Integrand size = 21, antiderivative size = 288 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {3 d e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c}-\frac {d \left (c d^2-3 a e^2\right ) (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}+\frac {e^3 x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c (1+n)}-\frac {e \left (3 c d^2-a e^2\right ) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n (1+n)} \]
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Time = 0.17 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1451, 1445, 1432, 251, 371} \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=-\frac {e (1-n) x^{n+1} \left (3 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n (n+1)}-\frac {d (1-2 n) x \left (c d^2-3 a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}+\frac {x \left (e x^n \left (3 c d^2-a e^2\right )+d \left (c d^2-3 a e^2\right )\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {3 d e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c}+\frac {e^3 x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c (n+1)} \]
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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^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{c \left (a+c x^{2 n}\right )^2}+\frac {e^2 \left (3 d+e x^n\right )}{c \left (a+c x^{2 n}\right )}\right ) \, dx \\ & = \frac {\int \frac {c d^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{c}+\frac {e^2 \int \frac {3 d+e x^n}{a+c x^{2 n}} \, dx}{c} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {\left (3 d e^2\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{c}+\frac {e^3 \int \frac {x^n}{a+c x^{2 n}} \, dx}{c}-\frac {\int \frac {\left (c d^3-3 a d e^2\right ) (1-2 n)+\left (3 c d^2 e-a e^3\right ) (1-n) x^n}{a+c x^{2 n}} \, dx}{2 a c n} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {3 d e^2 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 )}{a c}+\frac {e^3 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 )}{a c (1+n)}-\frac {\left (d \left (c d^2-3 a e^2\right ) (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{2 a c n}-\frac {\left (e \left (3 c d^2-a e^2\right ) (1-n)\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{2 a c n} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {3 d e^2 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 )}{a c}-\frac {d \left (c d^2-3 a e^2\right ) (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 )}{2 a^2 c n}+\frac {e^3 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 )}{a c (1+n)}-\frac {e \left (3 c d^2-a e^2\right ) (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 )}{2 a^2 c n (1+n)} \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.65 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\frac {x \left (3 a d e^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\frac {a e^3 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+n}+d \left (c d^2-3 a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\frac {e \left (3 c d^2-a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+n}\right )}{a^2 c} \]
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\[\int \frac {\left (d +e \,x^{n}\right )^{3}}{\left (a +c \,x^{2 n}\right )^{2}}d x\]
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\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\int \frac {\left (d + e x^{n}\right )^{3}}{\left (a + c x^{2 n}\right )^{2}}\, dx \]
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\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^2} \, dx=\int \frac {{\left (d+e\,x^n\right )}^3}{{\left (a+c\,x^{2\,n}\right )}^2} \,d x \]
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