Integrand size = 32, antiderivative size = 1654 \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {g x^4 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {4+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {4+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {2 b c^2 e (2-n) x^{2+n} \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (2+n)}+\frac {2 b c^2 e (2-n) x^{2+n} \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (2+n)}-\frac {2 b c^2 f (3-n) x^{3+n} \operatorname {Hypergeometric2F1}\left (1,\frac {3+n}{n},2+\frac {3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (3+n)}+\frac {2 b c^2 f (3-n) x^{3+n} \operatorname {Hypergeometric2F1}\left (1,\frac {3+n}{n},2+\frac {3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (3+n)}-\frac {2 b c^2 g (4-n) x^{4+n} \operatorname {Hypergeometric2F1}\left (1,\frac {4+n}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (4+n)}+\frac {2 b c^2 g (4-n) x^{4+n} \operatorname {Hypergeometric2F1}\left (1,\frac {4+n}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (4+n)} \]
[Out]
Time = 1.99 (sec) , antiderivative size = 1654, normalized size of antiderivative = 1.00, number of steps used = 33, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1810, 1359, 1436, 251, 1398, 1574, 1397, 371} \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+2)}+\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+2)}-\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+3)}+\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+3)}-\frac {2 b c^2 g (4-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+4}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+4}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+4)}+\frac {2 b c^2 g (4-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+4}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+4}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+4)}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^4}{2 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^4}{2 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {g \left (b c x^n+b^2-2 a c\right ) x^4}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {f \left (b c x^n+b^2-2 a c\right ) x^3}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {e \left (b c x^n+b^2-2 a c\right ) x^2}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c d \left (-\left ((1-n) b^2\right )-\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c d \left (-\left ((1-n) b^2\right )+\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {d \left (b c x^n+b^2-2 a c\right ) x}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )} \]
[In]
[Out]
Rule 251
Rule 371
Rule 1359
Rule 1397
Rule 1398
Rule 1436
Rule 1574
Rule 1810
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {e x}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {f x^2}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {g x^3}{\left (a+b x^n+c x^{2 n}\right )^2}\right ) \, dx \\ & = d \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx+e \int \frac {x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx+f \int \frac {x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx+g \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx \\ & = \frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {g x^4 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {d \int \frac {b^2-2 a c-\left (b^2-4 a c\right ) n+b c (1-n) x^n}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {e \int \frac {x \left (-4 a c (1-n)+b^2 (2-n)+b c (2-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {f \int \frac {x^2 \left (-2 a c (3-2 n)+b^2 (3-n)+b c (3-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {g \int \frac {x^3 \left (-4 a c (2-n)+b^2 (4-n)+b c (4-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n} \\ & = \frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {g x^4 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {e \int \left (-\frac {b^2 \left (1-\frac {4 a c (-1+n)}{b^2 (-2+n)}\right ) (-2+n) x}{a+b x^n+c x^{2 n}}-\frac {b c (-2+n) x^{1+n}}{a+b x^n+c x^{2 n}}\right ) \, dx}{a \left (b^2-4 a c\right ) n}-\frac {f \int \left (-\frac {b^2 (-3+n) \left (1-\frac {2 a c (-3+2 n)}{b^2 (-3+n)}\right ) x^2}{a+b x^n+c x^{2 n}}-\frac {b c (-3+n) x^{2+n}}{a+b x^n+c x^{2 n}}\right ) \, dx}{a \left (b^2-4 a c\right ) n}-\frac {g \int \left (-\frac {b^2 \left (1-\frac {4 a c (-2+n)}{b^2 (-4+n)}\right ) (-4+n) x^3}{a+b x^n+c x^{2 n}}-\frac {b c (-4+n) x^{3+n}}{a+b x^n+c x^{2 n}}\right ) \, dx}{a \left (b^2-4 a c\right ) n}+\frac {\left (c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a \left (b^2-4 a c\right )^{3/2} n} \\ & = \frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {g x^4 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {\left (e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}+\frac {\left (f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}+\frac {\left (g \left (4 a c (2-n)-b^2 (4-n)\right )\right ) \int \frac {x^3}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {(b c e (2-n)) \int \frac {x^{1+n}}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {(b c f (3-n)) \int \frac {x^{2+n}}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {(b c g (4-n)) \int \frac {x^{3+n}}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n} \\ & = \frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {g x^4 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {\left (2 c e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 c e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 c g \left (4 a c (2-n)-b^2 (4-n)\right )\right ) \int \frac {x^3}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 c g \left (4 a c (2-n)-b^2 (4-n)\right )\right ) \int \frac {x^3}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 b c^2 e (2-n)\right ) \int \frac {x^{1+n}}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 b c^2 e (2-n)\right ) \int \frac {x^{1+n}}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 b c^2 f (3-n)\right ) \int \frac {x^{2+n}}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 b c^2 f (3-n)\right ) \int \frac {x^{2+n}}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 b c^2 g (4-n)\right ) \int \frac {x^{3+n}}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 b c^2 g (4-n)\right ) \int \frac {x^{3+n}}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n} \\ & = \text {Too large to display} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(8737\) vs. \(2(1654)=3308\).
Time = 6.73 (sec) , antiderivative size = 8737, normalized size of antiderivative = 5.28 \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \]
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\[\int \frac {g \,x^{3}+f \,x^{2}+e x +d}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]
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\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]
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\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {g\,x^3+f\,x^2+e\,x+d}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]
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