Integrand size = 28, antiderivative size = 56 \[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\frac {\left (\left (-4+x^2\right )^4\right )^{7/8} \left (\frac {\sqrt {-4+x^2}}{8 x^2}+\frac {1}{16} \arctan \left (\frac {1}{2} \sqrt {-4+x^2}\right )\right )}{\left (-4+x^2\right )^{7/2}} \]
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Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6820, 1973, 272, 44, 65, 212} \[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=-\frac {\sqrt {4-x^2} \text {arctanh}\left (\sqrt {1-\frac {x^2}{4}}\right )}{16 \sqrt [8]{\left (x^2-4\right )^4}}-\frac {4-x^2}{8 x^2 \sqrt [8]{\left (x^2-4\right )^4}} \]
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Rule 44
Rule 65
Rule 212
Rule 272
Rule 1973
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^3 \sqrt [8]{\left (-4+x^2\right )^4}} \, dx \\ & = \frac {\sqrt {1-\frac {x^2}{4}} \int \frac {1}{x^3 \sqrt {1-\frac {x^2}{4}}} \, dx}{\sqrt [8]{\left (-4+x^2\right )^4}} \\ & = \frac {\sqrt {1-\frac {x^2}{4}} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x}{4}} x^2} \, dx,x,x^2\right )}{2 \sqrt [8]{\left (-4+x^2\right )^4}} \\ & = -\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}+\frac {\sqrt {1-\frac {x^2}{4}} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x}{4}} x} \, dx,x,x^2\right )}{16 \sqrt [8]{\left (-4+x^2\right )^4}} \\ & = -\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}-\frac {\sqrt {1-\frac {x^2}{4}} \text {Subst}\left (\int \frac {1}{4-4 x^2} \, dx,x,\sqrt {1-\frac {x^2}{4}}\right )}{2 \sqrt [8]{\left (-4+x^2\right )^4}} \\ & = -\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}-\frac {\sqrt {4-x^2} \text {arctanh}\left (\sqrt {1-\frac {x^2}{4}}\right )}{16 \sqrt [8]{\left (-4+x^2\right )^4}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\frac {-8+2 x^2+x^2 \sqrt {-4+x^2} \arctan \left (\frac {1}{2} \sqrt {-4+x^2}\right )}{16 x^2 \sqrt [8]{\left (-4+x^2\right )^4}} \]
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Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x^{2}-4}{8 x^{2} {\left (\left (x^{2}-4\right )^{4}\right )}^{\frac {1}{8}}}-\frac {\arctan \left (\frac {2}{\sqrt {x^{2}-4}}\right ) \sqrt {x^{2}-4}}{16 {\left (\left (x^{2}-4\right )^{4}\right )}^{\frac {1}{8}}}\) | \(49\) |
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Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\frac {x^{2} \arctan \left (-\frac {1}{2} \, x + \frac {1}{2} \, {\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}}\right ) + {\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}}}{8 \, x^{2}} \]
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\[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\int \frac {1}{x^{3} \sqrt [8]{\left (x - 2\right )^{4} \left (x + 2\right )^{4}}}\, dx \]
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\[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\int { \frac {1}{{\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}} x^{3}} \,d x } \]
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\[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\int { \frac {1}{{\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx=\int \frac {1}{x^3\,{\left (x^8-16\,x^6+96\,x^4-256\,x^2+256\right )}^{1/8}} \,d x \]
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