Integrand size = 20, antiderivative size = 332 \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \]
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Time = 0.31 (sec) , antiderivative size = 332, 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, 1643, 70} \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {(d+e x)^{n+1}}{c^2 e (n+1)} \]
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Rule 70
Rule 1643
Rule 1663
Rubi steps \begin{align*} \text {integral}& = \frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {(d+e x)^n \left (\frac {a^2 \left (c d^2+a e^2 (1+n)\right )}{c^2}+\frac {a^2 d e n x}{c}-2 a \left (d^2+\frac {a e^2}{c}\right ) x^2\right )}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )} \\ & = \frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (-\frac {2 a \left (c d^2+a e^2\right ) (d+e x)^n}{c^2}+\frac {\left (-\frac {a^3 d e n}{c^{3/2}}+\sqrt {-a} \left (\frac {3 a^2 d^2}{c}+\frac {3 a^3 e^2}{c^2}+\frac {a^3 e^2 n}{c^2}\right )\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\frac {a^3 d e n}{c^{3/2}}+\sqrt {-a} \left (\frac {3 a^2 d^2}{c}+\frac {3 a^3 e^2}{c^2}+\frac {a^3 e^2 n}{c^2}\right )\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 {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 c^2 \left (c d^2+a e^2\right )}-\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 c^2 \left (c d^2+a e^2\right )} \\ & = \frac {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+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 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+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 c^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.24 \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n} \left (\frac {4}{e+e n}+\frac {2 a (a e+c d x)}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {4 \sqrt {-a} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {4 \sqrt {-a} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}+\frac {a \left (\frac {\left (c d^2-a e^2 (-1+n)+\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}-\frac {\left (c d^2-a e^2 (-1+n)-\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} \left (c d^2+a e^2\right ) (1+n)}\right )}{4 c^2} \]
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\[\int \frac {x^{4} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]
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\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
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\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {x^4\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]
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