3.8 \(\int \frac{d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

Optimal. Leaf size=1194 \[ \text{result too large to display} \]

[Out]

(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e*x
^2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (f*x^3
*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) - (c*d*(4*
a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1
, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2
- 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) + b*Sqr
t[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b
+ Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n)
- (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-
2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 -
4*a*c])*n) - (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2
+ n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*
Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hypergeometri
c2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*
(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*
x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3
*a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (2*b*c^2*e*(2 - n)*x^(
2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b - Sqrt[b^2
- 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*n*(2 + n)) + (2*b*c^2
*e*(2 - n)*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/
(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*n*(2 +
n)) - (2*b*c^2*f*(3 - n)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, (-2*
c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*
n*(3 + n)) + (2*b*c^2*f*(3 - n)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/
n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4
*a*c])*n*(3 + n))

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Rubi [A]  time = 5.01394, antiderivative size = 1194, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296 \[ -\frac{2 b c^2 e (2-n) \, _2F_1\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) \, _2F_1\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) \, _2F_1\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) \, _2F_1\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 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \, _2F_1\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 ) \, _2F_1\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 ) \, _2F_1\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 ) \, _2F_1\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 (-(1-n) b^2-\sqrt{b^2-4 a c} (1-n) b+4 a c (1-2 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 ) 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 (-(1-n) b^2+\sqrt{b^2-4 a c} (1-n) b+4 a c (1-2 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 ) 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 )} \]

Antiderivative was successfully verified.

[In]  Int[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e*x
^2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (f*x^3
*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) - (c*d*(4*
a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1
, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2
- 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) + b*Sqr
t[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b
+ Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n)
- (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-
2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 -
4*a*c])*n) - (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2
+ n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*
Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hypergeometri
c2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*
(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*
x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3
*a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (2*b*c^2*e*(2 - n)*x^(
2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b - Sqrt[b^2
- 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*n*(2 + n)) + (2*b*c^2
*e*(2 - n)*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/
(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*n*(2 +
n)) - (2*b*c^2*f*(3 - n)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, (-2*
c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*
n*(3 + n)) + (2*b*c^2*f*(3 - n)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/
n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4
*a*c])*n*(3 + n))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x**2+e*x+d)/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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Mathematica [B]  time = 6.57774, size = 6525, normalized size = 5.46 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]  Integrate[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

Result too large to show

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Maple [F]  time = 0.097, size = 0, normalized size = 0. \[ \int{\frac{f{x}^{2}+ex+d}{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="maxima")

[Out]

((b^2*f - 2*a*c*f)*x^3 + (b^2*e - 2*a*c*e)*x^2 + (b*c*f*x^3 + b*c*e*x^2 + b*c*d*
x)*x^n + (b^2*d - 2*a*c*d)*x)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)
*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n) - integrate((2*a*c*d*(2*n - 1) - b^2*d*(
n - 1) + (2*a*c*f*(2*n - 3) - b^2*f*(n - 3))*x^2 - (b*c*f*(n - 3)*x^2 + b*c*e*(n
- 2)*x + b*c*d*(n - 1))*x^n + (4*a*c*e*(n - 1) - b^2*e*(n - 2))*x)/(a^2*b^2*n -
4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n), x
)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{f x^{2} + e x + d}{c^{2} x^{4 \, n} + 2 \, a b x^{n} + a^{2} +{\left (2 \, b c x^{n} + b^{2} + 2 \, a c\right )} x^{2 \, n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="fricas")

[Out]

integral((f*x^2 + e*x + d)/(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((f*x**2+e*x+d)/(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. \[ \int \frac{f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="giac")

[Out]

integrate((f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2, x)