Integrand size = 17, antiderivative size = 55 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\frac {3 x}{2 c^2}-\frac {x^3}{2 c \left (b+c x^2\right )}-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}} \]
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Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1598, 294, 327, 211} \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}}-\frac {x^3}{2 c \left (b+c x^2\right )}+\frac {3 x}{2 c^2} \]
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Rule 211
Rule 294
Rule 327
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^4}{\left (b+c x^2\right )^2} \, dx \\ & = -\frac {x^3}{2 c \left (b+c x^2\right )}+\frac {3 \int \frac {x^2}{b+c x^2} \, dx}{2 c} \\ & = \frac {3 x}{2 c^2}-\frac {x^3}{2 c \left (b+c x^2\right )}-\frac {(3 b) \int \frac {1}{b+c x^2} \, dx}{2 c^2} \\ & = \frac {3 x}{2 c^2}-\frac {x^3}{2 c \left (b+c x^2\right )}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\frac {x}{c^2}+\frac {b x}{2 c^2 \left (b+c x^2\right )}-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}} \]
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Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76
method | result | size |
default | \(\frac {x}{c^{2}}-\frac {b \left (-\frac {x}{2 \left (c \,x^{2}+b \right )}+\frac {3 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{c^{2}}\) | \(42\) |
risch | \(\frac {x}{c^{2}}+\frac {b x}{2 \left (c \,x^{2}+b \right ) c^{2}}+\frac {3 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right )}{4 c^{3}}-\frac {3 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right )}{4 c^{3}}\) | \(72\) |
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Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.47 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\left [\frac {4 \, c x^{3} + 3 \, {\left (c x^{2} + b\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 6 \, b x}{4 \, {\left (c^{3} x^{2} + b c^{2}\right )}}, \frac {2 \, c x^{3} - 3 \, {\left (c x^{2} + b\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 3 \, b x}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}}\right ] \]
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Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\frac {b x}{2 b c^{2} + 2 c^{3} x^{2}} + \frac {3 \sqrt {- \frac {b}{c^{5}}} \log {\left (- c^{2} \sqrt {- \frac {b}{c^{5}}} + x \right )}}{4} - \frac {3 \sqrt {- \frac {b}{c^{5}}} \log {\left (c^{2} \sqrt {- \frac {b}{c^{5}}} + x \right )}}{4} + \frac {x}{c^{2}} \]
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Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\frac {b x}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} - \frac {3 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} + \frac {x}{c^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {3 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} + \frac {b x}{2 \, {\left (c x^{2} + b\right )} c^{2}} + \frac {x}{c^{2}} \]
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Time = 12.89 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx=\frac {x}{c^2}+\frac {b\,x}{2\,\left (c^3\,x^2+b\,c^2\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,c^{5/2}} \]
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