Integrand size = 19, antiderivative size = 45 \[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=-\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {c x^2}{b}\right )}{b (1-m) x} \]
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Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1156, 1598, 371} \[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=-\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {m-1}{2},\frac {m+1}{2},-\frac {c x^2}{b}\right )}{b (1-m) x} \]
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Rule 371
Rule 1156
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \left (x^{-m} (c x)^m\right ) \text {Subst}\left (\int \frac {x^m}{b x^2+c x^4} \, dx,x,x\right ) \\ & = \left (x^{-m} (c x)^m\right ) \text {Subst}\left (\int \frac {x^{-2+m}}{b+c x^2} \, dx,x,x\right ) \\ & = -\frac {(c x)^m \, _2F_1\left (1,\frac {1}{2} (-1+m);\frac {1+m}{2};-\frac {c x^2}{b}\right )}{b (1-m) x} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93 \[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {c x^2}{b}\right )}{b (-1+m) x} \]
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\[\int \frac {\left (c x \right )^{m}}{c \,x^{4}+b \,x^{2}}d x\]
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\[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\int { \frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}} \,d x } \]
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\[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\int \frac {\left (c x\right )^{m}}{x^{2} \left (b + c x^{2}\right )}\, dx \]
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\[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\int { \frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}} \,d x } \]
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\[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\int { \frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(c x)^m}{b x^2+c x^4} \, dx=\int \frac {{\left (c\,x\right )}^m}{c\,x^4+b\,x^2} \,d x \]
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