Optimal. Leaf size=726 \[ -\frac {c e^2 x \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {c e^2 x \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {c x \left ((1-n) \left (2 a c e+b^2 (-e)+b c d\right )+\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^3 (-e) (1-n)+b^2 c d (1-n)}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) \left (a e^2-b d e+c d^2\right )}+\frac {x \left (c x^n \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d-b^3 e+b^2 c d\right )}{a n \left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac {c x \left (b^2 (1-n) \left (e \sqrt {b^2-4 a c}+c d\right )+b c \left (2 a e (2-3 n)-d (1-n) \sqrt {b^2-4 a c}\right )-2 a c \left (e (1-n) \sqrt {b^2-4 a c}+2 c d (1-2 n)\right )+b^3 (-e) (1-n)\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac {e^4 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^2} \]
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Rubi [A] time = 1.93, antiderivative size = 726, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1436, 245, 1430, 1422} \[ -\frac {c x \left (\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)+b^3 (-e) (1-n)}{\sqrt {b^2-4 a c}}+(1-n) \left (2 a c e+b^2 (-e)+b c d\right )\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) \left (a e^2-b d e+c d^2\right )}+\frac {x \left (c x^n \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d+b^2 c d-b^3 e\right )}{a n \left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac {c e^2 x \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {c e^2 x \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac {c x \left (b^2 (1-n) \left (e \sqrt {b^2-4 a c}+c d\right )+b c \left (2 a e (2-3 n)-d (1-n) \sqrt {b^2-4 a c}\right )-2 a c \left (e (1-n) \sqrt {b^2-4 a c}+2 c d (1-2 n)\right )+b^3 (-e) (1-n)\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac {e^4 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 245
Rule 1422
Rule 1430
Rule 1436
Rubi steps
\begin {align*} \int \frac {1}{\left (d+e x^n\right ) \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 )}+\frac {c d-b e-c e x^n}{\left (c d^2-b d e+a e^2\right ) \left (a+b x^n+c x^{2 n}\right )^2}-\frac {e^2 \left (-c d+b e+c e x^n\right )}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x^n+c x^{2 n}\right )}\right ) \, dx\\ &=-\frac {e^2 \int \frac {-c d+b e+c e x^n}{a+b x^n+c x^{2 n}} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac {e^4 \int \frac {1}{d+e x^n} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac {\int \frac {c d-b e-c e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{c d^2-b d e+a e^2}\\ &=\frac {x \left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e^4 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 )^2}-\frac {\left (c e^2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}-\frac {\left (c e^2 \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}-\frac {\int \frac {a b c e-2 a c (c d-b e) (1-2 n)+b^2 (c d-b e) (1-n)+c \left (b c d-b^2 e+2 a c e\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 ) n}\\ &=\frac {x \left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c e^2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\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 )}{\left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {c e^2 \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\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 )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {e^4 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 )^2}-\frac {\left (c \left (\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)-b^3 e (1-n)}{\sqrt {b^2-4 a c}}+\left (b c d-b^2 e+2 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 ) \left (c d^2-b d e+a e^2\right ) n}-\frac {\left (c \left (\left (b c d-b^2 e+2 a c e\right ) (1-n)-\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)-b^3 (e-e n)}{\sqrt {b^2-4 a c}}\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 ) \left (c d^2-b d e+a e^2\right ) n}\\ &=\frac {x \left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c e^2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\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 )}{\left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {c \left (\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)-b^3 e (1-n)}{\sqrt {b^2-4 a c}}+\left (b c d-b^2 e+2 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 ) \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) n}-\frac {c e^2 \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\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 )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {c \left (\left (b c d-b^2 e+2 a c e\right ) (1-n)-\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)-b^3 (e-e n)}{\sqrt {b^2-4 a c}}\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+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) n}+\frac {e^4 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 )^2}\\ \end {align*}
