\(\int \frac {1}{x^2 (a+b x^n+c x^{2 n})} \, dx\) [566]

Optimal result

Integrand size = 20, antiderivative size = 142 \[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {2 c \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) x}+\frac {2 c \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1397, 371} \[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {2 c \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{x \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {2 c \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{x \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1397

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*} \text {integral}& = \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] (verified)

Time = 0.37 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.69 \[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {2^{1+\frac {1}{n}} c \left (\frac {\left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},1+\frac {1}{n},2+\frac {1}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{-b+\sqrt {b^2-4 a c}-2 c x^n}+\frac {x^{-n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{1+\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},1+\frac {1}{n},2+\frac {1}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{c}\right )}{\sqrt {b^2-4 a c} (1+n) x} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{x^{2} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]

[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)

Fricas [F]

\[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{2}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {1}{x^2\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \]

[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)