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

Optimal. Leaf size=126 \[ -\frac{\left (b^2-a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^3 n}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^3 n \sqrt{b^2-4 a c}}+\frac{\log (x) \left (b^2-a c\right )}{a^3}+\frac{b x^{-n}}{a^2 n}-\frac{x^{-2 n}}{2 a n} \]

[Out]

-1/(2*a*n*x^(2*n)) + b/(a^2*n*x^n) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt[b^2
- 4*a*c]*n) + ((b^2 - a*c)*Log[x])/a^3 - ((b^2 - a*c)*Log[a + b*x^n + c*x^(2*n)])/(2*a^3*n)

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Rubi [A]  time = 0.170971, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {1357, 709, 800, 634, 618, 206, 628} \[ -\frac{\left (b^2-a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^3 n}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^3 n \sqrt{b^2-4 a c}}+\frac{\log (x) \left (b^2-a c\right )}{a^3}+\frac{b x^{-n}}{a^2 n}-\frac{x^{-2 n}}{2 a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*a*n*x^(2*n)) + b/(a^2*n*x^n) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt[b^2
- 4*a*c]*n) + ((b^2 - a*c)*Log[x])/a^3 - ((b^2 - a*c)*Log[a + b*x^n + c*x^(2*n)])/(2*a^3*n)

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 709

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^{-1-2 n}}{a+b x^n+c x^{2 n}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \left (a+b x+c x^2\right )} \, dx,x,x^n\right )}{n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{\operatorname{Subst}\left (\int \frac{-b-c x}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^n\right )}{a n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{\operatorname{Subst}\left (\int \left (-\frac{b}{a x^2}+\frac{b^2-a c}{a^2 x}+\frac{-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^n\right )}{a n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{b x^{-n}}{a^2 n}+\frac{\left (b^2-a c\right ) \log (x)}{a^3}+\frac{\operatorname{Subst}\left (\int \frac{-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x}{a+b x+c x^2} \, dx,x,x^n\right )}{a^3 n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{b x^{-n}}{a^2 n}+\frac{\left (b^2-a c\right ) \log (x)}{a^3}-\frac{\left (b \left (b^2-3 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^n\right )}{2 a^3 n}-\frac{\left (b^2-a c\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^n\right )}{2 a^3 n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{b x^{-n}}{a^2 n}+\frac{\left (b^2-a c\right ) \log (x)}{a^3}-\frac{\left (b^2-a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^3 n}+\frac{\left (b \left (b^2-3 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^n\right )}{a^3 n}\\ &=-\frac{x^{-2 n}}{2 a n}+\frac{b x^{-n}}{a^2 n}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^3 \sqrt{b^2-4 a c} n}+\frac{\left (b^2-a c\right ) \log (x)}{a^3}-\frac{\left (b^2-a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^3 n}\\ \end{align*}

