\(\int \frac {1}{a+b x^n+c x^{2 n}} \, dx\) [564]

Optimal result

Integrand size = 16, antiderivative size = 124 \[ \int \frac {1}{a+b x^n+c x^{2 n}} \, dx=-\frac {2 c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {2 c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}} \]

[Out]

-2*c*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))-2*c*x*hype
rgeom([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))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1361, 251} \[ \int \frac {1}{a+b x^n+c x^{2 n}} \, dx=-\frac {2 c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {2 c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2} \]

[In]

Int[(a + b*x^n + c*x^(2*n))^(-1),x]

[Out]

(-2*c*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*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])

Rule 251

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 1361

Int[((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, In
t[1/(b/2 - q/2 + c*x^n), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c}, x] && EqQ[n
2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {c \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {c \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{\sqrt {b^2-4 a c}} \\ & = -\frac {2 c 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 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}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(261\) vs. \(2(124)=248\).

Time = 0.46 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.10 \[ \int \frac {1}{a+b x^n+c x^{2 n}} \, dx=-2 c x \left (\frac {1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {1-2^{-1/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right ) \]

[In]

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

[Out]

-2*c*x*((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)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^n^(-1))/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + (1 - Hyp
ergeometric2F1[-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)))/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(1/(a + b*x^n + c*x^(2*n)), x)