3.555 \(\int \frac{x^{-1-3 n}}{a+b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^4 n \sqrt{b^2-4 a c}}-\frac{x^{-n} \left (b^2-a c\right )}{a^3 n}+\frac{b \left (b^2-2 a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^4 n}-\frac{b \log (x) \left (b^2-2 a c\right )}{a^4}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{x^{-3 n}}{3 a n} \]

[Out]

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

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Rubi [A]  time = 0.231263, antiderivative size = 164, 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} \[ -\frac{\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^4 n \sqrt{b^2-4 a c}}-\frac{x^{-n} \left (b^2-a c\right )}{a^3 n}+\frac{b \left (b^2-2 a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^4 n}-\frac{b \log (x) \left (b^2-2 a c\right )}{a^4}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{x^{-3 n}}{3 a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*a*n*x^(3*n)) + b/(2*a^2*n*x^(2*n)) - (b^2 - a*c)/(a^3*n*x^n) - ((b^4 - 4*a*b^2*c + 2*a^2*c^2)*ArcTanh[(b
+ 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(a^4*Sqrt[b^2 - 4*a*c]*n) - (b*(b^2 - 2*a*c)*Log[x])/a^4 + (b*(b^2 - 2*a*c)*Lo
g[a + b*x^n + c*x^(2*n)])/(2*a^4*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^{-1-3 n}}{a+b x^n+c x^{2 n}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x+c x^2\right )} \, dx,x,x^n\right )}{n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{\operatorname{Subst}\left (\int \frac{-b-c x}{x^3 \left (a+b x+c x^2\right )} \, dx,x,x^n\right )}{a n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{\operatorname{Subst}\left (\int \left (-\frac{b}{a x^3}+\frac{b^2-a c}{a^2 x^2}+\frac{-b^3+2 a b c}{a^3 x}+\frac{b^4-3 a b^2 c+a^2 c^2+b c \left (b^2-2 a c\right ) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^n\right )}{a n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{\left (b^2-a c\right ) x^{-n}}{a^3 n}-\frac{b \left (b^2-2 a c\right ) \log (x)}{a^4}+\frac{\operatorname{Subst}\left (\int \frac{b^4-3 a b^2 c+a^2 c^2+b c \left (b^2-2 a c\right ) x}{a+b x+c x^2} \, dx,x,x^n\right )}{a^4 n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{\left (b^2-a c\right ) x^{-n}}{a^3 n}-\frac{b \left (b^2-2 a c\right ) \log (x)}{a^4}+\frac{\left (b \left (b^2-2 a c\right )\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^n\right )}{2 a^4 n}+\frac{\left (b^4-4 a b^2 c+2 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^n\right )}{2 a^4 n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{\left (b^2-a c\right ) x^{-n}}{a^3 n}-\frac{b \left (b^2-2 a c\right ) \log (x)}{a^4}+\frac{b \left (b^2-2 a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^4 n}-\frac{\left (b^4-4 a b^2 c+2 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^n\right )}{a^4 n}\\ &=-\frac{x^{-3 n}}{3 a n}+\frac{b x^{-2 n}}{2 a^2 n}-\frac{\left (b^2-a c\right ) x^{-n}}{a^3 n}-\frac{\left (b^4-4 a b^2 c+2 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^4 \sqrt{b^2-4 a c} n}-\frac{b \left (b^2-2 a c\right ) \log (x)}{a^4}+\frac{b \left (b^2-2 a c\right ) \log \left (a+b x^n+c x^{2 n}\right )}{2 a^4 n}\\ \end{align*}

Mathematica [A]  time = 0.457669, size = 143, normalized size = 0.87 \[ \frac{-\frac{6 \left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}+3 a^2 b x^{-2 n}-2 a^3 x^{-3 n}+6 a x^{-n} \left (a c-b^2\right )+3 b \left (b^2-2 a c\right ) \log \left (a+x^n \left (b+c x^n\right )\right )-6 b n \log (x) \left (b^2-2 a c\right )}{6 a^4 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a^3)/x^(3*n) + (3*a^2*b)/x^(2*n) + (6*a*(-b^2 + a*c))/x^n - (6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*ArcTanh[(b +
2*c*x^n)/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c] - 6*b*(b^2 - 2*a*c)*n*Log[x] + 3*b*(b^2 - 2*a*c)*Log[a + x^n*(
b + c*x^n)])/(6*a^4*n)

