Integrand size = 81, antiderivative size = 31 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \]
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Result contains complex when optimal does not.
Time = 0.87 (sec) , antiderivative size = 172, normalized size of antiderivative = 5.55, number of steps used = 64, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {12, 6820, 6857, 6860, 648, 632, 210, 642, 2608, 2513, 815, 2512, 266, 2463, 2441, 2440, 2438} \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=-\frac {1}{8} e^{3-e^{\log ^2(5)}} \arctan \left (\frac {x+1}{2}\right )+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 (-2 x-(2-4 i))}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 (2 x+(2+4 i))}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (x^2+2 x+5\right )+\left (\frac {3}{16}-\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log (x+(1-2 i))+\left (\frac {3}{16}+\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log (x+(1+2 i)) \]
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Rule 12
Rule 210
Rule 266
Rule 632
Rule 642
Rule 648
Rule 815
Rule 2438
Rule 2440
Rule 2441
Rule 2463
Rule 2512
Rule 2513
Rule 2608
Rule 6820
Rule 6857
Rule 6860
Rubi steps \begin{align*} \text {integral}& = e^{3-e^{\log ^2(5)}} \int \frac {10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx \\ & = e^{3-e^{\log ^2(5)}} \int \frac {-2 x \left (5+2 x+x^2\right )+2 (-1+x) (1+x)^2 \log \left (1-x^2\right )}{\left (1-x^2\right ) \left (5+2 x+x^2\right )^2} \, dx \\ & = e^{3-e^{\log ^2(5)}} \int \left (\frac {2 x}{(-1+x) (1+x) \left (5+2 x+x^2\right )}-\frac {2 (1+x) \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}\right ) \, dx \\ & = \left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{(-1+x) (1+x) \left (5+2 x+x^2\right )} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {(1+x) \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx \\ & = \left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {1}{16 (-1+x)}+\frac {1}{8 (1+x)}+\frac {-5-3 x}{16 \left (5+2 x+x^2\right )}\right ) \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}+\frac {x \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2}\right ) \, dx \\ & = \frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {1}{8} e^{3-e^{\log ^2(5)}} \int \frac {-5-3 x}{5+2 x+x^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \frac {x \log \left (1-x^2\right )}{\left (5+2 x+x^2\right )^2} \, dx \\ & = \frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)-\frac {1}{16} \left (3 e^{3-e^{\log ^2(5)}}\right ) \int \frac {2+2 x}{5+2 x+x^2} \, dx-\frac {1}{4} e^{3-e^{\log ^2(5)}} \int \frac {1}{5+2 x+x^2} \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\left (\frac {1}{4}-\frac {i}{2}\right ) \log \left (1-x^2\right )}{((-2+4 i)-2 x)^2}-\frac {i \log \left (1-x^2\right )}{16 ((-2+4 i)-2 x)}+\frac {\left (\frac {1}{4}+\frac {i}{2}\right ) \log \left (1-x^2\right )}{((2+4 i)+2 x)^2}-\frac {i \log \left (1-x^2\right )}{16 ((2+4 i)+2 x)}\right ) \, dx-\left (2 e^{3-e^{\log ^2(5)}}\right ) \int \left (-\frac {\log \left (1-x^2\right )}{4 ((-2+4 i)-2 x)^2}+\frac {i \log \left (1-x^2\right )}{16 ((-2+4 i)-2 x)}-\frac {\log \left (1-x^2\right )}{4 ((2+4 i)+2 x)^2}+\frac {i \log \left (1-x^2\right )}{16 ((2+4 i)+2 x)}\right ) \, dx \\ & = \frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {\log \left (1-x^2\right )}{((-2+4 i)-2 x)^2} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {\log \left (1-x^2\right )}{((2+4 i)+2 x)^2} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \text {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,2+2 x\right )-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{((-2+4 i)-2 x)^2} \, dx-\left (\left (\frac {1}{2}+i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {\log \left (1-x^2\right )}{((2+4 