Integrand size = 11, antiderivative size = 46 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {x^2}{2}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {75}{2 \left (5-x^2\right )}+\frac {15}{2} \log \left (5-x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 46, 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, 45} \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {x^2}{2}+\frac {75}{2 \left (5-x^2\right )}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {15}{2} \log \left (5-x^2\right ) \]
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Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^3}{(-5+x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (1+\frac {125}{(-5+x)^3}+\frac {75}{(-5+x)^2}+\frac {15}{-5+x}\right ) \, dx,x,x^2\right ) \\ & = \frac {x^2}{2}-\frac {125}{4 \left (5-x^2\right )^2}+\frac {75}{2 \left (5-x^2\right )}+\frac {15}{2} \log \left (5-x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {1}{4} \left (2 x^2-\frac {125}{\left (-5+x^2\right )^2}-\frac {150}{-5+x^2}+30 \log \left (-5+x^2\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65
method | result | size |
norman | \(\frac {-75 x^{2}+\frac {1}{2} x^{6}+\frac {1125}{4}}{\left (x^{2}-5\right )^{2}}+\frac {15 \ln \left (x^{2}-5\right )}{2}\) | \(30\) |
risch | \(\frac {x^{2}}{2}+\frac {-\frac {75 x^{2}}{2}+\frac {625}{4}}{\left (x^{2}-5\right )^{2}}+\frac {15 \ln \left (x^{2}-5\right )}{2}\) | \(30\) |
default | \(\frac {x^{2}}{2}+\frac {15 \ln \left (x^{2}-5\right )}{2}-\frac {125}{4 \left (x^{2}-5\right )^{2}}-\frac {75}{2 \left (x^{2}-5\right )}\) | \(33\) |
meijerg | \(\frac {x^{2} \left (\frac {4}{25} x^{4}-\frac {18}{5} x^{2}+12\right )}{8 \left (-\frac {x^{2}}{5}+1\right )^{2}}+\frac {15 \ln \left (-\frac {x^{2}}{5}+1\right )}{2}\) | \(38\) |
parallelrisch | \(\frac {2 x^{6}+30 \ln \left (x^{2}-5\right ) x^{4}+1125-300 \ln \left (x^{2}-5\right ) x^{2}-300 x^{2}+750 \ln \left (x^{2}-5\right )}{4 \left (x^{2}-5\right )^{2}}\) | \(52\) |
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Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {2 \, x^{6} - 20 \, x^{4} - 100 \, x^{2} + 30 \, {\left (x^{4} - 10 \, x^{2} + 25\right )} \log \left (x^{2} - 5\right ) + 625}{4 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {x^{2}}{2} + \frac {625 - 150 x^{2}}{4 x^{4} - 40 x^{2} + 100} + \frac {15 \log {\left (x^{2} - 5 \right )}}{2} \]
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Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {1}{2} \, x^{2} - \frac {25 \, {\left (6 \, x^{2} - 25\right )}}{4 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {15}{2} \, \log \left (x^{2} - 5\right ) \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {1}{2} \, x^{2} - \frac {5 \, {\left (9 \, x^{4} - 60 \, x^{2} + 100\right )}}{4 \, {\left (x^{2} - 5\right )}^{2}} + \frac {15}{2} \, \log \left ({\left | x^{2} - 5 \right |}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {x^7}{\left (-5+x^2\right )^3} \, dx=\frac {15\,\ln \left (x^2-5\right )}{2}-\frac {\frac {75\,x^2}{2}-\frac {625}{4}}{x^4-10\,x^2+25}+\frac {x^2}{2} \]
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