Integrand size = 28, antiderivative size = 212 \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {a^3 x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )}+\frac {b^4 x^{3 (1+n)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 (1+n) \left (a b+b^2 x^n\right )}+\frac {3 a^2 b^2 x^{3+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(3+n) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{3+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(3+2 n) \left (a b+b^2 x^n\right )} \]
1/3*a^3*x^3*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(a+b*x^n)+1/3*b^4*x^(3+3*n)* (a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(1+n)/(a*b+b^2*x^n)+3*a^2*b^2*x^(3+n)*(a ^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(3+n)/(a*b+b^2*x^n)+3*a*b^3*x^(3+2*n)*(a^2 +2*a*b*x^n+b^2*x^(2*n))^(1/2)/(3+2*n)/(a*b+b^2*x^n)
Time = 0.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.58 \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x^3 \sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (9+18 n+11 n^2+2 n^3\right )+9 a^2 b \left (3+5 n+2 n^2\right ) x^n+9 a b^2 \left (3+4 n+n^2\right ) x^{2 n}+b^3 \left (9+9 n+2 n^2\right ) x^{3 n}\right )}{3 (1+n) (3+n) (3+2 n) \left (a+b x^n\right )} \]
(x^3*Sqrt[(a + b*x^n)^2]*(a^3*(9 + 18*n + 11*n^2 + 2*n^3) + 9*a^2*b*(3 + 5 *n + 2*n^2)*x^n + 9*a*b^2*(3 + 4*n + n^2)*x^(2*n) + b^3*(9 + 9*n + 2*n^2)* x^(3*n)))/(3*(1 + n)*(3 + n)*(3 + 2*n)*(a + b*x^n))
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.52, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1384, 802, 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.
\(\displaystyle \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x^2 \left (b^2 x^n+a b\right )^3dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (3 a b^5 x^{2 (n+1)}+3 a^2 b^4 x^{n+2}+b^6 x^{3 n+2}+a^3 b^3 x^2\right )dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \left (\frac {1}{3} a^3 b^3 x^3+\frac {3 a^2 b^4 x^{n+3}}{n+3}+\frac {3 a b^5 x^{2 n+3}}{2 n+3}+\frac {b^6 x^{3 (n+1)}}{3 (n+1)}\right )}{a b^3+b^4 x^n}\) |
(Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*((a^3*b^3*x^3)/3 + (b^6*x^(3*(1 + n)) )/(3*(1 + n)) + (3*a^2*b^4*x^(3 + n))/(3 + n) + (3*a*b^5*x^(3 + 2*n))/(3 + 2*n)))/(a*b^3 + b^4*x^n)
3.6.23.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
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 = 146, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} x^{3}}{3 a +3 b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{3} x^{3 n}}{3 \left (a +b \,x^{n}\right ) \left (1+n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} a \,x^{3} x^{2 n}}{\left (a +b \,x^{n}\right ) \left (3+2 n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b \,x^{3} x^{n}}{\left (a +b \,x^{n}\right ) \left (3+n \right )}\) | \(146\) |
1/3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^3*x^3+1/3*((a+b*x^n)^2)^(1/2)/(a+b*x^n )*b^3*x^3/(1+n)*(x^n)^3+3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^2*a/(3+2*n)*x^3* (x^n)^2+3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^2*b/(3+n)*x^3*x^n
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (2 \, b^{3} n^{2} + 9 \, b^{3} n + 9 \, b^{3}\right )} x^{3} x^{3 \, n} + 9 \, {\left (a b^{2} n^{2} + 4 \, a b^{2} n + 3 \, a b^{2}\right )} x^{3} x^{2 \, n} + 9 \, {\left (2 \, a^{2} b n^{2} + 5 \, a^{2} b n + 3 \, a^{2} b\right )} x^{3} x^{n} + {\left (2 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 18 \, a^{3} n + 9 \, a^{3}\right )} x^{3}}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]
1/3*((2*b^3*n^2 + 9*b^3*n + 9*b^3)*x^3*x^(3*n) + 9*(a*b^2*n^2 + 4*a*b^2*n + 3*a*b^2)*x^3*x^(2*n) + 9*(2*a^2*b*n^2 + 5*a^2*b*n + 3*a^2*b)*x^3*x^n + ( 2*a^3*n^3 + 11*a^3*n^2 + 18*a^3*n + 9*a^3)*x^3)/(2*n^3 + 11*n^2 + 18*n + 9 )
\[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int x^{2} \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.51 \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (2 \, n^{2} + 9 \, n + 9\right )} b^{3} x^{3} x^{3 \, n} + 9 \, {\left (n^{2} + 4 \, n + 3\right )} a b^{2} x^{3} x^{2 \, n} + 9 \, {\left (2 \, n^{2} + 5 \, n + 3\right )} a^{2} b x^{3} x^{n} + {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )} a^{3} x^{3}}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]
1/3*((2*n^2 + 9*n + 9)*b^3*x^3*x^(3*n) + 9*(n^2 + 4*n + 3)*a*b^2*x^3*x^(2* n) + 9*(2*n^2 + 5*n + 3)*a^2*b*x^3*x^n + (2*n^3 + 11*n^2 + 18*n + 9)*a^3*x ^3)/(2*n^3 + 11*n^2 + 18*n + 9)
Time = 0.31 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.38 \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {2 \, b^{3} n^{2} x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 2 \, a^{3} n^{3} x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, b^{3} n x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 36 \, a b^{2} n x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 45 \, a^{2} b n x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, b^{3} x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 27 \, a b^{2} x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 27 \, a^{2} b x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{3} n x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a^{3} x^{3} \mathrm {sgn}\left (b x^{n} + a\right )}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]
1/3*(2*b^3*n^2*x^3*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2*n^2*x^3*x^(2*n)*sgn(b* x^n + a) + 18*a^2*b*n^2*x^3*x^n*sgn(b*x^n + a) + 2*a^3*n^3*x^3*sgn(b*x^n + a) + 9*b^3*n*x^3*x^(3*n)*sgn(b*x^n + a) + 36*a*b^2*n*x^3*x^(2*n)*sgn(b*x^ n + a) + 45*a^2*b*n*x^3*x^n*sgn(b*x^n + a) + 11*a^3*n^2*x^3*sgn(b*x^n + a) + 9*b^3*x^3*x^(3*n)*sgn(b*x^n + a) + 27*a*b^2*x^3*x^(2*n)*sgn(b*x^n + a) + 27*a^2*b*x^3*x^n*sgn(b*x^n + a) + 18*a^3*n*x^3*sgn(b*x^n + a) + 9*a^3*x^ 3*sgn(b*x^n + a))/(2*n^3 + 11*n^2 + 18*n + 9)
Timed out. \[ \int x^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int x^2\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,d x \]