3.17 \(\int \frac{A+B x^n+C x^{2 n}+D x^{3 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

Optimal. Leaf size=494 \[ \frac{x \left (x^n \left (b c (a C+A c)-a b^2 D-2 a c (B c-a D)\right )+A c \left (b^2-2 a c\right )-a (a b D-2 a c C+b B c)\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac{x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right ) \left (\frac{A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt{b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right ) \left (-\frac{A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt{b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )} \]

[Out]

(x*(A*c*(b^2 - 2*a*c) - a*(b*B*c - 2*a*c*C + a*b*D) + (b*c*(A*c + a*C) - a*b^2*D
- 2*a*c*(B*c - a*D))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + ((a*
b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(1 + n)) + (A*c^2*(4*
a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 - n) - 2*b*c*(B
*c*n + a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-
1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*
c])*n) + ((a*b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(1 + n))
- (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 -
n) - 2*b*c*(B*c*n + a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(
-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b + Sq
rt[b^2 - 4*a*c])*n)

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Rubi [A]  time = 5.38918, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079 \[ \frac{x \left (x^n \left (b c (a C+A c)-a b^2 D-2 a c (B c-a D)\right )+A c \left (b^2-2 a c\right )-a (a b D-2 a c C+b B c)\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac{x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right ) \left (\frac{A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt{b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right ) \left (-\frac{A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt{b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

[In]  Int[(A + B*x^n + C*x^(2*n) + D*x^(3*n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

(x*(A*c*(b^2 - 2*a*c) - a*(b*B*c - 2*a*c*C + a*b*D) + (b*c*(A*c + a*C) - a*b^2*D
- 2*a*c*(B*c - a*D))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + ((a*
b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(1 + n)) + (A*c^2*(4*
a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 - n) - 2*b*c*(B
*c*n + a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-
1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*
c])*n) + ((a*b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(1 + n))
- (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 -
n) - 2*b*c*(B*c*n + a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(
-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b + Sq
rt[b^2 - 4*a*c])*n)

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{A + B x^{n} + C x^{2 n} + D x^{3 n}}{\left (a + b x^{n} + c x^{2 n}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((A+B*x**n+C*x**(2*n)+D*x**(3*n))/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Integral((A + B*x**n + C*x**(2*n) + D*x**(3*n))/(a + b*x**n + c*x**(2*n))**2, x)

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Mathematica [B]  time = 6.51999, size = 5439, normalized size = 11.01 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]  Integrate[(A + B*x^n + C*x^(2*n) + D*x^(3*n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

Result too large to show

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \[ \int{\frac{A+B{x}^{n}+C{x}^{2\,n}+D{x}^{3\,n}}{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(a+b*x^n+c*x^(2*n))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (C a b c - 2 \, B a c^{2} + A b c^{2} -{\left (a b^{2} - 2 \, a^{2} c\right )} D\right )} x x^{n} -{\left (D a^{2} b - 2 \, C a^{2} c + B a b c -{\left (b^{2} c - 2 \, a c^{2}\right )} A\right )} x}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n +{\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} +{\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}} - \int -\frac{D a^{2} b - 2 \, C a^{2} c + B a b c -{\left (2 \, a c^{2}{\left (2 \, n - 1\right )} - b^{2} c{\left (n - 1\right )}\right )} A +{\left (C a b c{\left (n - 1\right )} - 2 \, B a c^{2}{\left (n - 1\right )} + A b c^{2}{\left (n - 1\right )} -{\left (2 \, a^{2} c{\left (n + 1\right )} - a b^{2}\right )} D\right )} x^{n}}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n +{\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} +{\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="maxima")

[Out]

((C*a*b*c - 2*B*a*c^2 + A*b*c^2 - (a*b^2 - 2*a^2*c)*D)*x*x^n - (D*a^2*b - 2*C*a^
2*c + B*a*b*c - (b^2*c - 2*a*c^2)*A)*x)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*
n - 4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a^2*b*c^2*n)*x^n) - integrate(-(D*a^2*
b - 2*C*a^2*c + B*a*b*c - (2*a*c^2*(2*n - 1) - b^2*c*(n - 1))*A + (C*a*b*c*(n -
1) - 2*B*a*c^2*(n - 1) + A*b*c^2*(n - 1) - (2*a^2*c*(n + 1) - a*b^2)*D)*x^n)/(a^
2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a
^2*b*c^2*n)*x^n), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{D x^{3 \, n} + C x^{2 \, n} + B x^{n} + A}{c^{2} x^{4 \, n} + 2 \, a b x^{n} + a^{2} +{\left (2 \, b c x^{n} + b^{2} + 2 \, a c\right )} x^{2 \, n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="fricas")

[Out]

integral((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c^2*x^(4*n) + 2*a*b*x^n + a^2 + (2
*b*c*x^n + b^2 + 2*a*c)*x^(2*n)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((A+B*x**n+C*x**(2*n)+D*x**(3*n))/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{D x^{3 \, n} + C x^{2 \, n} + B x^{n} + A}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="giac")

[Out]

integrate((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c*x^(2*n) + b*x^n + a)^2, x)