Integrand size = 22, antiderivative size = 160 \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {(d x)^{1+m} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {3}{2},\frac {3}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d (1+m) \sqrt {a+b x^2+c x^4}} \]
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Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1155, 524} \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {(d x)^{m+1} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {3}{2},\frac {3}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d (m+1) \sqrt {a+b x^2+c x^4}} \]
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Rule 524
Rule 1155
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {(d x)^m}{\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2}} \, dx}{a \sqrt {a+b x^2+c x^4}} \\ & = \frac {(d x)^{1+m} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {1+m}{2};\frac {3}{2},\frac {3}{2};\frac {3+m}{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 )}{a d (1+m) \sqrt {a+b x^2+c x^4}} \\ \end{align*}
Time = 10.53 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.38 \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {x (d x)^m \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {3}{2},\frac {3}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )}{\left (-b+\sqrt {b^2-4 a c}\right ) (1+m) \left (a+b x^2+c x^4\right )^{3/2}} \]
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\[\int \frac {\left (d x \right )^{m}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]
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\[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\left (d x\right )^{m}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]
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