Integrand size = 21, antiderivative size = 272 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}-\frac {x \left (\left (c d^2-a e^2\right ) (1-4 n)+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {\left (c d^2-a e^2\right ) (1-4 n) (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}+\frac {d e (1-3 n) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (1+n)}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 c} \]
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Time = 0.16 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1451, 1445, 1432, 251, 371} \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {(1-4 n) (1-2 n) x \left (c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}+\frac {d e (1-3 n) (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (n+1)}-\frac {x \left ((1-4 n) \left (c d^2-a e^2\right )+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 c}+\frac {x \left (-a e^2+c d^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2} \]
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Rule 251
Rule 371
Rule 1432
Rule 1445
Rule 1451
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c d^2-a e^2+2 c d e x^n}{c \left (a+c x^{2 n}\right )^3}+\frac {e^2}{c \left (a+c x^{2 n}\right )^2}\right ) \, dx \\ & = \frac {\int \frac {c d^2-a e^2+2 c d e x^n}{\left (a+c x^{2 n}\right )^3} \, dx}{c}+\frac {e^2 \int \frac {1}{\left (a+c x^{2 n}\right )^2} \, dx}{c} \\ & = \frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \, _2F_1\left (2,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 c}-\frac {\int \frac {\left (c d^2-a e^2\right ) (1-4 n)+2 c d e (1-3 n) x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{4 a c n} \\ & = \frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}-\frac {x \left (\left (c d^2-a e^2\right ) (1-4 n)+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (2,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 c}+\frac {\int \frac {\left (c d^2-a e^2\right ) (1-4 n) (1-2 n)+2 c d e (1-3 n) (1-n) x^n}{a+c x^{2 n}} \, dx}{8 a^2 c n^2} \\ & = \frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}-\frac {x \left (\left (c d^2-a e^2\right ) (1-4 n)+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (2,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 c}+\frac {\left (\left (c d^2-a e^2\right ) (1-4 n) (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{8 a^2 c n^2}+\frac {(d e (1-3 n) (1-n)) \int \frac {x^n}{a+c x^{2 n}} \, dx}{4 a^2 n^2} \\ & = \frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}-\frac {x \left (\left (c d^2-a e^2\right ) (1-4 n)+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {\left (c d^2-a e^2\right ) (1-4 n) (1-2 n) x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}+\frac {d e (1-3 n) (1-n) x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (1+n)}+\frac {e^2 x \, _2F_1\left (2,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 c} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.50 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (a e^2 (1+n) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\left (c d^2-a e^2\right ) (1+n) \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+2 c d e x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )}{a^3 c (1+n)} \]
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\[\int \frac {\left (d +e \,x^{n}\right )^{2}}{\left (a +c \,x^{2 n}\right )^{3}}d x\]
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\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \]
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\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int \frac {{\left (d+e\,x^n\right )}^2}{{\left (a+c\,x^{2\,n}\right )}^3} \,d x \]
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