Integrand size = 24, antiderivative size = 68 \[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} n}+\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 c n} \]
1/2*ln(a+b*x^n+c*x^(2*n))/c/n+b*arctanh((b+2*c*x^n)/(-4*a*c+b^2)^(1/2))/c/ n/(-4*a*c+b^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\log \left (a+x^n \left (b+c x^n\right )\right )}{2 c n} \]
((-2*b*ArcTan[(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)
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1693, 1142, 1083, 219, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{2 n-1}}{a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {\int \frac {x^n}{b x^n+c x^{2 n}+a}dx^n}{n}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\frac {\int \frac {2 c x^n+b}{b x^n+c x^{2 n}+a}dx^n}{2 c}-\frac {b \int \frac {1}{b x^n+c x^{2 n}+a}dx^n}{2 c}}{n}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {b \int \frac {1}{-x^{2 n}+b^2-4 a c}d\left (2 c x^n+b\right )}{c}+\frac {\int \frac {2 c x^n+b}{b x^n+c x^{2 n}+a}dx^n}{2 c}}{n}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\int \frac {2 c x^n+b}{b x^n+c x^{2 n}+a}dx^n}{2 c}+\frac {b \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}}{n}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {b \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 c}}{n}\) |
((b*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c]) + Log[ a + b*x^n + c*x^(2*n)]/(2*c))/n
3.6.51.3.1 Defintions of rubi rules used
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(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] && IntegerQ [Simplify[(m + 1)/n]]
Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(62)=124\).
Time = 0.24 (sec) , antiderivative size = 402, normalized size of antiderivative = 5.91
method | result | size |
risch | \(\frac {\ln \left (x \right )}{c}-\frac {4 n^{2} \ln \left (x \right ) a c}{4 a \,c^{2} n^{2}-b^{2} c \,n^{2}}+\frac {n^{2} \ln \left (x \right ) b^{2}}{4 a \,c^{2} n^{2}-b^{2} c \,n^{2}}+\frac {2 \ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) a}{\left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 c \left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 c \left (4 a c -b^{2}\right ) n}+\frac {2 \ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) a}{\left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 c \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 c \left (4 a c -b^{2}\right ) n}\) | \(402\) |
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*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)
Time = 0.26 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.40 \[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} x^{n} + \sqrt {b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} n}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c x^{n} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} n}\right ] \]
[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*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))/((b^2*c - 4*a *c^2)*n)]
Timed out. \[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{2 \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
\[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{2 \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx=\int \frac {x^{2\,n-1}}{a+b\,x^n+c\,x^{2\,n}} \,d x \]