3.566 \(\int \frac{1}{x^2 (a+b x^n+c x^{2 n})} \, dx\)

Optimal. Leaf size=142 \[ \frac{2 c \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{x \left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right )}+\frac{2 c \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{x \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right )} \]

[Out]

(2*c*Hypergeometric2F1[1, -n^(-1), -((1 - 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])*x) + (2*c*Hypergeometric2F1[1, -n^(-1), -((1 - 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])*x)

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Rubi [A]  time = 0.0462026, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1383, 364} \[ \frac{2 c \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{x \left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right )}+\frac{2 c \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{x \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^n + c*x^(2*n))),x]

[Out]

(2*c*Hypergeometric2F1[1, -n^(-1), -((1 - 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])*x) + (2*c*Hypergeometric2F1[1, -n^(-1), -((1 - 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])*x)

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.122172, size = 129, normalized size = 0.91 \[ -\frac{\left (\sqrt{b^2-4 a c}+b\right ) \, _2F_1\left (1,-\frac{1}{n};\frac{n-1}{n};\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )+\left (\sqrt{b^2-4 a c}-b\right ) \, _2F_1\left (1,-\frac{1}{n};\frac{n-1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 a x \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^n + c*x^(2*n))),x]

[Out]

-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, -n^(-1), (-1 + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + (-b
+ Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, -n^(-1), (-1 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*a*Sqrt[
b^2 - 4*a*c]*x)

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a+b*x^n+c*x^(2*n)),x)

[Out]

int(1/x^2/(a+b*x^n+c*x^(2*n)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

integrate(1/((c*x^(2*n) + b*x^n + a)*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{c x^{2} x^{2 \, n} + b x^{2} x^{n} + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

integral(1/(c*x^2*x^(2*n) + b*x^2*x^n + a*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a+b*x**n+c*x**(2*n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

integrate(1/((c*x^(2*n) + b*x^n + a)*x^2), x)