Integrand size = 20, antiderivative size = 306 \[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=-\frac {(a e+c d x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (c d^2-\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} c \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (c d^2+\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 \sqrt {-a} c \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \]
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Time = 0.34 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1663, 845, 70} \[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{n+1} \left (-\sqrt {-a} \sqrt {c} d e n+a e^2 (n+1)+c d^2\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} c (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{n+1} \left (\sqrt {-a} \sqrt {c} d e n+a e^2 (n+1)+c d^2\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 \sqrt {-a} c (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{n+1} (a e+c d x)}{2 c \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]
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Rule 70
Rule 845
Rule 1663
Rubi steps \begin{align*} \text {integral}& = -\frac {(a e+c d x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {(d+e x)^n \left (-\frac {a \left (c d^2+a e^2 (1+n)\right )}{c}-a d e n x\right )}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )} \\ & = -\frac {(a e+c d x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (\frac {\left (\frac {a^2 d e n}{\sqrt {c}}-\frac {\sqrt {-a} a \left (c d^2+a e^2 (1+n)\right )}{c}\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (-\frac {a^2 d e n}{\sqrt {c}}-\frac {\sqrt {-a} a \left (c d^2+a e^2 (1+n)\right )}{c}\right ) (d+e x)^n}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a \left (c d^2+a e^2\right )} \\ & = -\frac {(a e+c d x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (c d^2-\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 \sqrt {-a} c \left (c d^2+a e^2\right )}-\frac {\left (c d^2+\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 \sqrt {-a} c \left (c d^2+a e^2\right )} \\ & = -\frac {(a e+c d x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (c d^2-\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} c \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (c d^2+\sqrt {-a} \sqrt {c} d e n+a e^2 (1+n)\right ) (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 \sqrt {-a} c \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.32 \[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n} \left (-\frac {2 (a e+c d x)}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}+\frac {a \left (\frac {\left (c d^2-a e^2 (-1+n)+\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}-\frac {\left (c d^2-a e^2 (-1+n)-\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{(-a)^{3/2} \left (c d^2+a e^2\right ) (1+n)}\right )}{4 c} \]
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\[\int \frac {x^{2} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]
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\[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
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\[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^2 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {x^2\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]
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