Optimal. Leaf size=1129 \[ \text{result too large to display} \]
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Rubi [A] time = 3.34142, antiderivative size = 1129, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1436, 245, 1430, 1422} \[ \frac{2 (2 c d-b e) x \, _2F_1\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 \, _2F_1\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 \, _2F_1\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 \, _2F_1\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 \, _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 ) \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 \, _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 ) \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 )} \]
Antiderivative was successfully verified.
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Rule 1436
Rule 245
Rule 1430
Rule 1422
Rubi steps
\begin{align*} \int \frac{1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\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*}
Mathematica [B] time = 8.10529, size = 11657, normalized size = 10.33 \[ \text{Result too large to show} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.221, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d+e{x}^{n} \right ) ^{2} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{2} e^{2} x^{4 \, n} + a^{2} d^{2} +{\left (c^{2} e^{2} x^{2 \, n} + 2 \, c^{2} d e x^{n} + c^{2} d^{2}\right )} x^{4 \, n} + 2 \,{\left (b^{2} d e + a b e^{2}\right )} x^{3 \, n} + 2 \,{\left (b c e^{2} x^{3 \, n} + a c d^{2} +{\left (2 \, b c d e + a c e^{2}\right )} x^{2 \, n} +{\left (b c d^{2} + 2 \, a c d e\right )} x^{n}\right )} x^{2 \, n} +{\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{2 \, n} + 2 \,{\left (a b d^{2} + a^{2} d e\right )} x^{n}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}{\left (e x^{n} + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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