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Maple [B]  time = 0.16, size = 1300, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2/n/(x^n)*c-1/a^3/n/(x^n)*b^2+1/2*b/a^2/n/(x^n)^2-1/3/a/n/(x^n)^3+8/(4*a^5*c*n^2-a^4*b^2*n^2)*n^2*ln(x)*a^
2*b*c^2-6/(4*a^5*c*n^2-a^4*b^2*n^2)*n^2*ln(x)*a*b^3*c+1/(4*a^5*c*n^2-a^4*b^2*n^2)*n^2*ln(x)*b^5-4/a^2/(4*a*c-b
^2)/n*ln(x^n+1/2*(2*a^2*b*c^2-4*a*b^3*c+b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8
*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*b*c^2+3/a^3/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a^2*b*c^2-4*a*b^3*c+b^5+(
-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*
b^3*c-1/2/a^4/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a^2*b*c^2-4*a*b^3*c+b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+5
2*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*b^5+1/2/a^4/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a^2
*b*c^2-4*a*b^3*c+b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2))/c/(2*a^
2*c^2-4*a*b^2*c+b^4))*(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2)-4/a^2/(
4*a*c-b^2)/n*ln(x^n-1/2*(-2*a^2*b*c^2+4*a*b^3*c-b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-
12*a*b^8*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*b*c^2+3/a^3/(4*a*c-b^2)/n*ln(x^n-1/2*(-2*a^2*b*c^2+4*a*b^
3*c-b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2
*c+b^4))*b^3*c-1/2/a^4/(4*a*c-b^2)/n*ln(x^n-1/2*(-2*a^2*b*c^2+4*a*b^3*c-b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3
*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*b^5-1/2/a^4/(4*a*c-b^2)/n*ln(x^n-
1/2*(-2*a^2*b*c^2+4*a*b^3*c-b^5+(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1/
2))/c/(2*a^2*c^2-4*a*b^2*c+b^4))*(-16*a^5*c^5+68*a^4*b^2*c^4-96*a^3*b^4*c^3+52*a^2*b^6*c^2-12*a*b^8*c+b^10)^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 \, a b x^{n} - 2 \, a^{2} - 6 \,{\left (b^{2} - a c\right )} x^{2 \, n}}{6 \, a^{3} n x^{3 \, n}} + \int -\frac{b^{3} - 2 \, a b c +{\left (b^{2} c - a c^{2}\right )} x^{n}}{a^{3} c x x^{2 \, n} + a^{3} b x x^{n} + a^{4} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a*b*x^n - 2*a^2 - 6*(b^2 - a*c)*x^(2*n))/(a^3*n*x^(3*n)) + integrate(-(b^3 - 2*a*b*c + (b^2*c - a*c^2)*
x^n)/(a^3*c*x*x^(2*n) + a^3*b*x*x^n + a^4*x), x)

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Fricas [A]  time = 1.70359, size = 1130, normalized size = 6.89 \begin{align*} \left [-\frac{2 \, a^{3} b^{2} - 8 \, a^{4} c + 6 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} n x^{3 \, n} \log \left (x\right ) - 3 \,{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt{b^{2} - 4 \, a c} x^{3 \, 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 ) - 3 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3 \, n} \log \left (c x^{2 \, n} + b x^{n} + a\right ) + 6 \,{\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2 \, n} - 3 \,{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{n}}{6 \,{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} n x^{3 \, n}}, -\frac{2 \, a^{3} b^{2} - 8 \, a^{4} c + 6 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} n x^{3 \, n} \log \left (x\right ) + 6 \,{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c} x^{3 \, 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 ) - 3 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3 \, n} \log \left (c x^{2 \, n} + b x^{n} + a\right ) + 6 \,{\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2 \, n} - 3 \,{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{n}}{6 \,{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} n x^{3 \, n}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(2*a^3*b^2 - 8*a^4*c + 6*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*n*x^(3*n)*log(x) - 3*(b^4 - 4*a*b^2*c + 2*a^2*c
^2)*sqrt(b^2 - 4*a*c)*x^(3*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)) - 3*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*x^(3*n)*log(c*x^(2*n) + b*x^n + a) +
6*(a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*x^(2*n) - 3*(a^2*b^3 - 4*a^3*b*c)*x^n)/((a^4*b^2 - 4*a^5*c)*n*x^(3*n)), -1
/6*(2*a^3*b^2 - 8*a^4*c + 6*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*n*x^(3*n)*log(x) + 6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)
*sqrt(-b^2 + 4*a*c)*x^(3*n)*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) - 3*(b^
5 - 6*a*b^3*c + 8*a^2*b*c^2)*x^(3*n)*log(c*x^(2*n) + b*x^n + a) + 6*(a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*x^(2*n)
- 3*(a^2*b^3 - 4*a^3*b*c)*x^n)/((a^4*b^2 - 4*a^5*c)*n*x^(3*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**(-1-3*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^{-3 \, 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^(-1-3*n)/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

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