Integrand size = 280, antiderivative size = 32 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {5 x}{3 \left (-e^x+\frac {-1+x}{x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}\right )} \]
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\[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{3 \left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx \\ & = \frac {5}{3} \int \frac {x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx \\ & = \frac {5}{3} \int \left (\frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx \\ & = \frac {5}{3} \int \frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx \\ & = \frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4+\left (-9-9 x+8 x^2-x^3+x^4\right ) \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {5}{3} \int \left (-\frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx \\ & = -\left (\frac {5}{3} \int \frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx\right )+\frac {5}{3} \int \frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \left (\frac {63 x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {27 x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {40 x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {5 x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {5 x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {8 x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx \\ & = -\left (\frac {5}{3} \int \frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\right )+\frac {5}{3} \int \frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {5}{3} \int \left (-\frac {5 x}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {5}{3} \int \left (-\frac {5 x^2}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {25}{3} \int \frac {x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {25}{3} \int \frac {x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {40}{3} \int \frac {x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+45 \int \frac {x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {200}{3} \int \frac {x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+105 \int \frac {x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{3 \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(197\) vs. \(2(29)=58\).
Time = 2.89 (sec) , antiderivative size = 198, normalized size of antiderivative = 6.19
method | result | size |
parallelrisch | \(\frac {-2025+2025 x +10125 \,{\mathrm e}^{x} x -810 x \ln \left (x \right )+405 \ln \left (\frac {x^{2}+9}{x^{2}}\right )-405 \ln \left (x^{2}+9\right )-4050 x \,{\mathrm e}^{x} \ln \left (x \right )+810 \ln \left (x \right )+450 x^{2}-4050 x \,{\mathrm e}^{x} \ln \left (\frac {x^{2}+9}{x^{2}}\right )+405 x \,{\mathrm e}^{x} \ln \left (\frac {x^{2}+9}{x^{2}}\right )^{2}-90 \ln \left (\frac {x^{2}+9}{x^{2}}\right ) x^{2}-405 \ln \left (\frac {x^{2}+9}{x^{2}}\right ) x +2025 x \,{\mathrm e}^{x} \ln \left (x^{2}+9\right )+810 \ln \left (x \right ) x \,{\mathrm e}^{x} \ln \left (\frac {x^{2}+9}{x^{2}}\right )-405 \ln \left (x^{2}+9\right ) x \,{\mathrm e}^{x} \ln \left (\frac {x^{2}+9}{x^{2}}\right )+405 \ln \left (x^{2}+9\right ) x}{54 x \,{\mathrm e}^{x} \ln \left (\frac {x^{2}+9}{x^{2}}\right )-270 \,{\mathrm e}^{x} x -54 x +54}\) | \(198\) |
risch | \(-\frac {5 x \,{\mathrm e}^{-x}}{3}+\frac {10 i x \left (-1+x \right ) {\mathrm e}^{-x}}{3 \left (\pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}-2 \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\pi x \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}-\pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+9\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right ) {\mathrm e}^{x}+\pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{2} {\mathrm e}^{x}+\pi x \,\operatorname {csgn}\left (i \left (x^{2}+9\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{2} {\mathrm e}^{x}-\pi x {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{3} {\mathrm e}^{x}+4 i x \,{\mathrm e}^{x} \ln \left (x \right )-2 i x \,{\mathrm e}^{x} \ln \left (x^{2}+9\right )+10 i x \,{\mathrm e}^{x}+2 i x -2 i\right )}\) | \(216\) |
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Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \]
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Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {- 5 x^{2} \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} + 25 x^{2}}{- 3 x + \left (3 x \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} - 15 x\right ) e^{x} + 3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (29) = 58\).
Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.16 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} e^{x} \log \left (x^{2} + 9\right ) - {\left (2 \, x^{2} \log \left (x\right ) + 5 \, x^{2}\right )} e^{x}\right )}}{3 \, {\left (x e^{\left (2 \, x\right )} \log \left (x^{2} + 9\right ) - {\left (2 \, x \log \left (x\right ) + 5 \, x\right )} e^{\left (2 \, x\right )} - {\left (x - 1\right )} e^{x}\right )}} \]
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Time = 0.63 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \]
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Time = 13.66 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {5\,x^2\,\left (\ln \left (\frac {x^2+9}{x^2}\right )-5\right )}{3\,\left (x+5\,x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\ln \left (\frac {x^2+9}{x^2}\right )-1\right )} \]
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