Integrand size = 26, antiderivative size = 1129 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {x \left (2 b^3 c d e-6 a b c^2 d e-b^4 e^2-b^2 c \left (c d^2-4 a e^2\right )+2 a c^2 \left (c d^2-a e^2\right )+c \left (2 b^2 c d e-4 a c^2 d e-b^3 e^2-b c \left (c d^2-3 a e^2\right )\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 n \left (a+b x^n+c x^{2 n}\right )}-\frac {2 c e^2 \left (3 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d+2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {c \left (4 a c^2 \left (e \left (a e (1-2 n)+\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)+2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)+\sqrt {b^2-4 a c} d (1-n)\right )-3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d-\sqrt {b^2-4 a c} e\right ) (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 ) \left (c d^2-b d e+a e^2\right )^2 n}-\frac {2 c e^2 \left (3 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d-2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {c \left (4 a c^2 \left (e \left (a e (1-2 n)-\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)-2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)-\sqrt {b^2-4 a c} d (1-n)\right )+3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d+\sqrt {b^2-4 a c} e\right ) (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 ) \left (c d^2-b d e+a e^2\right )^2 n}+\frac {2 e^4 (2 c d-b e) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^4 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2} \]
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Time = 2.11 (sec) , antiderivative size = 1129, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1450, 251, 1444, 1436} \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {2 (2 c d-b e) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right ) e^4}{d \left (c d^2-b e d+a e^2\right )^3}+\frac {x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right ) e^4}{d^2 \left (c d^2-b e d+a e^2\right )^2}-\frac {2 c \left (3 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d+2 \sqrt {b^2-4 a c} d+a e\right )\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 ) e^2}{\left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) \left (c d^2-b e d+a e^2\right )^3}-\frac {2 c \left (3 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d-2 \sqrt {b^2-4 a c} d+a e\right )\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 ) e^2}{\left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) \left (c d^2-b e d+a e^2\right )^3}+\frac {c \left (e^2 (1-n) b^4-e \left (2 c d-\sqrt {b^2-4 a c} e\right ) (1-n) b^3-c \left (e \left (a e (5-7 n)+2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right ) b^2+c \left (c d \left (4 a e (2-3 n)+\sqrt {b^2-4 a c} d (1-n)\right )-3 a \sqrt {b^2-4 a c} e^2 (1-n)\right ) b+4 a c^2 \left (e \left (a e (1-2 n)+\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )\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-\sqrt {b^2-4 a c} b-4 a c\right ) \left (c d^2-b e d+a e^2\right )^2 n}+\frac {c \left (e^2 (1-n) b^4-e \left (2 c d+\sqrt {b^2-4 a c} e\right ) (1-n) b^3-c \left (e \left (a e (5-7 n)-2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right ) b^2+c \left (3 a \sqrt {b^2-4 a c} (1-n) e^2+c d \left (4 a e (2-3 n)-\sqrt {b^2-4 a c} d (1-n)\right )\right ) b+4 a c^2 \left (e \left (a e (1-2 n)-\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )\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+\sqrt {b^2-4 a c} b-4 a c\right ) \left (c d^2-b e d+a e^2\right )^2 n}-\frac {x \left (c \left (-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e\right ) x^n-b^4 e^2-6 a b c^2 d e+2 b^3 c d e-b^2 c \left (c d^2-4 a e^2\right )+2 a c^2 \left (c d^2-a e^2\right )\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b e d+a e^2\right )^2 n \left (b x^n+c x^{2 n}+a\right )} \]
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Rule 251
Rule 1436
Rule 1444
Rule 1450
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^4}{\left (c d^2-b d e+a e^2\right )^2 \left (d+e x^n\right )^2}-\frac {2 e^4 (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right )^3 \left (d+e x^n\right )}+\frac {c^2 d^2-2 b c d e+b^2 e^2-a c e^2-\left (2 c^2 d e-b c e^2\right ) x^n}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x^n+c x^{2 n}\right )^2}+\frac {e^2 \left (3 c^2 d^2-5 b c d e+2 b^2 e^2-a c e^2+\left (-4 c^2 d e+2 b c e^2\right ) x^n\right )}{\left (c d^2-b d e+a e^2\right )^3 \left (a+b x^n+c x^{2 n}\right )}\right ) \, dx \\ & = \frac {e^2 \int \frac {3 c^2 d^2-5 b c d e+2 b^2 e^2-a c e^2+\left (-4 c^2 d e+2 b c e^2\right ) x^n}{a+b x^n+c x^{2 n}} \, dx}{\left (c d^2-b d e+a e^2\right )^3}+\frac {\left (2 e^4 (2 c d-b e)\right ) \int \frac {1}{d+e x^n} \, dx}{\left (c d^2-b d e+a e^2\right )^3}+\frac {\int \frac {c^2 d^2-2 b c d e+b^2 e^2-a c e^2-\left (2 c^2 d e-b c e^2\right ) x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac {e^4 \int \frac {1}{\left (d+e x^n\right )^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2} \\ & = -\frac {x \left (2 b^3 c d e-6 a b c^2 d e-b^4 e^2-b^2 c \left (c d^2-4 a e^2\right )+2 a c^2 \left (c d^2-a e^2\right )+c \left (2 b^2 c d e-4 a c^2 d e-b^3 e^2-b c \left (c d^2-3 a e^2\right )\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 n \left (a+b x^n+c x^{2 n}\right )}+\frac {2 e^4 (2 c d-b e) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^4 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2}-\frac {\left (c e^2 \left (3 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d-2 \sqrt {b^2-4 a c} d+a e\right )\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}+\frac {\left (c e^2 \left (3 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d+2 \sqrt {b^2-4 a c} d+a e\right )\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}-\frac {\int \frac {-b^2 c \left (a e^2 (4-5 n)-c d^2 (1-n)\right )+2 a b c^2 d e (3-4 n)-2 a c^2 \left (c d^2-a e^2\right ) (1-2 n)-2 b^3 c d e (1-n)+b^4 e^2 (1-n)-c \left (2 b^2 c d e-4 a c^2 d e-b^3 e^2-b c \left (c d^2-3 a e^2\right )\right ) (1-n) x^n}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 n} \\ & = -\frac {x \left (2 b^3 c d e-6 a b c^2 d e-b^4 e^2-b^2 c \left (c d^2-4 a e^2\right )+2 a c^2 \left (c d^2-a e^2\right )+c \left (2 b^2 c d e-4 a c^2 d e-b^3 e^2-b c \left (c d^2-3 a e^2\right )\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 n \left (a+b x^n+c x^{2 n}\right )}+\frac {2 c e^2 \left (3 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d+2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}-\frac {2 c e^2 \left (3 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d-2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {2 e^4 (2 c d-b e) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^4 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2}-\frac {\left (c \left (4 a c^2 \left (e \left (a e (1-2 n)+\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)+2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)+\sqrt {b^2-4 a c} d (1-n)\right )-3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d-\sqrt {b^2-4 a c} e\right ) (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} \left (c d^2-b d e+a e^2\right )^2 n}+\frac {\left (c \left (4 a c^2 \left (e \left (a e (1-2 n)-\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)-2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)-\sqrt {b^2-4 a c} d (1-n)\right )+3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d+\sqrt {b^2-4 a c} e\right ) (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} \left (c d^2-b d e+a e^2\right )^2 n} \\ & = -\frac {x \left (2 b^3 c d e-6 a b c^2 d e-b^4 e^2-b^2 c \left (c d^2-4 a e^2\right )+2 a c^2 \left (c d^2-a e^2\right )+c \left (2 b^2 c d e-4 a c^2 d e-b^3 e^2-b c \left (c d^2-3 a e^2\right )\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 n \left (a+b x^n+c x^{2 n}\right )}+\frac {2 c e^2 \left (3 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d+2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}-\frac {c \left (4 a c^2 \left (e \left (a e (1-2 n)+\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)+2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)+\sqrt {b^2-4 a c} d (1-n)\right )-3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d-\sqrt {b^2-4 a c} e\right ) (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 ) \left (c d^2-b d e+a e^2\right )^2 n}-\frac {2 c e^2 \left (3 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-c e \left (3 b d-2 \sqrt {b^2-4 a c} d+a e\right )\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 )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {c \left (4 a c^2 \left (e \left (a e (1-2 n)-\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )-b^2 c \left (e \left (a e (5-7 n)-2 \sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-n)\right )+b c \left (c d \left (4 a e (2-3 n)-\sqrt {b^2-4 a c} d (1-n)\right )+3 a \sqrt {b^2-4 a c} e^2 (1-n)\right )+b^4 e^2 (1-n)-b^3 e \left (2 c d+\sqrt {b^2-4 a c} e\right ) (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 ) \left (c d^2-b d e+a e^2\right )^2 n}+\frac {2 e^4 (2 c d-b e) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^4 x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(16855\) vs. \(2(1129)=2258\).
Time = 7.87 (sec) , antiderivative size = 16855, normalized size of antiderivative = 14.93 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \]
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\[\int \frac {1}{\left (d +e \,x^{n}\right )^{2} \left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]
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\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
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\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2} {\left (e x^{n} + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {1}{{\left (d+e\,x^n\right )}^2\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]
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