Integrand size = 30, antiderivative size = 368 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {2 B c (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 (2+m)}-\frac {2 B c (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 (2+m)} \]
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Time = 0.42 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1676, 1299, 371, 12, 1145} \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {(d x)^{m+1} \left (\frac {2 A c-b C}{\sqrt {b^2-4 a c}}+C\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(d x)^{m+1} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {2 B c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d^2 (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 B c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d^2 (m+2) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )} \]
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Rule 12
Rule 371
Rule 1145
Rule 1299
Rule 1676
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {B (d x)^{1+m}}{a+b x^2+c x^4} \, dx}{d}+\int \frac {(d x)^m \left (A+C x^2\right )}{a+b x^2+c x^4} \, dx \\ & = \frac {1}{2} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {(d x)^m}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx+\frac {1}{2} \left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {(d x)^m}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx+\frac {B \int \frac {(d x)^{1+m}}{a+b x^2+c x^4} \, dx}{d} \\ & = \frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {(B c) \int \frac {(d x)^{1+m}}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx}{\sqrt {b^2-4 a c} d}-\frac {(B c) \int \frac {(d x)^{1+m}}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx}{\sqrt {b^2-4 a c} d} \\ & = \frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 (2+m)}-\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 (2+m)} \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 2.26 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.19 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {(d x)^m \left (A \left (2+3 m+m^2\right ) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+B (2+m) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x+\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}+m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+C \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]\right )}{2 m (1+m) (2+m)} \]
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\[\int \frac {\left (d x \right )^{m} \left (C \,x^{2}+B x +A \right )}{c \,x^{4}+b \,x^{2}+a}d x\]
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\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
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\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {\left (d x\right )^{m} \left (A + B x + C x^{2}\right )}{a + b x^{2} + c x^{4}}\, dx \]
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\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
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\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
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Timed out. \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {{\left (d\,x\right )}^m\,\left (C\,x^2+B\,x+A\right )}{c\,x^4+b\,x^2+a} \,d x \]
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