Integrand size = 32, antiderivative size = 195 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=-\frac {a^3 x^{-n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a+b x^n\right )}+\frac {3 a b^3 x^n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac {b^4 x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac {3 a^2 b^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \log (x)}{a b+b^2 x^n} \] Output:
-a^3*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/n/(x^n)/(a+b*x^n)+3*a*b^3*x^n*(a^2+ 2*a*b*x^n+b^2*x^(2*n))^(1/2)/n/(a*b+b^2*x^n)+1/2*b^4*x^(2*n)*(a^2+2*a*b*x^ n+b^2*x^(2*n))^(1/2)/n/(a*b+b^2*x^n)+3*a^2*b^2*(a^2+2*a*b*x^n+b^2*x^(2*n)) ^(1/2)*ln(x)/(a*b+b^2*x^n)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.37 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x^{-n} \sqrt {\left (a+b x^n\right )^2} \left (-2 a^3+6 a b^2 x^{2 n}+b^3 x^{3 n}+6 a^2 b x^n \log \left (x^n\right )\right )}{2 n \left (a+b x^n\right )} \] Input:
Integrate[x^(-1 - n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]
Output:
(Sqrt[(a + b*x^n)^2]*(-2*a^3 + 6*a*b^2*x^(2*n) + b^3*x^(3*n) + 6*a^2*b*x^n *Log[x^n]))/(2*n*x^n*(a + b*x^n))
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.45, 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, 49, 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.
\(\displaystyle \int x^{-n-1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x^{-n-1} \left (b^2 x^n+a b\right )^3dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int b^3 x^{-2 n} \left (b x^n+a\right )^3dx^n}{n \left (a b^3+b^4 x^n\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x^{-2 n} \left (b x^n+a\right )^3dx^n}{n \left (a b^3+b^4 x^n\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {b^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (a^3 x^{-2 n}+3 a^2 b x^{-n}+b^3 x^n+3 a b^2\right )dx^n}{n \left (a b^3+b^4 x^n\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \left (-a^3 x^{-n}+3 a^2 b \log \left (x^n\right )+3 a b^2 x^n+\frac {1}{2} b^3 x^{2 n}\right )}{n \left (a b^3+b^4 x^n\right )}\) |
Input:
Int[x^(-1 - n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]
Output:
(b^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*(-(a^3/x^n) + 3*a*b^2*x^n + (b^3* x^(2*n))/2 + 3*a^2*b*Log[x^n]))/(n*(a*b^3 + b^4*x^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 [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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 = 0.03 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.66
method | result | size |
risch | \(\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,a^{2} \ln \left (x \right )}{a +b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{2 n}}{2 \left (a +b \,x^{n}\right ) n}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} a \,x^{n}}{\left (a +b \,x^{n}\right ) n}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} x^{-n}}{\left (a +b \,x^{n}\right ) n}\) | \(128\) |
Input:
int(x^(-1-n)*(a^2+2*x^n*a*b+b^2*x^(2*n))^(3/2),x,method=_RETURNVERBOSE)
Output:
3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b*a^2*ln(x)+1/2*((a+b*x^n)^2)^(1/2)/(a+b*x ^n)*b^3/n*(x^n)^2+3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^2*a/n*x^n-((a+b*x^n)^2 )^(1/2)/(a+b*x^n)*a^3/n/(x^n)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.25 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {6 \, a^{2} b n x^{n} \log \left (x\right ) + b^{3} x^{3 \, n} + 6 \, a b^{2} x^{2 \, n} - 2 \, a^{3}}{2 \, n x^{n}} \] Input:
integrate(x^(-1-n)*(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="fricas" )
Output:
1/2*(6*a^2*b*n*x^n*log(x) + b^3*x^(3*n) + 6*a*b^2*x^(2*n) - 2*a^3)/(n*x^n)
\[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int x^{- n - 1} \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**(-1-n)*(a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2),x)
Output:
Integral(x**(-n - 1)*((a + b*x**n)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.23 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=3 \, a^{2} b \log \left (x\right ) + \frac {b^{3} x^{3 \, n} + 6 \, a b^{2} x^{2 \, n} - 2 \, a^{3}}{2 \, n x^{n}} \] Input:
integrate(x^(-1-n)*(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="maxima" )
Output:
3*a^2*b*log(x) + 1/2*(b^3*x^(3*n) + 6*a*b^2*x^(2*n) - 2*a^3)/(n*x^n)
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.41 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {6 \, a^{2} b n x^{n} \log \left (x\right ) \mathrm {sgn}\left (b x^{n} + a\right ) + b^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 6 \, a b^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) - 2 \, a^{3} \mathrm {sgn}\left (b x^{n} + a\right )}{2 \, n x^{n}} \] Input:
integrate(x^(-1-n)*(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="giac")
Output:
1/2*(6*a^2*b*n*x^n*log(x)*sgn(b*x^n + a) + b^3*x^(3*n)*sgn(b*x^n + a) + 6* a*b^2*x^(2*n)*sgn(b*x^n + a) - 2*a^3*sgn(b*x^n + a))/(n*x^n)
Timed out. \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int \frac {{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}}{x^{n+1}} \,d x \] Input:
int((a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2)/x^(n + 1),x)
Output:
int((a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2)/x^(n + 1), x)
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.25 \[ \int x^{-1-n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x^{3 n} b^{3}+6 x^{2 n} a \,b^{2}+6 x^{n} \mathrm {log}\left (x \right ) a^{2} b n -2 a^{3}}{2 x^{n} n} \] Input:
int(x^(-1-n)*(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x)
Output:
(x**(3*n)*b**3 + 6*x**(2*n)*a*b**2 + 6*x**n*log(x)*a**2*b*n - 2*a**3)/(2*x **n*n)