i)+2 x)^2} \, dx \\ & = -\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )-\left (\left (-\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{((2+4 i)+2 x) \left (1-x^2\right )} \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {x}{((-2+4 i)-2 x) \left (1-x^2\right )} \, dx-\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \frac {x}{((2+4 i)+2 x) \left (1-x^2\right )} \, dx-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \frac {x}{((-2+4 i)-2 x) \left (1-x^2\right )} \, dx \\ & = -\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\frac {1}{8} e^{3-e^{\log ^2(5)}} \log (1-x)+\frac {1}{4} e^{3-e^{\log ^2(5)}} \log (1+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right )-\left (\left (-\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \left (-\frac {\frac {1}{16}-\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}+\frac {\frac {1}{16}-\frac {3 i}{16}}{(1+2 i)+x}\right ) \, dx+\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \left (\frac {\frac {1}{16}+\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}-\frac {\frac {1}{16}+\frac {3 i}{16}}{(1-2 i)+x}\right ) \, dx-\frac {1}{2} e^{3-e^{\log ^2(5)}} \int \left (-\frac {\frac {1}{16}-\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}+\frac {\frac {1}{16}-\frac {3 i}{16}}{(1+2 i)+x}\right ) \, dx-\left (\left (\frac {1}{2}-i\right ) e^{3-e^{\log ^2(5)}}\right ) \int \left (\frac {\frac {1}{16}+\frac {i}{16}}{-1+x}+\frac {i}{8 (1+x)}-\frac {\frac {1}{16}+\frac {3 i}{16}}{(1-2 i)+x}\right ) \, dx \\ & = -\frac {1}{8} e^{3-e^{\log ^2(5)}} \tan ^{-1}\left (\frac {1+x}{2}\right )+\left (\frac {3}{16}-\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log ((1-2 i)+x)+\left (\frac {3}{16}+\frac {i}{16}\right ) e^{3-e^{\log ^2(5)}} \log ((1+2 i)+x)+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((-2+4 i)-2 x)}+\frac {i e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{2 ((2+4 i)+2 x)}-\frac {3}{16} e^{3-e^{\log ^2(5)}} \log \left (5+2 x+x^2\right ) \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \]
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Time = 2.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {\ln \left (-x^{2}+1\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{x^{2}+2 x +5}\) | \(30\) |
parallelrisch | \(\frac {\ln \left (-x^{2}+1\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{x^{2}+2 x +5}\) | \(30\) |
norman | \(\frac {{\mathrm e}^{-{\mathrm e}^{\ln \left (5\right )^{2}}} {\mathrm e}^{3} \ln \left (-x^{2}+1\right )}{x^{2}+2 x +5}\) | \(32\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \]
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Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3} \log {\left (1 - x^{2} \right )}}{x^{2} e^{e^{\log {\left (5 \right )}^{2}}} + 2 x e^{e^{\log {\left (5 \right )}^{2}}} + 5 e^{e^{\log {\left (5 \right )}^{2}}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (29) = 58\).
Time = 0.72 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.71 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=-\frac {1}{32} \, {\left (\frac {4 \, {\left (2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + {\left (x^{2} + 2 \, x - 3\right )} \log \left (-x + 1\right )\right )}}{x^{2} + 2 \, x + 5} + \frac {9 \, x + 35}{x^{2} + 2 \, x + 5} - \frac {2 \, {\left (7 \, x + 5\right )}}{x^{2} + 2 \, x + 5} + \frac {5 \, {\left (x - 5\right )}}{x^{2} + 2 \, x + 5} - 8 \, \log \left (x + 1\right ) - 4 \, \log \left (x - 1\right )\right )} e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \]
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Time = 9.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {\ln \left (1-x^2\right )\,{\mathrm {e}}^{3-{\mathrm {e}}^{{\ln \left (5\right )}^2}}}{x^2+2\,x+5} \]
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