Integrand size = 32, antiderivative size = 190 \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=-\frac {x^{-2 n} \left (a+b x^n\right )}{2 a n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {x^{-n} \left (a b+b^2 x^n\right )}{a^2 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {\left (a b^2+b^3 x^n\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a b^2+b^3 x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \] Output:
-1/2*(a+b*x^n)/a/n/(x^(2*n))/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)+(a*b+b^2*x^ n)/a^2/n/(x^n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)+(a*b^2+b^3*x^n)*ln(x)/a^3 /(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)-(a*b^2+b^3*x^n)*ln(a+b*x^n)/a^3/n/(a^2+ 2*a*b*x^n+b^2*x^(2*n))^(1/2)
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.41 \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=-\frac {x^{-2 n} \left (a+b x^n\right ) \left (a \left (a-2 b x^n\right )-2 b^2 x^{2 n} \log \left (x^n\right )+2 b^2 x^{2 n} \log \left (a+b x^n\right )\right )}{2 a^3 n \sqrt {\left (a+b x^n\right )^2}} \] Input:
Integrate[x^(-1 - 2*n)/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]
Output:
-1/2*((a + b*x^n)*(a*(a - 2*b*x^n) - 2*b^2*x^(2*n)*Log[x^n] + 2*b^2*x^(2*n )*Log[a + b*x^n]))/(a^3*n*x^(2*n)*Sqrt[(a + b*x^n)^2])
Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.48, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1384, 798, 27, 54, 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.
\(\displaystyle \int \frac {x^{-2 n-1}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \int \frac {x^{-2 n-1}}{b^2 x^n+a b}dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \int \frac {x^{-3 n}}{b \left (b x^n+a\right )}dx^n}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \int \frac {x^{-3 n}}{b x^n+a}dx^n}{b n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \int \left (\frac {x^{-3 n}}{a}-\frac {b x^{-2 n}}{a^2}+\frac {b^2 x^{-n}}{a^3}-\frac {b^3}{a^3 \left (b x^n+a\right )}\right )dx^n}{b n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \left (\frac {b^2 \log \left (x^n\right )}{a^3}-\frac {b^2 \log \left (a+b x^n\right )}{a^3}+\frac {b x^{-n}}{a^2}-\frac {x^{-2 n}}{2 a}\right )}{b n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
Input:
Int[x^(-1 - 2*n)/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]
Output:
((a*b + b^2*x^n)*(-1/2*1/(a*x^(2*n)) + b/(a^2*x^n) + (b^2*Log[x^n])/a^3 - (b^2*Log[a + b*x^n])/a^3))/(b*n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,x^{-n}}{\left (a +b \,x^{n}\right ) a^{2} n}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, x^{-2 n}}{2 \left (a +b \,x^{n}\right ) a n}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} \ln \left (x \right )}{\left (a +b \,x^{n}\right ) a^{3}}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} \ln \left (x^{n}+\frac {a}{b}\right )}{\left (a +b \,x^{n}\right ) a^{3} n}\) | \(138\) |
Input:
int(x^(-1-2*n)/(a^2+2*x^n*a*b+b^2*x^(2*n))^(1/2),x,method=_RETURNVERBOSE)
Output:
((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b/a^2/n/(x^n)-1/2*((a+b*x^n)^2)^(1/2)/(a+b*x ^n)/a/n/(x^n)^2+((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^2/a^3*ln(x)-((a+b*x^n)^2)^ (1/2)/(a+b*x^n)*b^2/a^3/n*ln(x^n+a/b)
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.31 \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {2 \, b^{2} n x^{2 \, n} \log \left (x\right ) - 2 \, b^{2} x^{2 \, n} \log \left (b x^{n} + a\right ) + 2 \, a b x^{n} - a^{2}}{2 \, a^{3} n x^{2 \, n}} \] Input:
integrate(x^(-1-2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="frica s")
Output:
1/2*(2*b^2*n*x^(2*n)*log(x) - 2*b^2*x^(2*n)*log(b*x^n + a) + 2*a*b*x^n - a ^2)/(a^3*n*x^(2*n))
\[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {x^{- 2 n - 1}}{\sqrt {\left (a + b x^{n}\right )^{2}}}\, dx \] Input:
integrate(x**(-1-2*n)/(a**2+2*a*b*x**n+b**2*x**(2*n))**(1/2),x)
Output:
Integral(x**(-2*n - 1)/sqrt((a + b*x**n)**2), x)
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.31 \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {b^{2} \log \left (x\right )}{a^{3}} - \frac {b^{2} \log \left (\frac {b x^{n} + a}{b}\right )}{a^{3} n} + \frac {2 \, b x^{n} - a}{2 \, a^{2} n x^{2 \, n}} \] Input:
integrate(x^(-1-2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="maxim a")
Output:
b^2*log(x)/a^3 - b^2*log((b*x^n + a)/b)/(a^3*n) + 1/2*(2*b*x^n - a)/(a^2*n *x^(2*n))
\[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {x^{-2 \, n - 1}}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \] Input:
integrate(x^(-1-2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="giac" )
Output:
integrate(x^(-2*n - 1)/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
Timed out. \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {1}{x^{2\,n+1}\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n}} \,d x \] Input:
int(1/(x^(2*n + 1)*(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(1/2)),x)
Output:
int(1/(x^(2*n + 1)*(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(1/2)), x)
Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.31 \[ \int \frac {x^{-1-2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {-2 x^{2 n} \mathrm {log}\left (x^{n} b +a \right ) b^{2}+2 x^{2 n} \mathrm {log}\left (x \right ) b^{2} n +2 x^{n} a b -a^{2}}{2 x^{2 n} a^{3} n} \] Input:
int(x^(-1-2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x)
Output:
( - 2*x**(2*n)*log(x**n*b + a)*b**2 + 2*x**(2*n)*log(x)*b**2*n + 2*x**n*a* b - a**2)/(2*x**(2*n)*a**3*n)