Integrand size = 21, antiderivative size = 701 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\frac {c x \left (c d^2-a e^2-2 c d e x^n\right )}{4 a \left (c d^2+a e^2\right )^2 n \left (a+c x^{2 n}\right )^2}+\frac {c e^2 x \left (3 c d^2-a e^2-4 c d e x^n\right )}{2 a \left (c d^2+a e^2\right )^3 n \left (a+c x^{2 n}\right )}-\frac {c 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 \left (c d^2+a e^2\right )^2 n^2 \left (a+c x^{2 n}\right )}+\frac {c e^4 \left (5 c d^2-a e^2\right ) 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 \left (c d^2+a e^2\right )^4}+\frac {c \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 \left (c d^2+a e^2\right )^2 n^2}-\frac {c e^2 \left (3 c d^2-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 \left (c d^2+a e^2\right )^3 n}+\frac {6 c e^6 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{\left (c d^2+a e^2\right )^4}-\frac {6 c^2 d e^5 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 \left (c d^2+a e^2\right )^4 (1+n)}-\frac {c^2 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 \left (c d^2+a e^2\right )^2 n^2 (1+n)}+\frac {2 c^2 d e^3 (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 )}{a^2 \left (c d^2+a e^2\right )^3 n (1+n)}+\frac {e^6 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3} \]
[Out]
Time = 0.47 (sec) , antiderivative size = 701, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1451, 251, 1445, 1432, 371} \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=-\frac {c^2 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) \left (a e^2+c d^2\right )^2}+\frac {c (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 n^2 \left (a e^2+c d^2\right )^2}+\frac {2 c^2 d e^3 (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 )}{a^2 n (n+1) \left (a e^2+c d^2\right )^3}-\frac {c e^2 (1-2 n) x \left (3 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 )}{2 a^2 n \left (a e^2+c d^2\right )^3}-\frac {c 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 n^2 \left (a e^2+c d^2\right )^2 \left (a+c x^{2 n}\right )}-\frac {6 c^2 d e^5 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 (n+1) \left (a e^2+c d^2\right )^4}+\frac {c e^2 x \left (-a e^2+3 c d^2-4 c d e x^n\right )}{2 a n \left (a e^2+c d^2\right )^3 \left (a+c x^{2 n}\right )}+\frac {c x \left (-a e^2+c d^2-2 c d e x^n\right )}{4 a n \left (a e^2+c d^2\right )^2 \left (a+c x^{2 n}\right )^2}+\frac {6 c e^6 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{\left (a e^2+c d^2\right )^4}+\frac {e^6 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (a e^2+c d^2\right )^3}+\frac {c e^4 x \left (5 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 )}{a \left (a e^2+c d^2\right )^4} \]
[In]
[Out]
Rule 251
Rule 371
Rule 1432
Rule 1445
Rule 1451
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^6}{\left (c d^2+a e^2\right )^3 \left (d+e x^n\right )^2}+\frac {6 c d e^6}{\left (c d^2+a e^2\right )^4 \left (d+e x^n\right )}-\frac {c \left (-c d^2+a e^2+2 c d e x^n\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^{2 n}\right )^3}-\frac {c e^2 \left (-3 c d^2+a e^2+4 c d e x^n\right )}{\left (c d^2+a e^2\right )^3 \left (a+c x^{2 n}\right )^2}-\frac {c e^4 \left (-5 c d^2+a e^2+6 c d e x^n\right )}{\left (c d^2+a e^2\right )^4 \left (a+c x^{2 n}\right )}\right ) \, dx \\ & = -\frac {\left (c e^4\right ) \int \frac {-5 c d^2+a e^2+6 c d e x^n}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^4}+\frac {\left (6 c d e^6\right ) \int \frac {1}{d+e x^n} \, dx}{\left (c d^2+a e^2\right )^4}-\frac {\left (c e^2\right ) \int \frac {-3 c d^2+a e^2+4 c d e x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac {e^6 \int \frac {1}{\left (d+e x^n\right )^2} \, dx}{\left (c d^2+a e^2\right )^3}-\frac {c \int \frac {-c d^2+a e^2+2 c d e x^n}{\left (a+c x^{2 n}\right )^3} \, dx}{\left (c d^2+a e^2\right )^2} \\ & = \frac {c x \left (c d^2-a e^2-2 c d e x^n\right )}{4 a \left (c d^2+a e^2\right )^2 n \left (a+c x^{2 n}\right )^2}+\frac {c e^2 x \left (3 c d^2-a e^2-4 c d e x^n\right )}{2 a \left (c d^2+a e^2\right )^3 n \left (a+c x^{2 n}\right )}+\frac {6 c e^6 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{\left (c d^2+a e^2\right )^4}+\frac {e^6 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3}-\frac {\left (6 c^2 d e^5\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^4}+\frac {\left (c e^4 \left (5 c d^2-a e^2\right )\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^4}+\frac {\left (c e^2\right ) \int \frac {\left (-3 c d^2+a e^2\right ) (1-2 n)+4 c d e (1-n) x^n}{a+c x^{2 n}} \, dx}{2 a \left (c d^2+a e^2\right )^3 n}+\frac {c \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 \left (c d^2+a e^2\right )^2 n} \\ & = \frac {c x \left (c d^2-a e^2-2 c d e x^n\right )}{4 a \left (c d^2+a e^2\right )^2 n \left (a+c x^{2 n}\right )^2}+\frac {c e^2 x \left (3 c d^2-a e^2-4 c d e x^n\right )}{2 a \left (c d^2+a e^2\right )^3 n \left (a+c x^{2 n}\right )}-\frac {c 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 \left (c d^2+a e^2\right )^2 n^2 \left (a+c x^{2 n}\right )}+\frac {c e^4 \left (5 c d^2-a e^2\right ) 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 \left (c d^2+a e^2\right )^4}+\frac {6 c e^6 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{\left (c d^2+a e^2\right )^4}-\frac {6 c^2 d e^5 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 \left (c d^2+a e^2\right )^4 (1+n)}+\frac {e^6 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3}-\frac {c \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 \left (c d^2+a e^2\right )^2 n^2}-\frac {\left (c e^2 \left (3 c d^2-a e^2\right ) (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{2 a \left (c d^2+a e^2\right )^3 n}+\frac {\left (2 c^2 d e^3 (1-n)\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{a \left (c d^2+a e^2\right )^3 n} \\ & = \frac {c x \left (c d^2-a e^2-2 c d e x^n\right )}{4 a \left (c d^2+a e^2\right )^2 n \left (a+c x^{2 n}\right )^2}+\frac {c e^2 x \left (3 c d^2-a e^2-4 c d e x^n\right )}{2 a \left (c d^2+a e^2\right )^3 n \left (a+c x^{2 n}\right )}-\frac {c 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 \left (c d^2+a e^2\right )^2 n^2 \left (a+c x^{2 n}\right )}+\frac {c e^4 \left (5 c d^2-a e^2\right ) 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 \left (c d^2+a e^2\right )^4}-\frac {c e^2 \left (3 c d^2-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 \left (c d^2+a e^2\right )^3 n}+\frac {6 c e^6 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{\left (c d^2+a e^2\right )^4}-\frac {6 c^2 d e^5 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 \left (c d^2+a e^2\right )^4 (1+n)}+\frac {2 c^2 d e^3 (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 )}{a^2 \left (c d^2+a e^2\right )^3 n (1+n)}+\frac {e^6 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3}+\frac {\left (c \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 \left (c d^2+a e^2\right )^2 n^2}-\frac {\left (c^2 d e (1-3 n) (1-n)\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{4 a^2 \left (c d^2+a e^2\right )^2 n^2} \\ & = \frac {c x \left (c d^2-a e^2-2 c d e x^n\right )}{4 a \left (c d^2+a e^2\right )^2 n \left (a+c x^{2 n}\right )^2}+\frac {c e^2 x \left (3 c d^2-a e^2-4 c d e x^n\right )}{2 a \left (c d^2+a e^2\right )^3 n \left (a+c x^{2 n}\right )}-\frac {c 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 \left (c d^2+a e^2\right )^2 n^2 \left (a+c x^{2 n}\right )}+\frac {c e^4 \left (5 c d^2-a e^2\right ) 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 \left (c d^2+a e^2\right )^4}+\frac {c \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 \left (c d^2+a e^2\right )^2 n^2}-\frac {c e^2 \left (3 c d^2-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 \left (c d^2+a e^2\right )^3 n}+\frac {6 c e^6 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{\left (c d^2+a e^2\right )^4}-\frac {6 c^2 d e^5 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 \left (c d^2+a e^2\right )^4 (1+n)}-\frac {c^2 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 \left (c d^2+a e^2\right )^2 n^2 (1+n)}+\frac {2 c^2 d e^3 (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 )}{a^2 \left (c d^2+a e^2\right )^3 n (1+n)}+\frac {e^6 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 426, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (\frac {c e^4 \left (5 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a}+6 c e^6 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )-\frac {6 c^2 d e^5 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (1+n)}+\frac {c e^2 \left (3 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2}+\frac {e^6 \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}-\frac {4 c^2 d e^3 \left (c d^2+a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 (1+n)}+\frac {c \left (c d^2-a e^2\right ) \left (c d^2+a e^2\right )^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3}-\frac {2 c^2 d e \left (c d^2+a e^2\right )^2 x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3 (1+n)}\right )}{\left (c d^2+a e^2\right )^4} \]
[In]
[Out]
\[\int \frac {1}{\left (d +e \,x^{n}\right )^{2} \left (a +c \,x^{2 n}\right )^{3}}d x\]
[In]
[Out]
\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int \frac {1}{{\left (a+c\,x^{2\,n}\right )}^3\,{\left (d+e\,x^n\right )}^2} \,d x \]
[In]
[Out]