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

Optimal. Leaf size=74 \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a n \sqrt{b^2-4 a c}}-\frac{\log \left (a+b x^n+c x^{2 n}\right )}{2 a n}+\frac{\log (x)}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0658304, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1357, 705, 29, 634, 618, 206, 628} \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^n}{\sqrt{b^2-4 a c}}\right )}{a n \sqrt{b^2-4 a c}}-\frac{\log \left (a+b x^n+c x^{2 n}\right )}{2 a n}+\frac{\log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^n + c*x^(2*n))),x]

[Out]

(b*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(a*Sqrt[b^2 - 4*a*c]*n) + Log[x]/a - Log[a + b*x^n + c*x^(2*n)]/(
2*a*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 705

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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

Mathematica [A]  time = 0.115239, size = 74, normalized size = 1. \[ -\frac{\frac{2 b \tan ^{-1}\left (\frac{b+2 c x^n}{\sqrt{4 a c-b^2}}\right )}{n \sqrt{4 a c-b^2}}+\frac{\log \left (a+x^n \left (b+c x^n\right )\right )}{n}-2 \log (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.066, size = 397, normalized size = 5.4 \begin{align*} 4\,{\frac{{n}^{2}\ln \left ( x \right ) ac}{4\,{a}^{2}c{n}^{2}-a{b}^{2}{n}^{2}}}-{\frac{{n}^{2}\ln \left ( x \right ){b}^{2}}{4\,{a}^{2}c{n}^{2}-a{b}^{2}{n}^{2}}}-2\,{\frac{c}{ \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\,a \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\,a \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{c}{ \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\,a \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\,a \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(1/x/(a+b*x^n+c*x^(2*n)),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.60941, size = 590, normalized size = 7.97 \begin{align*} \left [\frac{2 \,{\left (b^{2} - 4 \, a c\right )} n \log \left (x\right ) + \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 (a b^{2} - 4 \, a^{2} c\right )} n}, \frac{2 \,{\left (b^{2} - 4 \, a c\right )} n \log \left (x\right ) + 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 (a b^{2} - 4 \, a^{2} c\right )} n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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