Integrand size = 22, antiderivative size = 171 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2 x^{1+m}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (a d (3-m)+b c (5+m)) x^{1+m}}{8 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (2 a b c d \left (1-m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )+b^2 c^2 \left (3+4 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 d^2 (1+m)} \]
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Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {474, 468, 371} \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {x^{m+1} \left (\frac {(1-m) \left (4 a^2 d^2-(m+1) (b c-a d)^2\right )}{c^2 (m+1)}+4 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{8 c d^2}-\frac {x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{8 c^2 d^2 \left (c+d x^2\right )}+\frac {x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]
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Rule 371
Rule 468
Rule 474
Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^{1+m}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int \frac {x^m \left (-4 a^2 d^2+(b c-a d)^2 (1+m)-4 b^2 c d x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d^2} \\ & = \frac {(b c-a d)^2 x^{1+m}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (a d (3-m)+b c (5+m)) x^{1+m}}{8 c^2 d^2 \left (c+d x^2\right )}+-\frac {\left (-4 b^2 c^2 d (1+m)-d (-1+m) \left (-4 a^2 d^2+(b c-a d)^2 (1+m)\right )\right ) \int \frac {x^m}{c+d x^2} \, dx}{8 c^2 d^3} \\ & = \frac {(b c-a d)^2 x^{1+m}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (a d (3-m)+b c (5+m)) x^{1+m}}{8 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (4 b^2 c^2 (1+m)+(1-m) \left (4 a^2 d^2-(b c-a d)^2 (1+m)\right )\right ) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 d^2 (1+m)} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {x^{1+m} \left (b^2 c^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-(b c-a d) \left (2 b c \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+(-b c+a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )\right )}{c^3 d^2 (1+m)} \]
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\[\int \frac {x^{m} \left (b \,x^{2}+a \right )^{2}}{\left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int \frac {x^{m} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{3}}\, dx \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {x^m \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\int \frac {x^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^3} \,d x \]
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