Integrand size = 24, antiderivative size = 98 \[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=-\frac {x^{-n}}{a n}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} n}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 a^2 n} \]
-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)*ar ctanh((b+2*c*x^n)/(-4*a*c+b^2)^(1/2))/a^2/n/(-4*a*c+b^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\frac {-2 a x^{-n}+\frac {2 \left (b^2-2 a c\right ) \arctan \left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 b \log \left (x^n\right )+b \log \left (a+x^n \left (b+c x^n\right )\right )}{2 a^2 n} \]
((-2*a)/x^n + (2*(b^2 - 2*a*c)*ArcTan[(b + 2*c*x^n)/Sqrt[-b^2 + 4*a*c]])/S qrt[-b^2 + 4*a*c] - 2*b*Log[x^n] + b*Log[a + x^n*(b + c*x^n)])/(2*a^2*n)
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1693, 1145, 25, 1200, 2009}
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.
\(\displaystyle \int \frac {x^{-n-1}}{a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {\int \frac {x^{-2 n}}{b x^n+c x^{2 n}+a}dx^n}{n}\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {\frac {\int -\frac {x^{-n} \left (c x^n+b\right )}{b x^n+c x^{2 n}+a}dx^n}{a}-\frac {x^{-n}}{a}}{n}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {x^{-n} \left (c x^n+b\right )}{b x^n+c x^{2 n}+a}dx^n}{a}-\frac {x^{-n}}{a}}{n}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {-\frac {\int \left (\frac {b x^{-n}}{a}+\frac {-b c x^n-b^2+a c}{a \left (b x^n+c x^{2 n}+a\right )}\right )dx^n}{a}-\frac {x^{-n}}{a}}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 a}+\frac {b \log \left (x^n\right )}{a}}{a}-\frac {x^{-n}}{a}}{n}\) |
(-(1/(a*x^n)) - (((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/ (a*Sqrt[b^2 - 4*a*c]) + (b*Log[x^n])/a - (b*Log[a + b*x^n + c*x^(2*n)])/(2 *a))/a)/n
3.6.53.3.1 Defintions of rubi rules used
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] + Simp [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}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
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. \(657\) vs. \(2(92)=184\).
Time = 0.25 (sec) , antiderivative size = 658, normalized size of antiderivative = 6.71
method | result | size |
risch | \(-\frac {x^{-n}}{a n}-\frac {4 n^{2} \ln \left (x \right ) a b c}{4 a^{3} c \,n^{2}-a^{2} b^{2} n^{2}}+\frac {n^{2} \ln \left (x \right ) b^{3}}{4 a^{3} c \,n^{2}-a^{2} b^{2} n^{2}}+\frac {2 \ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b c}{a \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 a^{2} \left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 a^{2} \left (4 a c -b^{2}\right ) n}+\frac {2 \ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b c}{a \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 a^{2} \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 a^{2} \left (4 a c -b^{2}\right ) n}\) | \(658\) |
-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)
Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.40 \[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\left [-\frac {2 \, {\left (b^{3} - 4 \, a b c\right )} n x^{n} \log \left (x\right ) + {\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x^{n} \log \left (\frac {2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} x^{n} + \sqrt {b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x^{n} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} n x^{n}}, -\frac {2 \, {\left (b^{3} - 4 \, a b c\right )} n x^{n} \log \left (x\right ) + 2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x^{n} \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c x^{n} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x^{n} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} n x^{n}}\right ] \]
[-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 + s qrt(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)]
Timed out. \[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{-n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
\[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{-n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {x^{-1-n}}{a+b x^n+c x^{2 n}} \, dx=\int \frac {1}{x^{n+1}\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \]