\(\int \frac {1}{x (a+b x^n+c x^{2 n})} \, dx\) [565]

Optimal result

Integrand size = 20, antiderivative size = 74 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} n}+\frac {\log (x)}{a}-\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 a n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1371, 719, 29, 648, 632, 212, 642} \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{a n \sqrt {b^2-4 a c}}-\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 a n}+\frac {\log (x)}{a} \]

[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 29

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

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 719

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 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \left (a+b x+c x^2\right )} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,x^n\right )}{a n}+\frac {\text {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 {\text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^n\right )}{2 a n}-\frac {b \text {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 \text {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] (verified)

Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=-\frac {\frac {2 b \arctan \left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 \log \left (x^n\right )+\log \left (a+x^n \left (b+c x^n\right )\right )}{2 a n} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(396\) vs. \(2(68)=136\).

Time = 0.19 (sec) , antiderivative size = 397, normalized size of antiderivative = 5.36

method	result	size
risch	\(\frac {4 n^{2} \ln \left (x \right ) a c}{4 a^{2} c \,n^{2}-a \,b^{2} n^{2}}-\frac {n^{2} \ln \left (x \right ) b^{2}}{4 a^{2} c \,n^{2}-a \,b^{2} n^{2}}-\frac {2 \ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) c}{\left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 a \left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 a \left (4 a c -b^{2}\right ) n}-\frac {2 \ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) c}{\left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 a \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 a \left (4 a c -b^{2}\right ) n}\)	\(397\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.50 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\left [\frac {2 \, {\left (b^{2} - 4 \, a c\right )} n \log \left (x\right ) + \sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} x^{n} + \sqrt {b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} n}, \frac {2 \, {\left (b^{2} - 4 \, a c\right )} n \log \left (x\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c x^{n} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} n}\right ] \]

[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)]

Timed out. \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 8.49 (sec) , antiderivative size = 224, normalized size of antiderivative = 3.03 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {\ln \left (-\frac {1}{c\,x}-\frac {\left (2\,a\,n+b\,n\,x^n\right )\,\left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2\right )}{2\,c\,x\,\left (a\,b^2\,n-4\,a^2\,c\,n\right )}\right )\,\left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2\right )}{2\,\left (a\,b^2\,n-4\,a^2\,c\,n\right )}-\frac {\ln \left (\frac {\left (2\,a\,n+b\,n\,x^n\right )\,\left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2\right )}{2\,c\,x\,\left (a\,b^2\,n-4\,a^2\,c\,n\right )}-\frac {1}{c\,x}\right )\,\left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2\right )}{2\,\left (a\,b^2\,n-4\,a^2\,c\,n\right )}+\frac {\ln \left (x\right )\,\left (n-1\right )}{a\,n} \]

[In]

int(1/(x*(a + b*x^n + c*x^(2*n))),x)

[Out]

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