Mathematica [A]  time = 0.339076, size = 112, normalized size = 0.89 \[ \frac{-a^2 x^{-2 n}-\left (b^2-a c\right ) \log \left (a+x^n \left (b+c x^n\right )\right )+\frac{2 b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}+2 n \log (x) \left (b^2-a c\right )+2 a b x^{-n}}{2 a^3 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.135, size = 958, normalized size = 7.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/a^2/n/(x^n)-1/2/a/n/(x^n)^2-4/(4*a^4*c*n^2-a^3*b^2*n^2)*n^2*ln(x)*a^2*c^2+5/(4*a^4*c*n^2-a^3*b^2*n^2)*n^2*ln
(x)*a*b^2*c-1/(4*a^4*c*n^2-a^3*b^2*n^2)*n^2*ln(x)*b^4+2/a/(4*a*c-b^2)/n*ln(x^n+1/2*(3*a*b^2*c-b^4+(-36*a^3*b^2
*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*c^2-5/2/a^2/(4*a*c-b^2)/n*ln(x^n+1/2*(3*a*b^2*c-b^
4+(-36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*b^2*c+1/2/a^3/(4*a*c-b^2)/n*ln(x^n+1
/2*(3*a*b^2*c-b^4+(-36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*b^4+1/2/a^3/(4*a*c-b
^2)/n*ln(x^n+1/2*(3*a*b^2*c-b^4+(-36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*(-36*a
^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2)+2/a/(4*a*c-b^2)/n*ln(x^n-1/2*(-3*a*b^2*c+b^4+(-36*a^3*b^2*c^3+
33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*c^2-5/2/a^2/(4*a*c-b^2)/n*ln(x^n-1/2*(-3*a*b^2*c+b^4+(-
36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*b^2*c+1/2/a^3/(4*a*c-b^2)/n*ln(x^n-1/2*(
-3*a*b^2*c+b^4+(-36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*b^4-1/2/a^3/(4*a*c-b^2)
/n*ln(x^n-1/2*(-3*a*b^2*c+b^4+(-36*a^3*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2))/c/b/(3*a*c-b^2))*(-36*a^3
*b^2*c^3+33*a^2*b^4*c^2-10*a*b^6*c+b^8)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, b x^{n} - a}{2 \, a^{2} n x^{2 \, n}} + \int \frac{b c x^{n} + b^{2} - a c}{a^{2} c x x^{2 \, n} + a^{2} b x x^{n} + a^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^n - a)/(a^2*n*x^(2*n)) + integrate((b*c*x^n + b^2 - a*c)/(a^2*c*x*x^(2*n) + a^2*b*x*x^n + a^3*x), x
)

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Fricas [A]  time = 1.67542, size = 938, normalized size = 7.44 \begin{align*} \left [-\frac{a^{2} b^{2} - 4 \, a^{3} c - 2 \,{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} n x^{2 \, n} \log \left (x\right ) +{\left (b^{3} - 3 \, a b c\right )} \sqrt{b^{2} - 4 \, a c} x^{2 \, n} \log \left (\frac{2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \,{\left (b c - \sqrt{b^{2} - 4 \, a c} c\right )} x^{n} - \sqrt{b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) +{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2 \, n} \log \left (c x^{2 \, n} + b x^{n} + a\right ) - 2 \,{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{n}}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n x^{2 \, n}}, -\frac{a^{2} b^{2} - 4 \, a^{3} c - 2 \,{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} n x^{2 \, n} \log \left (x\right ) - 2 \,{\left (b^{3} - 3 \, a b c\right )} \sqrt{-b^{2} + 4 \, a c} x^{2 \, n} \arctan \left (-\frac{2 \, \sqrt{-b^{2} + 4 \, a c} c x^{n} + \sqrt{-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) +{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2 \, n} \log \left (c x^{2 \, n} + b x^{n} + a\right ) - 2 \,{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{n}}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n x^{2 \, n}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^2*b^2 - 4*a^3*c - 2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*n*x^(2*n)*log(x) + (b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c
)*x^(2*n)*log((2*c^2*x^(2*n) + b^2 - 2*a*c + 2*(b*c - sqrt(b^2 - 4*a*c)*c)*x^n - sqrt(b^2 - 4*a*c)*b)/(c*x^(2*
n) + b*x^n + a)) + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*x^(2*n)*log(c*x^(2*n) + b*x^n + a) - 2*(a*b^3 - 4*a^2*b*c)*x^
n)/((a^3*b^2 - 4*a^4*c)*n*x^(2*n)), -1/2*(a^2*b^2 - 4*a^3*c - 2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*n*x^(2*n)*log(x)
- 2*(b^3 - 3*a*b*c)*sqrt(-b^2 + 4*a*c)*x^(2*n)*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n + sqrt(-b^2 + 4*a*c)*b)/(b
^2 - 4*a*c)) + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*x^(2*n)*log(c*x^(2*n) + b*x^n + a) - 2*(a*b^3 - 4*a^2*b*c)*x^n)/(
(a^3*b^2 - 4*a^4*c)*n*x^(2*n))]

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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(x**(-1-2*n)/(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{x^{-2 \, n - 1}}{c x^{2 \, n} + b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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