Integrand size = 150, antiderivative size = 32 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\log \left (\log (4) \left (e^{12 x}-x^2+\frac {5}{\log \left (x-\frac {4+x}{1+x}\right )}\right )\right ) \]
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\[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (4+2 x+x^2\right )-2 \left (6 e^{12 x}-x\right ) \left (-4-4 x+x^2+x^3\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (4+4 x-x^2-x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx \\ & = \int \left (12-\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}\right ) \, dx \\ & = 12 x-\int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx \\ & = 12 x-\int \frac {-5 \left (4+2 x+x^2\right )-60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2-x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx \\ & = 12 x-\int \left (\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{12 (-2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}-\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{3 (1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}+\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{4 (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}\right ) \, dx \\ & = 12 x-\frac {1}{12} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx-\frac {1}{4} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx+\frac {1}{3} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx \\ & = 12 x-\frac {1}{12} \int \frac {-5 \left (4+2 x+x^2\right )-60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2-x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx-\frac {1}{4} \int \frac {5 \left (4+2 x+x^2\right )+60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )-2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx+\frac {1}{3} \int \frac {5 \left (4+2 x+x^2\right )+60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )-2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 1.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=-\log \left (\log \left (\frac {-4+x^2}{1+x}\right )\right )+\log \left (5+e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )-x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right ) \]
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Time = 8.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (\frac {x^{2}-4}{1+x}\right )\right )+\ln \left (\ln \left (\frac {x^{2}-4}{1+x}\right ) x^{2}-{\mathrm e}^{12 x} \ln \left (\frac {x^{2}-4}{1+x}\right )-5\right )\) | \(54\) |
risch | \(\ln \left (-x^{2}+{\mathrm e}^{12 x}\right )+\ln \left (\ln \left (x^{2}-4\right )-\frac {i \left (\pi \,x^{2} \operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )-\pi \,x^{2} \operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2}-\pi \,x^{2} \operatorname {csgn}\left (\frac {i}{1+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2}+\pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{3}-\pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right ) {\mathrm e}^{12 x}+\pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2} {\mathrm e}^{12 x}+\pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2} {\mathrm e}^{12 x}-\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{3} {\mathrm e}^{12 x}-2 i x^{2} \ln \left (1+x \right )+2 i {\mathrm e}^{12 x} \ln \left (1+x \right )-10 i\right )}{2 \left (x^{2}-{\mathrm e}^{12 x}\right )}\right )-\ln \left (\ln \left (x^{2}-4\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )-\pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{1+x}\right )}^{3}-2 i \ln \left (1+x \right )\right )}{2}\right )\) | \(433\) |
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (32) = 64\).
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.06 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\log \left (-x^{2} + e^{\left (12 \, x\right )}\right ) + \log \left (\frac {{\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (\frac {x^{2} - 4}{x + 1}\right ) - 5}{x^{2} - e^{\left (12 \, x\right )}}\right ) - \log \left (\log \left (\frac {x^{2} - 4}{x + 1}\right )\right ) \]
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Time = 0.62 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\log {\left (\frac {- x^{2} \log {\left (\frac {x^{2} - 4}{x + 1} \right )} + 5}{\log {\left (\frac {x^{2} - 4}{x + 1} \right )}} + e^{12 x} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (32) = 64\).
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.03 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\log \left (x + e^{\left (6 \, x\right )}\right ) + \log \left (-x + e^{\left (6 \, x\right )}\right ) + \log \left (\frac {{\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x + 2\right ) - {\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x + 1\right ) + {\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x - 2\right ) - 5}{x^{2} - e^{\left (12 \, x\right )}}\right ) - \log \left (\log \left (x + 2\right ) - \log \left (x + 1\right ) + \log \left (x - 2\right )\right ) \]
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Time = 0.77 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.66 \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\log \left (-x^{2} \log \left (\frac {x^{2} - 4}{x + 1}\right ) + e^{\left (12 \, x\right )} \log \left (\frac {x^{2} - 4}{x + 1}\right ) + 5\right ) - \log \left (\log \left (\frac {x^{2} - 4}{x + 1}\right )\right ) \]
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Timed out. \[ \int \frac {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx=\int \frac {10\,x+{\ln \left (\frac {x^2-4}{x+1}\right )}^2\,\left ({\mathrm {e}}^{12\,x}\,\left (-12\,x^3-12\,x^2+48\,x+48\right )-8\,x-8\,x^2+2\,x^3+2\,x^4\right )+5\,x^2+20}{\left ({\mathrm {e}}^{12\,x}\,\left (-x^3-x^2+4\,x+4\right )-4\,x^2-4\,x^3+x^4+x^5\right )\,{\ln \left (\frac {x^2-4}{x+1}\right )}^2+\left (-5\,x^3-5\,x^2+20\,x+20\right )\,\ln \left (\frac {x^2-4}{x+1}\right )} \,d x \]
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