + x^{2} - 64 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{5} + 64 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 16 \, {\left (8 \, {\left (x + 1\right )} e^{10} \log \left (3\right ) + 32 \, {\left (x^{2} + 2 \, x + 1\right )} e^{10} - {\left (x^{3} + 2 \, x^{2} + x\right )} e^{5}\right )} e^{\left (x^{4} + 2 \, x^{2} + 1\right )} + 16 \, {\left (32 \, {\left (x + 1\right )} e^{10} + {\left (x + 1\right )} e^{5}\right )} \log \left (3\right )}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (24) = 48\).
Time = 0.35 (sec) , antiderivative size = 136, normalized size of antiderivative = 4.25 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\frac {x^{2}}{64} - x e^{5} + \frac {x \left (e^{5} \log {\left (3 \right )} + 32 e^{10} \log {\left (3 \right )}\right ) + e^{5} \log {\left (3 \right )} + 4 e^{10} \log {\left (3 \right )}^{2} + 32 e^{10} \log {\left (3 \right )}}{4 x^{2} + 8 x + 4} + \frac {\left (4 x e^{10} + 4 e^{10}\right ) e^{2 x^{4} + 4 x^{2} + 2} + \left (x^{2} e^{5} - 32 x e^{10} + x e^{5} - 32 e^{10} - 8 e^{10} \log {\left (3 \right )}\right ) e^{x^{4} + 2 x^{2} + 1}}{4 x + 4} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (32) = 64\).
Time = 0.36 (sec) , antiderivative size = 323, normalized size of antiderivative = 10.09 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\frac {1}{64} \, x^{2} - \frac {1}{2} \, {\left (2 \, x - \frac {6 \, x + 5}{x^{2} + 2 \, x + 1} - 6 \, \log \left (x + 1\right )\right )} e^{5} - \frac {3}{2} \, {\left (\frac {4 \, x + 3}{x^{2} + 2 \, x + 1} + 2 \, \log \left (x + 1\right )\right )} e^{5} + \frac {4 \, {\left (2 \, x + 1\right )} e^{10} \log \left (3\right )}{x^{2} + 2 \, x + 1} + \frac {{\left (2 \, x + 1\right )} e^{5} \log \left (3\right )}{8 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {e^{10} \log \left (3\right )^{2}}{x^{2} + 2 \, x + 1} + \frac {3 \, {\left (2 \, x + 1\right )} e^{5}}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {4 \, e^{10} \log \left (3\right )}{x^{2} + 2 \, x + 1} + \frac {e^{5} \log \left (3\right )}{8 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {8 \, x + 7}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {3 \, {\left (6 \, x + 5\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 \, {\left (4 \, x + 3\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {2 \, x + 1}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {4 \, {\left (x e^{12} + e^{12}\right )} e^{\left (2 \, x^{4} + 4 \, x^{2}\right )} + {\left (x^{2} e^{6} - x {\left (32 \, e^{11} - e^{6}\right )} - 8 \, {\left (\log \left (3\right ) + 4\right )} e^{11}\right )} e^{\left (x^{4} + 2 \, x^{2}\right )}}{4 \, {\left (x + 1\right )}} + \frac {e^{5}}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (32) = 64\).
Time = 0.34 (sec) , antiderivative size = 246, normalized size of antiderivative = 7.69 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=\frac {x^{4} - 64 \, x^{3} e^{5} + 16 \, x^{3} e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 2 \, x^{3} - 128 \, x^{2} e^{5} + 64 \, x^{2} e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 512 \, x^{2} e^{\left (x^{4} + 2 \, x^{2} + 11\right )} + 32 \, x^{2} e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 512 \, x e^{10} \log \left (3\right ) + 16 \, x e^{5} \log \left (3\right ) - 128 \, x e^{\left (x^{4} + 2 \, x^{2} + 11\right )} \log \left (3\right ) + 64 \, e^{10} \log \left (3\right )^{2} + x^{2} - 64 \, x e^{5} + 128 \, x e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 1024 \, x e^{\left (x^{4} + 2 \, x^{2} + 11\right )} + 16 \, x e^{\left (x^{4} + 2 \, x^{2} + 6\right )} + 512 \, e^{10} \log \left (3\right ) + 16 \, e^{5} \log \left (3\right ) - 128 \, e^{\left (x^{4} + 2 \, x^{2} + 11\right )} \log \left (3\right ) + 64 \, e^{\left (2 \, x^{4} + 4 \, x^{2} + 12\right )} - 512 \, e^{\left (x^{4} + 2 \, x^{2} + 11\right )}}{64 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
[In]
[Out]
Time = 14.07 (sec) , antiderivative size = 218, normalized size of antiderivative = 6.81 \[ \int \frac {x+3 x^2+3 x^3+x^4+e^5 \left (-32-96 x-96 x^2-32 x^3\right )+e^{12+4 x^2+2 x^4} \left (256 x+768 x^2+1024 x^3+1024 x^4+768 x^5+256 x^6\right )+\left (e^{10} (-256-256 x)+e^5 (-8-8 x)\right ) \log (3)-64 e^{10} \log ^2(3)+e^{1+2 x^2+x^4} \left (e^{10} \left (-1024 x-3072 x^2-4096 x^3-4096 x^4-3072 x^5-1024 x^6\right )+e^5 \left (8+24 x+56 x^2+104 x^3+128 x^4+128 x^5+96 x^6+32 x^7\right )+e^{10} \left (64-192 x-512 x^2-512 x^3-512 x^4-256 x^5\right ) \log (3)\right )}{32+96 x+96 x^2+32 x^3} \, dx=-\frac {x^2\,\left (64\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \left (3\right )+256\,{\mathrm {e}}^{10}\,\ln \left (3\right )+32\,{\mathrm {e}}^{10}\,{\ln \left (3\right )}^2-\frac {1}{2}\right )-64\,x\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}-8\,x^3\,{\mathrm {e}}^{x^4+2\,x^2+6}-32\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}+x^3\,\left (32\,{\mathrm {e}}^5-1\right )+{\mathrm {e}}^{x^4+2\,x^2+11}\,\left (64\,\ln \left (3\right )+256\right )-32\,x^2\,{\mathrm {e}}^{2\,x^4+4\,x^2+12}-\frac {x^4}{2}+x\,\left (32\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \left (3\right )+256\,{\mathrm {e}}^{10}\,\ln \left (3\right )+64\,{\mathrm {e}}^{10}\,{\ln \left (3\right )}^2\right )+8\,x\,{\mathrm {e}}^{x^4+2\,x^2+6}\,\left (64\,{\mathrm {e}}^5+8\,{\mathrm {e}}^5\,\ln \left (3\right )-1\right )+16\,x^2\,{\mathrm {e}}^{x^4+2\,x^2+6}\,\left (16\,{\mathrm {e}}^5-1\right )}{32\,x^2+64\,x+32} \]
[In]
[Out]