Integrand size = 16, antiderivative size = 30 \[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{4},\frac {5+m}{4},x^4\right )}{1+m} \]
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Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 371} \[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{4},\frac {m+5}{4},x^4\right )}{m+1} \]
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Rule 28
Rule 371
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{4};\frac {5+m}{4};x^4\right )}{1+m} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{4},1+\frac {1+m}{4},x^4\right )}{1+m} \]
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\[\int \frac {x^{m}}{x^{8}-2 x^{4}+1}d x\]
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\[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} - 2 \, x^{4} + 1} \,d x } \]
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\[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\int \frac {x^{m}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2} \left (x^{2} + 1\right )^{2}}\, dx \]
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\[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} - 2 \, x^{4} + 1} \,d x } \]
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\[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} - 2 \, x^{4} + 1} \,d x } \]
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Timed out. \[ \int \frac {x^m}{1-2 x^4+x^8} \, dx=\int \frac {x^m}{x^8-2\,x^4+1} \,d x \]
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