3.562 \(\int \frac {x^2}{a+b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=140 \[ -\frac {2 c x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

[Out]

-2/3*c*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))-2/3*
c*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 = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1383, 364} \[ -\frac {2 c x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3 \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^n + c*x^(2*n)),x]

[Out]

(-2*c*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*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 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 1383

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

Rubi steps

\begin {align*} \int \frac {x^2}{a+b x^n+c x^{2 n}} \, dx &=\frac {(2 c) \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) \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 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 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 [A]  time = 0.65, size = 265, normalized size = 1.89 \[ -\frac {2}{3} c x^3 \left (\frac {1-\left (\frac {x^n}{x^n-\frac {\sqrt {b^2-4 a c}-b}{2 c}}\right )^{-3/n} \, _2F_1\left (-\frac {3}{n},-\frac {3}{n};\frac {n-3}{n};\frac {b-\sqrt {b^2-4 a c}}{2 c x^n+b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}+\frac {1-8^{-1/n} \left (\frac {c x^n}{\sqrt {b^2-4 a c}+b+2 c x^n}\right )^{-3/n} \, _2F_1\left (-\frac {3}{n},-\frac {3}{n};\frac {n-3}{n};\frac {b+\sqrt {b^2-4 a c}}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*x^3*((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 - Hyperge
ometric2F1[-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)))/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c]))))/3

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{c x^{2 \, n} + b x^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{a+b\,x^n+c\,x^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(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 \[ \int \frac {x^{2}}{a + b x^{n} + c x^{2 n}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b*x**n+c*x**(2*n)),x)

[Out]

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

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