Integrand size = 11, antiderivative size = 52 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {1}{6 x^6}+\frac {3}{4 x^4}-\frac {3}{x^2}-\frac {1}{4 \left (1+x^2\right )^2}-\frac {2}{1+x^2}-10 \log (x)+5 \log \left (1+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {272, 46} \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {1}{6 x^6}+\frac {3}{4 x^4}-\frac {2}{x^2+1}-\frac {3}{x^2}-\frac {1}{4 \left (x^2+1\right )^2}+5 \log \left (x^2+1\right )-10 \log (x) \]
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Rule 46
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 (1+x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x^4}-\frac {3}{x^3}+\frac {6}{x^2}-\frac {10}{x}+\frac {1}{(1+x)^3}+\frac {4}{(1+x)^2}+\frac {10}{1+x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {3}{4 x^4}-\frac {3}{x^2}-\frac {1}{4 \left (1+x^2\right )^2}-\frac {2}{1+x^2}-10 \log (x)+5 \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {2-5 x^2+20 x^4+90 x^6+60 x^8}{12 x^6 \left (1+x^2\right )^2}-10 \log (x)+5 \log \left (1+x^2\right ) \]
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Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {1}{6 x^{6}}+\frac {3}{4 x^{4}}-\frac {3}{x^{2}}-\frac {1}{4 \left (x^{2}+1\right )^{2}}-\frac {2}{x^{2}+1}-10 \ln \left (x \right )+5 \ln \left (x^{2}+1\right )\) | \(47\) |
norman | \(\frac {-\frac {1}{6}-5 x^{8}-\frac {15}{2} x^{6}+\frac {5}{12} x^{2}-\frac {5}{3} x^{4}}{x^{6} \left (x^{2}+1\right )^{2}}-10 \ln \left (x \right )+5 \ln \left (x^{2}+1\right )\) | \(47\) |
risch | \(\frac {-\frac {1}{6}-5 x^{8}-\frac {15}{2} x^{6}+\frac {5}{12} x^{2}-\frac {5}{3} x^{4}}{x^{6} \left (x^{2}+1\right )^{2}}-10 \ln \left (x \right )+5 \ln \left (x^{2}+1\right )\) | \(47\) |
meijerg | \(-\frac {1}{6 x^{6}}+\frac {3}{4 x^{4}}-\frac {3}{x^{2}}-\frac {9}{4}-10 \ln \left (x \right )+\frac {x^{2} \left (9 x^{2}+10\right )}{4 \left (x^{2}+1\right )^{2}}+5 \ln \left (x^{2}+1\right )\) | \(49\) |
parallelrisch | \(-\frac {120 x^{10} \ln \left (x \right )-60 \ln \left (x^{2}+1\right ) x^{10}+2+240 \ln \left (x \right ) x^{8}-120 \ln \left (x^{2}+1\right ) x^{8}+60 x^{8}+120 \ln \left (x \right ) x^{6}-60 x^{6} \ln \left (x^{2}+1\right )+90 x^{6}+20 x^{4}-5 x^{2}}{12 x^{6} \left (x^{2}+1\right )^{2}}\) | \(89\) |
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Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {60 \, x^{8} + 90 \, x^{6} + 20 \, x^{4} - 5 \, x^{2} - 60 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )} \log \left (x^{2} + 1\right ) + 120 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )} \log \left (x\right ) + 2}{12 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=- 10 \log {\left (x \right )} + 5 \log {\left (x^{2} + 1 \right )} + \frac {- 60 x^{8} - 90 x^{6} - 20 x^{4} + 5 x^{2} - 2}{12 x^{10} + 24 x^{8} + 12 x^{6}} \]
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Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {60 \, x^{8} + 90 \, x^{6} + 20 \, x^{4} - 5 \, x^{2} + 2}{12 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )}} + 5 \, \log \left (x^{2} + 1\right ) - 5 \, \log \left (x^{2}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=-\frac {30 \, x^{4} + 68 \, x^{2} + 39}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {110 \, x^{6} - 36 \, x^{4} + 9 \, x^{2} - 2}{12 \, x^{6}} + 5 \, \log \left (x^{2} + 1\right ) - 5 \, \log \left (x^{2}\right ) \]
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Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx=5\,\ln \left (x^2+1\right )-10\,\ln \left (x\right )-\frac {5\,x^8+\frac {15\,x^6}{2}+\frac {5\,x^4}{3}-\frac {5\,x^2}{12}+\frac {1}{6}}{x^{10}+2\,x^8+x^6} \]
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