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Mathematica [B] time = 7.22, size = 11767, normalized size = 16.21 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 2.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{2} e x^{3 \, n} + a^{2} d + {\left (c^{2} e x^{n} + c^{2} d\right )} x^{4 \, n} + 2 \, {\left (b c e x^{2 \, n} + a c d + {\left (b c d + a c e\right )} x^{n}\right )} x^{2 \, n} + {\left (b^{2} d + 2 \, a b e\right )} x^{2 \, n} + {\left (2 \, a b d + a^{2} e\right )} x^{n}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2} {\left (e x^{n} + d\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \,x^{n}+d \right ) \left (b \,x^{n}+c \,x^{2 n}+a \right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ e^{4} \int \frac {1}{c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + a^{2} d e^{4} + 2 \, {\left (c d^{3} e^{2} - b d^{2} e^{3}\right )} a + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + a^{2} e^{5} + 2 \, {\left (c d^{2} e^{3} - b d e^{4}\right )} a\right )} x^{n}}\,{d x} - \frac {{\left (b c^{2} d - b^{2} c e + 2 \, a c^{2} e\right )} x x^{n} + {\left (b^{2} c d - b^{3} e - {\left (2 \, c^{2} d - 3 \, b c e\right )} a\right )} x}{4 \, a^{4} c e^{2} n + {\left (4 \, c^{2} d^{2} n - 4 \, b c d e n - b^{2} e^{2} n\right )} a^{3} - {\left (b^{2} c d^{2} n - b^{3} d e n\right )} a^{2} + {\left (4 \, a^{3} c^{2} e^{2} n + {\left (4 \, c^{3} d^{2} n - 4 \, b c^{2} d e n - b^{2} c e^{2} n\right )} a^{2} - {\left (b^{2} c^{2} d^{2} n - b^{3} c d e n\right )} a\right )} x^{2 \, n} + {\left (4 \, a^{3} b c e^{2} n + {\left (4 \, b c^{2} d^{2} n - 4 \, b^{2} c d e n - b^{3} e^{2} n\right )} a^{2} - {\left (b^{3} c d^{2} n - b^{4} d e n\right )} a\right )} x^{n}} - \int \frac {b^{2} c^{2} d^{3} {\left (n - 1\right )} - 2 \, b^{3} c d^{2} e {\left (n - 1\right )} + b^{4} d e^{2} {\left (n - 1\right )} + {\left (b c e^{3} {\left (8 \, n - 3\right )} - 2 \, c^{2} d e^{2} {\left (4 \, n - 1\right )}\right )} a^{2} + {\left (b c^{2} d^{2} e {\left (8 \, n - 5\right )} - 2 \, c^{3} d^{3} {\left (2 \, n - 1\right )} - b^{3} e^{3} {\left (2 \, n - 1\right )} - 2 \, b^{2} c d e^{2} {\left (n - 1\right )}\right )} a + {\left (2 \, a^{2} c^{2} e^{3} {\left (3 \, n - 1\right )} + b c^{3} d^{3} {\left (n - 1\right )} - 2 \, b^{2} c^{2} d^{2} e {\left (n - 1\right )} + b^{3} c d e^{2} {\left (n - 1\right )} - {\left (b^{2} c e^{3} {\left (2 \, n - 1\right )} - 2 \, c^{3} d^{2} e {\left (n - 1\right )} + b c^{2} d e^{2} {\left (n - 1\right )}\right )} a\right )} x^{n}}{4 \, a^{5} c e^{4} n + {\left (8 \, c^{2} d^{2} e^{2} n - 8 \, b c d e^{3} n - b^{2} e^{4} n\right )} a^{4} + 2 \, {\left (2 \, c^{3} d^{4} n - 4 \, b c^{2} d^{3} e n + b^{2} c d^{2} e^{2} n + b^{3} d e^{3} n\right )} a^{3} - {\left (b^{2} c^{2} d^{4} n - 2 \, b^{3} c d^{3} e n + b^{4} d^{2} e^{2} n\right )} a^{2} + {\left (4 \, a^{4} c^{2} e^{4} n + {\left (8 \, c^{3} d^{2} e^{2} n - 8 \, b c^{2} d e^{3} n - b^{2} c e^{4} n\right )} a^{3} + 2 \, {\left (2 \, c^{4} d^{4} n - 4 \, b c^{3} d^{3} e n + b^{2} c^{2} d^{2} e^{2} n + b^{3} c d e^{3} n\right )} a^{2} - {\left (b^{2} c^{3} d^{4} n - 2 \, b^{3} c^{2} d^{3} e n + b^{4} c d^{2} e^{2} n\right )} a\right )} x^{2 \, n} + {\left (4 \, a^{4} b c e^{4} n + {\left (8 \, b c^{2} d^{2} e^{2} n - 8 \, b^{2} c d e^{3} n - b^{3} e^{4} n\right )} a^{3} + 2 \, {\left (2 \, b c^{3} d^{4} n - 4 \, b^{2} c^{2} d^{3} e n + b^{3} c d^{2} e^{2} n + b^{4} d e^{3} n\right )} a^{2} - {\left (b^{3} c^{2} d^{4} n - 2 \, b^{4} c d^{3} e n + b^{5} d^{2} e^{2} n\right )} a\right )} x^{n}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\left (d+e\,x^n\right )\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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