3.6.53 \(\int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx\) [553]

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

[Out]

-1/a/n/(x^n)-b*ln(x)/a^2+1/2*b*ln(a+b*x^n+c*x^(2*n))/a^2/n-(-2*a*c+b^2)*arctanh((b+2*c*x^n)/(-4*a*c+b^2)^(1/2)
)/a^2/n/(-4*a*c+b^2)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {1371, 723, 814, 648, 632, 212, 642} \begin {gather*} -\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} \end {gather*}

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 212

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

Rule 632

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 642

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]

Rule 648

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 723

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 814

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 1371

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]]

Rubi steps

\begin {align*} \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.20, size = 90, normalized size = 0.92 \begin {gather*} \frac {-2 a x^{-n}+\frac {2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 b \log \left (x^n\right )+b \log \left (a+x^n \left (b+c x^n\right )\right )}{2 a^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(657\) vs. \(2(92)=184\).
time = 0.14, size = 658, normalized size = 6.71

method	result	size
risch	\(-\frac {x^{-n}}{a n}-\frac {4 n^{2} \ln \left (x \right ) a b c}{4 a^{3} c \,n^{2}-a^{2} b^{2} n^{2}}+\frac {n^{2} \ln \left (x \right ) b^{3}}{4 a^{3} c \,n^{2}-a^{2} b^{2} n^{2}}+\frac {2 \ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b c}{a \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 a^{2} \left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 a^{2} \left (4 a c -b^{2}\right ) n}+\frac {2 \ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b c}{a \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 a^{2} \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 a^{2} \left (4 a c -b^{2}\right ) n}\)	\(658\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.38, size = 333, normalized size = 3.40 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{n+1}\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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