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

Optimal. Leaf size=68 \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{c n \sqrt{b^2-4 a c}}+\frac{\log \left (a+b x^n+c x^{2 n}\right )}{2 c n} \]

[Out]

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

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Rubi [A]  time = 0.0505273, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1357, 634, 618, 206, 628} \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{c n \sqrt{b^2-4 a c}}+\frac{\log \left (a+b x^n+c x^{2 n}\right )}{2 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0673643, size = 62, normalized size = 0.91 \[ \frac{\frac{2 b \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}+\log \left (a+x^n \left (b+c x^n\right )\right )}{2 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 402, normalized size = 5.9 \begin{align*}{\frac{\ln \left ( x \right ) }{c}}-4\,{\frac{{n}^{2}\ln \left ( x \right ) ac}{4\,a{c}^{2}{n}^{2}-{b}^{2}c{n}^{2}}}+{\frac{{n}^{2}\ln \left ( x \right ){b}^{2}}{4\,a{c}^{2}{n}^{2}-{b}^{2}c{n}^{2}}}+2\,{\frac{a}{ \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-1/2\,{\frac{-{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}}{bc}} \right ) }-{\frac{{b}^{2}}{2\,c \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-{\frac{1}{2\,bc} \left ( -{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) }+{\frac{1}{2\,c \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}-{\frac{1}{2\,bc} \left ( -{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}}+2\,{\frac{a}{ \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+1/2\,{\frac{{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}}{bc}} \right ) }-{\frac{{b}^{2}}{2\,c \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+{\frac{1}{2\,bc} \left ({b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) }-{\frac{1}{2\,c \left ( 4\,ac-{b}^{2} \right ) n}\ln \left ({x}^{n}+{\frac{1}{2\,bc} \left ({b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}} \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]

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

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

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

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Fricas [A]  time = 1.62732, size = 517, normalized size = 7.6 \begin{align*} \left [\frac{\sqrt{b^{2} - 4 \, a c} b \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^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} n}, \frac{2 \, \sqrt{-b^{2} + 4 \, a c} b \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^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} 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*(sqrt(b^2 - 4*a*c)*b*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^2 - 4*a*c)*log(c*x^(2*n) + b*x^n + a))/((b^2*c - 4*a*c^2)*n), 1/2*(2*sqr
t(-b^2 + 4*a*c)*b*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) + (b^2 - 4*a*c)*l
og(c*x^(2*n) + b*x^n + a))/((b^2*c - 4*a*c^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)