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*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-c*e*x^2*
hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2/3*c*f*x^3*hyp
ergeom([1, 3/n],[(3+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2*c*d*x*hypergeom(
[1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-c*e*x^2*hypergeom([1, 2/n],
[(2+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-2/3*c*f*x^3*hypergeom([1, 3/n],[(3
+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))

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Rubi [A]  time = 0.28, antiderivative size = 404, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1793, 1893, 245, 364} \[ -\frac {2 c d x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {2 c d x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {c e x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {c e x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {2 c f x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c f x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

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 - Sqrt[b^2 - 4*a*c])])/(3*(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 + Sqrt[b^2 - 4*a*c])])/(3*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]))

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1793

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(2*c
)/q, Int[Pq/(b - q + 2*c*x^n), x], x] - Dist[(2*c)/q, Int[Pq/(b + q + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, n},
x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2}{a+b x^n+c x^{2 n}} \, dx &=\frac {(2 c) \int \frac {d+e x+f x^2}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {d+e x+f x^2}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}\\ &=\frac {(2 c) \int \left (-\frac {d}{-b+\sqrt {b^2-4 a c}-2 c x^n}-\frac {e x}{-b+\sqrt {b^2-4 a c}-2 c x^n}-\frac {f x^2}{-b+\sqrt {b^2-4 a c}-2 c x^n}\right ) \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \left (\frac {d}{b+\sqrt {b^2-4 a c}+2 c x^n}+\frac {e x}{b+\sqrt {b^2-4 a c}+2 c x^n}+\frac {f x^2}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx}{\sqrt {b^2-4 a c}}\\ &=-\frac {(2 c d) \int \frac {1}{-b+\sqrt {b^2-4 a c}-2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c d) \int \frac {1}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c e) \int \frac {x}{-b+\sqrt {b^2-4 a c}-2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c e) \int \frac {x}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c f) \int \frac {x^2}{-b+\sqrt {b^2-4 a c}-2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c f) \int \frac {x^2}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}\\ &=-\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^2-4 a c-b \sqrt {b^2-4 a c}}-\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^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {c e x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c e x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {2 c f x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c f x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}\\ \end {align*}

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Mathematica [B]  time = 1.13, 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/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/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 - 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)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/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
/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/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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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {f \,x^{2}+e x +d}{b \,x^{n}+c \,x^{2 n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {f\,x^2+e\,x+d}{a+b\,x^n+c\,x^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x + f x^{2}}{a + b x^{n} + c x^{2 n}}\, dx \]

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]

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

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