Integrand size = 18, antiderivative size = 136 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=-\frac {\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-\frac {1}{2},-p,-p,\frac {1}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{x} \]
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Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1155, 524} \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=-\frac {\left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-\frac {1}{2},-p,-p,\frac {1}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{x} \]
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Rule 524
Rule 1155
Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p\right ) \int \frac {\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^p}{x^2} \, dx \\ & = -\frac {\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (-\frac {1}{2};-p,-p;\frac {1}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{x} \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=-\frac {\left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (-\frac {1}{2},-p,-p,\frac {1}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )}{x} \]
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\[\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{x^{2}}d x\]
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\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=\int { \frac {{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=\int \frac {\left (a + b x^{2} + c x^{4}\right )^{p}}{x^{2}}\, dx \]
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\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=\int { \frac {{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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\[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=\int { \frac {{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^p}{x^2} \, dx=\int \frac {{\left (c\,x^4+b\,x^2+a\right )}^p}{x^2} \,d x \]
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