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

Optimal. Leaf size=98 \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^2 n \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b x^n+c x^{2 n}\right )}{2 a^2 n}-\frac{b \log (x)}{a^2}-\frac{x^{-n}}{a n} \]

[Out]

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

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Rubi [A]  time = 0.126038, antiderivative size = 98, 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} \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a^2 n \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b x^n+c x^{2 n}\right )}{2 a^2 n}-\frac{b \log (x)}{a^2}-\frac{x^{-n}}{a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.614541, size = 135, normalized size = 1.38 \[ -\frac{-\frac{4 c^2 \log \left (x^{-n} \left (b-\sqrt{b^2-4 a c}\right )+2 c\right )}{\sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )^2}+\frac{4 c^2 \log \left (x^{-n} \left (\sqrt{b^2-4 a c}+b\right )+2 c\right )}{\sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )^2}+\frac{x^{-n}}{a}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.114, size = 658, normalized size = 6.7 \begin{align*} -{\frac{1}{an{x}^{n}}}-4\,{\frac{{n}^{2}\ln \left ( x \right ) abc}{4\,{a}^{3}c{n}^{2}-{a}^{2}{b}^{2}{n}^{2}}}+{\frac{{n}^{2}\ln \left ( x \right ){b}^{3}}{4\,{a}^{3}c{n}^{2}-{a}^{2}{b}^{2}{n}^{2}}}+2\,{\frac{bc}{a \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-1/2\,{\frac{-2\,abc+{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}}}{c \left ( 2\,ac-{b}^{2} \right ) }} \right ) }-{\frac{{b}^{3}}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-{\frac{1}{2\,c \left ( 2\,ac-{b}^{2} \right ) } \left ( -2\,abc+{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}} \right ) } \right ) }+{\frac{1}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-{\frac{1}{2\,c \left ( 2\,ac-{b}^{2} \right ) } \left ( -2\,abc+{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}} \right ) } \right ) \sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}}}+2\,{\frac{bc}{a \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+1/2\,{\frac{2\,abc-{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}}}{c \left ( 2\,ac-{b}^{2} \right ) }} \right ) }-{\frac{{b}^{3}}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+{\frac{1}{2\,c \left ( 2\,ac-{b}^{2} \right ) } \left ( 2\,abc-{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}} \right ) } \right ) }-{\frac{1}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+{\frac{1}{2\,c \left ( 2\,ac-{b}^{2} \right ) } \left ( 2\,abc-{b}^{3}+\sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}} \right ) } \right ) \sqrt{-16\,{a}^{3}{c}^{3}+20\,{a}^{2}{b}^{2}{c}^{2}-8\,a{b}^{4}c+{b}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/n/(x^n)-4/(4*a^3*c*n^2-a^2*b^2*n^2)*n^2*ln(x)*a*b*c+1/(4*a^3*c*n^2-a^2*b^2*n^2)*n^2*ln(x)*b^3+2/a/(4*a*c-
b^2)/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*b*c-1/2/a^2/(
4*a*c-b^2)/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*b^3+1/2
/a^2/(4*a*c-b^2)/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*(
-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2)+2/a/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a*b*c-b^3+(-16*a^3*c^3+20*a^2*
b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*b*c-1/2/a^2/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a*b*c-b^3+(-16*a^3*c^3+20
*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*b^3-1/2/a^2/(4*a*c-b^2)/n*ln(x^n+1/2*(2*a*b*c-b^3+(-16*a^3*c
^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70989, size = 738, normalized size = 7.53 \begin{align*} \left [-\frac{2 \,{\left (b^{3} - 4 \, a b c\right )} n x^{n} \log \left (x\right ) +{\left (b^{2} - 2 \, a c\right )} \sqrt{b^{2} - 4 \, a c} x^{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 ) + 2 \, a b^{2} - 8 \, a^{2} c -{\left (b^{3} - 4 \, a b c\right )} x^{n} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} n x^{n}}, -\frac{2 \,{\left (b^{3} - 4 \, a b c\right )} n x^{n} \log \left (x\right ) + 2 \,{\left (b^{2} - 2 \, a c\right )} \sqrt{-b^{2} + 4 \, a c} x^{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 ) + 2 \, a b^{2} - 8 \, a^{2} c -{\left (b^{3} - 4 \, a b c\right )} x^{n} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} n x^{n}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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