Integrand size = 24, antiderivative size = 87 \[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\frac {x^n}{c n}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} n}-\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 c^2 n} \]
x^n/c/n-1/2*b*ln(a+b*x^n+c*x^(2*n))/c^2/n-(-2*a*c+b^2)*arctanh((b+2*c*x^n) /(-4*a*c+b^2)^(1/2))/c^2/n/(-4*a*c+b^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\frac {2 c x^n+\frac {2 \left (b^2-2 a c\right ) \arctan \left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-b \log \left (a+x^n \left (b+c x^n\right )\right )}{2 c^2 n} \]
(2*c*x^n + (2*(b^2 - 2*a*c)*ArcTan[(b + 2*c*x^n)/Sqrt[-b^2 + 4*a*c]])/Sqrt [-b^2 + 4*a*c] - b*Log[a + x^n*(b + c*x^n)])/(2*c^2*n)
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1693, 1143, 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.
\(\displaystyle \int \frac {x^{3 n-1}}{a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {\int \frac {x^{2 n}}{b x^n+c x^{2 n}+a}dx^n}{n}\) |
\(\Big \downarrow \) 1143 |
\(\displaystyle \frac {\int \left (\frac {1}{c}-\frac {b x^n+a}{c \left (b x^n+c x^{2 n}+a\right )}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 c^2}+\frac {x^n}{c}}{n}\) |
(x^n/c - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqr t[b^2 - 4*a*c]) - (b*Log[a + b*x^n + c*x^(2*n)])/(2*c^2))/n
3.6.50.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 1]
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. \(663\) vs. \(2(81)=162\).
Time = 0.29 (sec) , antiderivative size = 664, normalized size of antiderivative = 7.63
method | result | size |
risch | \(-\frac {b \ln \left (x \right )}{c^{2}}+\frac {x^{n}}{c n}+\frac {4 n^{2} \ln \left (x \right ) a b c}{4 a \,c^{3} n^{2}-b^{2} c^{2} n^{2}}-\frac {n^{2} \ln \left (x \right ) b^{3}}{4 a \,c^{3} n^{2}-b^{2} c^{2} n^{2}}-\frac {2 \ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) a b}{\left (4 a c -b^{2}\right ) c n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 \left (4 a c -b^{2}\right ) c^{2} n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 \left (4 a c -b^{2}\right ) c^{2} n}-\frac {2 \ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) a b}{\left (4 a c -b^{2}\right ) c n}+\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 \left (4 a c -b^{2}\right ) c^{2} n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 c^{3} a^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 \left (4 a c -b^{2}\right ) c^{2} n}\) | \(664\) |
-b/c^2*ln(x)+x^n/c/n+4/(4*a*c^3*n^2-b^2*c^2*n^2)*n^2*ln(x)*a*b*c-1/(4*a*c^ 3*n^2-b^2*c^2*n^2)*n^2*ln(x)*b^3-2/(4*a*c-b^2)/c/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))*a*b+1/2 /(4*a*c-b^2)/c^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/(4*a*c-b^2)/c^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/(4*a*c-b^2)/c/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))*a*b+1/2/(4*a*c-b^2)/c^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/(4*a*c-b^2)/c^ 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 = 0.27 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.28 \[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} x^{n} + \sqrt {b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{n} + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} n}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c x^{n} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{n} + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} n}\right ] \]
[-1/2*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*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)) - 2*(b^2*c - 4*a*c^2)*x^n + (b^3 - 4*a*b*c)*log(c*x^(2*n) + b*x^n + a))/((b^2*c^2 - 4*a*c^3)*n), -1/2*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*ar ctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) - 2*(b^2*c - 4*a*c^2)*x^n + (b^3 - 4*a*b*c)*log(c*x^(2*n) + b*x^n + a))/((b ^2*c^2 - 4*a*c^3)*n)]
Timed out. \[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{3 \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
-b*log(x)/c^2 + x^n/(c*n) - integrate(-(a*b + (b^2 - a*c)*x^n)/(c^3*x*x^(2 *n) + b*c^2*x*x^n + a*c^2*x), x)
\[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{3 \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx=\int \frac {x^{3\,n-1}}{a+b\,x^n+c\,x^{2\,n}} \,d x \]