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

Optimal. Leaf size=404 \[ -\frac{2 c d x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c d x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c e x^2 \, _2F_1\left (1,\frac{2}{n};\frac{n+2}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c e x^2 \, _2F_1\left (1,\frac{2}{n};\frac{n+2}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c f x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{3 \left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right )}-\frac{2 c f x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{3 \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right )} \]

[Out]

(-2*c*d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*
a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (2*c*d*x*Hypergeometric2F1[1, n^(-
1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2 -
4*a*c]) - (c*e*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^
2 - 4*a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (c*e*x^2*Hypergeometric2F1[1
, 2/n, (2 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2
- 4*a*c]) - (2*c*f*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqr
t[b^2 - 4*a*c])])/(3*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])) - (2*c*f*x^3*Hypergeom
etric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3*(b^2 - 4*a*c
+ b*Sqrt[b^2 - 4*a*c]))

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Rubi [A]  time = 0.623988, antiderivative size = 404, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148 \[ -\frac{2 c d x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c d x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c e x^2 \, _2F_1\left (1,\frac{2}{n};\frac{n+2}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c e x^2 \, _2F_1\left (1,\frac{2}{n};\frac{n+2}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c f x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{3 \left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right )}-\frac{2 c f x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{3 \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(-2*c*d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*
a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (2*c*d*x*Hypergeometric2F1[1, n^(-
1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2 -
4*a*c]) - (c*e*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^
2 - 4*a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (c*e*x^2*Hypergeometric2F1[1
, 2/n, (2 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2
- 4*a*c]) - (2*c*f*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqr
t[b^2 - 4*a*c])])/(3*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])) - (2*c*f*x^3*Hypergeom
etric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3*(b^2 - 4*a*c
+ b*Sqrt[b^2 - 4*a*c]))

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Rubi in Sympy [A]  time = 94.0776, size = 357, normalized size = 0.88 \[ - \frac{2 c d x{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{1}{n} \\ 1 + \frac{1}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{- 4 a c + b^{2} + b \sqrt{- 4 a c + b^{2}}} - \frac{2 c d x{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{1}{n} \\ 1 + \frac{1}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{- 4 a c + b^{2} - b \sqrt{- 4 a c + b^{2}}} - \frac{c e x^{2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{2}{n} \\ \frac{n + 2}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{- 4 a c + b^{2} + b \sqrt{- 4 a c + b^{2}}} - \frac{c e x^{2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{2}{n} \\ \frac{n + 2}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{- 4 a c + b^{2} - b \sqrt{- 4 a c + b^{2}}} - \frac{2 c f x^{3}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{3}{n} \\ \frac{n + 3}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{3 \left (- 4 a c + b^{2} + b \sqrt{- 4 a c + b^{2}}\right )} - \frac{2 c f x^{3}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{3}{n} \\ \frac{n + 3}{n} \end{matrix}\middle |{- \frac{2 c x^{n}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{3 \left (- 4 a c + b^{2} - b \sqrt{- 4 a c + b^{2}}\right )} \]

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)),x)

[Out]

-2*c*d*x*hyper((1, 1/n), (1 + 1/n,), -2*c*x**n/(b + sqrt(-4*a*c + b**2)))/(-4*a*
c + b**2 + b*sqrt(-4*a*c + b**2)) - 2*c*d*x*hyper((1, 1/n), (1 + 1/n,), -2*c*x**
n/(b - sqrt(-4*a*c + b**2)))/(-4*a*c + b**2 - b*sqrt(-4*a*c + b**2)) - c*e*x**2*
hyper((1, 2/n), ((n + 2)/n,), -2*c*x**n/(b + sqrt(-4*a*c + b**2)))/(-4*a*c + b**
2 + b*sqrt(-4*a*c + b**2)) - c*e*x**2*hyper((1, 2/n), ((n + 2)/n,), -2*c*x**n/(b
- sqrt(-4*a*c + b**2)))/(-4*a*c + b**2 - b*sqrt(-4*a*c + b**2)) - 2*c*f*x**3*hy
per((1, 3/n), ((n + 3)/n,), -2*c*x**n/(b + sqrt(-4*a*c + b**2)))/(3*(-4*a*c + b*
*2 + b*sqrt(-4*a*c + b**2))) - 2*c*f*x**3*hyper((1, 3/n), ((n + 3)/n,), -2*c*x**
n/(b - sqrt(-4*a*c + b**2)))/(3*(-4*a*c + b**2 - b*sqrt(-4*a*c + b**2)))

