Integrand size = 139, antiderivative size = 32 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=\frac {4 x^2}{-x (3+x)+\log \left (e^{7/x} \left (5-x^2\right )^2\right )} \]
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\[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=\int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \left (35-22 x^2+4 x^3+3 x^4-2 x \left (-5+x^2\right ) \log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )}{\left (5-x^2\right ) \left (x (3+x)-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx \\ & = 4 \int \frac {35-22 x^2+4 x^3+3 x^4-2 x \left (-5+x^2\right ) \log \left (e^{7/x} \left (-5+x^2\right )^2\right )}{\left (5-x^2\right ) \left (x (3+x)-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx \\ & = 4 \int \left (\frac {-35-8 x^2-14 x^3+3 x^4+2 x^5}{\left (-5+x^2\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}-\frac {2 x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )}\right ) \, dx \\ & = 4 \int \frac {-35-8 x^2-14 x^3+3 x^4+2 x^5}{\left (-5+x^2\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-8 \int \frac {x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \, dx \\ & = 4 \int \left (\frac {7}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}-\frac {4 x}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}+\frac {3 x^2}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}+\frac {2 x^3}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}-\frac {20 x}{\left (-5+x^2\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}\right ) \, dx-8 \int \frac {x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \, dx \\ & = 8 \int \frac {x^3}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-8 \int \frac {x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \, dx+12 \int \frac {x^2}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-16 \int \frac {x}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx+28 \int \frac {1}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-80 \int \frac {x}{\left (-5+x^2\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx \\ & = 8 \int \frac {x^3}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-8 \int \frac {x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \, dx+12 \int \frac {x^2}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-16 \int \frac {x}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx+28 \int \frac {1}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-80 \int \left (-\frac {1}{2 \left (\sqrt {5}-x\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}+\frac {1}{2 \left (\sqrt {5}+x\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2}\right ) \, dx \\ & = 8 \int \frac {x^3}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-8 \int \frac {x}{3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \, dx+12 \int \frac {x^2}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-16 \int \frac {x}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx+28 \int \frac {1}{\left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx+40 \int \frac {1}{\left (\sqrt {5}-x\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx-40 \int \frac {1}{\left (\sqrt {5}+x\right ) \left (3 x+x^2-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )\right )^2} \, dx \\ \end{align*}
Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=-\frac {4 x^2}{x (3+x)-\log \left (e^{7/x} \left (-5+x^2\right )^2\right )} \]
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Time = 6.90 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(-\frac {4 x^{2}}{x^{2}+3 x -\ln \left (\left (x^{4}-10 x^{2}+25\right ) {\mathrm e}^{\frac {7}{x}}\right )}\) | \(35\) |
risch | \(-\frac {8 x^{2}}{i \pi {\operatorname {csgn}\left (i \left (x^{2}-5\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}-5\right )^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i \left (x^{2}-5\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}-5\right )^{2}\right )}^{2}+i \pi {\operatorname {csgn}\left (i \left (x^{2}-5\right )^{2}\right )}^{3}+i \pi \,\operatorname {csgn}\left (i \left (x^{2}-5\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}} \left (x^{2}-5\right )^{2}\right )-i \pi \,\operatorname {csgn}\left (i \left (x^{2}-5\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}} \left (x^{2}-5\right )^{2}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}} \left (x^{2}-5\right )^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\frac {7}{x}} \left (x^{2}-5\right )^{2}\right )^{3}+2 x^{2}+6 x -4 \ln \left (x^{2}-5\right )-2 \ln \left ({\mathrm e}^{\frac {7}{x}}\right )}\) | \(235\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=-\frac {4 \, x^{2}}{x^{2} + 3 \, x - \log \left ({\left (x^{4} - 10 \, x^{2} + 25\right )} e^{\frac {7}{x}}\right )} \]
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Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=\frac {4 x^{2}}{- x^{2} - 3 x + \log {\left (\left (x^{4} - 10 x^{2} + 25\right ) e^{\frac {7}{x}} \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=-\frac {4 \, x^{3}}{x^{3} + 3 \, x^{2} - 2 \, x \log \left (x^{2} - 5\right ) - 7} \]
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Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=-\frac {4 \, x^{3}}{x^{3} + 3 \, x^{2} - x \log \left (x^{4} - 10 \, x^{2} + 25\right ) - 7} \]
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Timed out. \[ \int \frac {-140+88 x^2-16 x^3-12 x^4+\left (-40 x+8 x^3\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )}{-45 x^2-30 x^3+4 x^4+6 x^5+x^6+\left (30 x+10 x^2-6 x^3-2 x^4\right ) \log \left (e^{7/x} \left (25-10 x^2+x^4\right )\right )+\left (-5+x^2\right ) \log ^2\left (e^{7/x} \left (25-10 x^2+x^4\right )\right )} \, dx=\int -\frac {\ln \left ({\mathrm {e}}^{7/x}\,\left (x^4-10\,x^2+25\right )\right )\,\left (40\,x-8\,x^3\right )-88\,x^2+16\,x^3+12\,x^4+140}{\ln \left ({\mathrm {e}}^{7/x}\,\left (x^4-10\,x^2+25\right )\right )\,\left (-2\,x^4-6\,x^3+10\,x^2+30\,x\right )+{\ln \left ({\mathrm {e}}^{7/x}\,\left (x^4-10\,x^2+25\right )\right )}^2\,\left (x^2-5\right )-45\,x^2-30\,x^3+4\,x^4+6\,x^5+x^6} \,d x \]
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