Integrand size = 22, antiderivative size = 120 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^2 x^{1+m}}{d^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m}}{2 c d^2 \left (c+d x^2\right )}-\frac {(b c-a d) (a d (1-m)+b c (3+m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{2 c^2 d^2 (1+m)} \]
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Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {474, 470, 371} \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=-\frac {x^{m+1} (b c-a d) (a d (1-m)+b c (m+3)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{2 c^2 d^2 (m+1)}+\frac {x^{m+1} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {b^2 x^{m+1}}{d^2 (m+1)} \]
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Rule 371
Rule 470
Rule 474
Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^{1+m}}{2 c d^2 \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (-2 a^2 d^2+(b c-a d)^2 (1+m)-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2} \\ & = \frac {b^2 x^{1+m}}{d^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m}}{2 c d^2 \left (c+d x^2\right )}-\frac {((b c-a d) (a d (1-m)+b c (3+m))) \int \frac {x^m}{c+d x^2} \, dx}{2 c d^2} \\ & = \frac {b^2 x^{1+m}}{d^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m}}{2 c d^2 \left (c+d x^2\right )}-\frac {(b c-a d) (a d (1-m)+b c (3+m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^2 (1+m)} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {x^{1+m} \left (b^2 c^2-2 b c (b c-a d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(b c-a d)^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )}{c^2 d^2 (1+m)} \]
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\[\int \frac {x^{m} \left (b \,x^{2}+a \right )^{2}}{\left (d \,x^{2}+c \right )^{2}}d x\]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\int \frac {x^{m} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\int \frac {x^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^2} \,d x \]
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