Integrand size = 94, antiderivative size = 32 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=e^{\frac {4 (3+x) \left (-(3-2 x)^2+x\right )}{x \left (\frac {1}{x}+x-\log (5)\right )}} \]
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\[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=\int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \left (2+\log ^2(5)\right )} \, dx \\ & = \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (224 x-16 x^4+12 (10-9 \log (5))+32 x^3 \log (5)-4 x^2 (42+\log (5))\right )}{1+x^4-2 x \log (5)-2 x^3 \log (5)+x^2 \left (2+\log ^2(5)\right )} \, dx \\ & = \int \left (-16 e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}}-\frac {4 e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (-34-8 x (7-\log (5))+27 \log (5)+x^2 \left (34+\log (5)-4 \log ^2(5)\right )\right )}{1+x^4-2 x \log (5)-2 x^3 \log (5)+x^2 \left (2+\log ^2(5)\right )}\right ) \, dx \\ & = -\left (4 \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (-34-8 x (7-\log (5))+27 \log (5)+x^2 \left (34+\log (5)-4 \log ^2(5)\right )\right )}{1+x^4-2 x \log (5)-2 x^3 \log (5)+x^2 \left (2+\log ^2(5)\right )} \, dx\right )-16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx \\ & = -\left (4 \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) \left (-34-8 x (7-\log (5))+27 \log (5)+x^2 \left (34+\log (5)-4 \log ^2(5)\right )\right )}{\left (1+x^2-x \log (5)\right )^2} \, dx\right )-16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx \\ & = -\left (4 \int \left (\frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) \left (34+\log (5)-4 \log ^2(5)\right )}{1+x^2-x \log (5)}+\frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) (2-\log (5)) \left (-2 (17+2 \log (5))-x \left (28-7 \log (5)-4 \log ^2(5)\right )\right )}{\left (1+x^2-x \log (5)\right )^2}\right ) \, dx\right )-16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx \\ & = -\left (16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx\right )-(4 (2-\log (5))) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) \left (-2 (17+2 \log (5))-x \left (28-7 \log (5)-4 \log ^2(5)\right )\right )}{\left (1+x^2-x \log (5)\right )^2} \, dx-\left (4 \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{1+x^2-x \log (5)} \, dx \\ & = -\left (16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx\right )-(4 (2-\log (5))) \int \left (\frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) x \left (-28+7 \log (5)+4 \log ^2(5)\right )}{\left (1+x^2-x \log (5)\right )^2}-\frac {2 \exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) (17+\log (25))}{\left (1+x^2-x \log (5)\right )^2}\right ) \, dx-\left (4 \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \left (\frac {2 i \exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{\sqrt {4-\log ^2(5)} \left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )}+\frac {2 i \exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{\sqrt {4-\log ^2(5)} \left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )}\right ) \, dx \\ & = -\left (16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx\right )+\left (4 (2-\log (5)) \left (28-7 \log (5)-4 \log ^2(5)\right )\right ) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right ) x}{\left (1+x^2-x \log (5)\right )^2} \, dx-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{2 x-\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{-2 x+\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}+(8 (2-\log (5)) (17+\log (25))) \int \frac {\exp \left (-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}\right )}{\left (1+x^2-x \log (5)\right )^2} \, dx \\ & = -\left (16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx\right )+\left (4 (2-\log (5)) \left (28-7 \log (5)-4 \log ^2(5)\right )\right ) \int \left (-\frac {2 e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}} \left (\log (5)-i \sqrt {4-\log ^2(5)}\right )}{\left (4-\log ^2(5)\right ) \left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )^2}+\frac {2 i e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}} \log (5)}{\left (4-\log ^2(5)\right )^{3/2} \left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )}-\frac {2 e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}} \left (\log (5)+i \sqrt {4-\log ^2(5)}\right )}{\left (4-\log ^2(5)\right ) \left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )^2}+\frac {2 i e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}} \log (5)}{\left (4-\log ^2(5)\right )^{3/2} \left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )}\right ) \, dx-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{2 x-\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{-2 x+\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}+(8 (2-\log (5)) (17+\log (25))) \int \left (-\frac {4 e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (4-\log ^2(5)\right ) \left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )^2}+\frac {4 i e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (4-\log ^2(5)\right )^{3/2} \left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )}-\frac {4 e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (4-\log ^2(5)\right ) \left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )^2}+\frac {4 i e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (4-\log ^2(5)\right )^{3/2} \left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )}\right ) \, dx \\ & = -\left (16 \int e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \, dx\right )+\frac {\left (8 i \log (5) \left (28-7 \log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{2 x-\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{(2+\log (5)) \sqrt {4-\log ^2(5)}}+\frac {\left (8 i \log (5) \left (28-7 \log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{-2 x+\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{(2+\log (5)) \sqrt {4-\log ^2(5)}}-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{2 x-\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}-\frac {\left (8 i \left (34+\log (5)-4 \log ^2(5)\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{-2 x+\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{\sqrt {4-\log ^2(5)}}-\frac {\left (8 \left (28-7 \log (5)-4 \log ^2(5)\right ) \left (\log (5)-i \sqrt {4-\log ^2(5)}\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )^2} \, dx}{2+\log (5)}-\frac {\left (8 \left (28-7 \log (5)-4 \log ^2(5)\right ) \left (\log (5)+i \sqrt {4-\log ^2(5)}\right )\right ) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )^2} \, dx}{2+\log (5)}-\frac {(32 (17+\log (25))) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (2 x-\log (5)+i \sqrt {4-\log ^2(5)}\right )^2} \, dx}{2+\log (5)}-\frac {(32 (17+\log (25))) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{\left (-2 x+\log (5)+i \sqrt {4-\log ^2(5)}\right )^2} \, dx}{2+\log (5)}+\frac {(32 i (17+\log (25))) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{2 x-\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{(2+\log (5)) \sqrt {4-\log ^2(5)}}+\frac {(32 i (17+\log (25))) \int \frac {e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}}}{-2 x+\log (5)+i \sqrt {4-\log ^2(5)}} \, dx}{(2+\log (5)) \sqrt {4-\log ^2(5)}} \\ \end{align*}
Time = 1.79 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=e^{-\frac {4 \left (27-30 x-x^2+4 x^3\right )}{1+x^2-x \log (5)}} \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
risch | \({\mathrm e}^{\frac {4 \left (-1+x \right ) \left (4 x -9\right ) \left (3+x \right )}{x \ln \left (5\right )-x^{2}-1}}\) | \(28\) |
gosper | \({\mathrm e}^{\frac {16 x^{3}-4 x^{2}-120 x +108}{x \ln \left (5\right )-x^{2}-1}}\) | \(32\) |
parallelrisch | \({\mathrm e}^{\frac {16 x^{3}-4 x^{2}-120 x +108}{x \ln \left (5\right )-x^{2}-1}}\) | \(32\) |
norman | \(\frac {x \ln \left (5\right ) {\mathrm e}^{\frac {16 x^{3}-4 x^{2}-120 x +108}{x \ln \left (5\right )-x^{2}-1}}-x^{2} {\mathrm e}^{\frac {16 x^{3}-4 x^{2}-120 x +108}{x \ln \left (5\right )-x^{2}-1}}-{\mathrm e}^{\frac {16 x^{3}-4 x^{2}-120 x +108}{x \ln \left (5\right )-x^{2}-1}}}{x \ln \left (5\right )-x^{2}-1}\) | \(117\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=e^{\left (-\frac {4 \, {\left (4 \, x^{3} - x^{2} - 30 \, x + 27\right )}}{x^{2} - x \log \left (5\right ) + 1}\right )} \]
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Time = 0.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=e^{\frac {16 x^{3} - 4 x^{2} - 120 x + 108}{- x^{2} + x \log {\left (5 \right )} - 1}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (31) = 62\).
Time = 0.44 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.78 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=\frac {1}{152587890625} \, e^{\left (-\frac {16 \, x \log \left (5\right )^{2}}{x^{2} - x \log \left (5\right ) + 1} - 16 \, x + \frac {4 \, x \log \left (5\right )}{x^{2} - x \log \left (5\right ) + 1} + \frac {136 \, x}{x^{2} - x \log \left (5\right ) + 1} + \frac {16 \, \log \left (5\right )}{x^{2} - x \log \left (5\right ) + 1} - \frac {112}{x^{2} - x \log \left (5\right ) + 1} + 4\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (31) = 62\).
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx=e^{\left (-\frac {16 \, x^{3}}{x^{2} - x \log \left (5\right ) + 1} + \frac {4 \, x^{2}}{x^{2} - x \log \left (5\right ) + 1} + \frac {120 \, x}{x^{2} - x \log \left (5\right ) + 1} - \frac {108}{x^{2} - x \log \left (5\right ) + 1}\right )} \]
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Time = 10.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12 \[ \int \frac {e^{\frac {108-120 x-4 x^2+16 x^3}{-1-x^2+x \log (5)}} \left (120+224 x-168 x^2-16 x^4+\left (-108-4 x^2+32 x^3\right ) \log (5)\right )}{1+2 x^2+x^4+\left (-2 x-2 x^3\right ) \log (5)+x^2 \log ^2(5)} \, dx={\mathrm {e}}^{\frac {120\,x}{x^2-\ln \left (5\right )\,x+1}}\,{\mathrm {e}}^{\frac {4\,x^2}{x^2-\ln \left (5\right )\,x+1}}\,{\mathrm {e}}^{-\frac {16\,x^3}{x^2-\ln \left (5\right )\,x+1}}\,{\mathrm {e}}^{-\frac {108}{x^2-\ln \left (5\right )\,x+1}} \]
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