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Mathematica [B]  time = 2.14359, size = 834, normalized size = 2.06 \[ \frac{x \left (2 f \left (\left (-b^2-\sqrt{b^2-4 a c} b+4 a c\right ) \left (1-\left (\frac{x^n}{x^n-\frac{\sqrt{b^2-4 a c}-b}{2 c}}\right )^{-3/n} \, _2F_1\left (-\frac{3}{n},-\frac{3}{n};\frac{n-3}{n};\frac{b-\sqrt{b^2-4 a c}}{2 c x^n+b-\sqrt{b^2-4 a c}}\right )\right )+\left (-b^2+\sqrt{b^2-4 a c} b+4 a c\right ) \left (1-8^{-1/n} \left (\frac{c x^n}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )^{-3/n} \, _2F_1\left (-\frac{3}{n},-\frac{3}{n};\frac{n-3}{n};\frac{b+\sqrt{b^2-4 a c}}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )\right )\right ) x^2+3 e \left (\left (-b^2-\sqrt{b^2-4 a c} b+4 a c\right ) \left (1-\left (\frac{x^n}{x^n-\frac{\sqrt{b^2-4 a c}-b}{2 c}}\right )^{-2/n} \, _2F_1\left (-\frac{2}{n},-\frac{2}{n};\frac{n-2}{n};\frac{b-\sqrt{b^2-4 a c}}{2 c x^n+b-\sqrt{b^2-4 a c}}\right )\right )+\left (-b^2+\sqrt{b^2-4 a c} b+4 a c\right ) \left (1-4^{-1/n} \left (\frac{c x^n}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )^{-2/n} \, _2F_1\left (-\frac{2}{n},-\frac{2}{n};\frac{n-2}{n};\frac{b+\sqrt{b^2-4 a c}}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )\right )\right ) x+6 d \left (\left (-b^2-\sqrt{b^2-4 a c} b+4 a c\right ) \left (1-\left (\frac{x^n}{x^n-\frac{\sqrt{b^2-4 a c}-b}{2 c}}\right )^{-1/n} \, _2F_1\left (-\frac{1}{n},-\frac{1}{n};\frac{n-1}{n};\frac{b-\sqrt{b^2-4 a c}}{2 c x^n+b-\sqrt{b^2-4 a c}}\right )\right )-2^{-1/n} \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}-b\right ) \left (\frac{c x^n}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )^{-1/n} \left (2^{\frac{1}{n}} \left (\frac{c x^n}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )^{\frac{1}{n}}-\, _2F_1\left (-\frac{1}{n},-\frac{1}{n};\frac{n-1}{n};\frac{b+\sqrt{b^2-4 a c}}{2 c x^n+b+\sqrt{b^2-4 a c}}\right )\right )\right )\right )}{12 a \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

(x*(2*f*x^2*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-3/n, -
3/n, (-3 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(x^n
/(-(-b + Sqrt[b^2 - 4*a*c])/(2*c) + x^n))^(3/n)) + (-b^2 + 4*a*c + b*Sqrt[b^2 -
4*a*c])*(1 - Hypergeometric2F1[-3/n, -3/n, (-3 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(
b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(8^n^(-1)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*
c*x^n))^(3/n)))) + 3*e*x*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometr
ic2F1[-2/n, -2/n, (-2 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2
*c*x^n)]/(x^n/(-(-b + Sqrt[b^2 - 4*a*c])/(2*c) + x^n))^(2/n)) + (-b^2 + 4*a*c +
b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, (b + Sqrt[b^
2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(4^n^(-1)*((c*x^n)/(b + Sqrt[b^2
- 4*a*c] + 2*c*x^n))^(2/n)))) + 6*d*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - H
ypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt
[b^2 - 4*a*c] + 2*c*x^n)]/(x^n/(-(-b + Sqrt[b^2 - 4*a*c])/(2*c) + x^n))^n^(-1))
- (Sqrt[b^2 - 4*a*c]*(-b + Sqrt[b^2 - 4*a*c])*(2^n^(-1)*((c*x^n)/(b + Sqrt[b^2 -
4*a*c] + 2*c*x^n))^n^(-1) - Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b
+ Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]))/(2^n^(-1)*((c*x^n)/(b
+ Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^(-1)))))/(12*a*(-b^2 + 4*a*c))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \[ \int{\frac{f{x}^{2}+ex+d}{a+b{x}^{n}+c{x}^{2\,n}}}\, dx \]

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)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a}\,{d x} \]

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),x, algorithm="maxima")

[Out]

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

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

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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a}\,{d x} \]

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),x, algorithm="giac")

[Out]

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