Integrand size = 26, antiderivative size = 211 \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {a^3 x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 \left (a+b x^n\right )}+\frac {3 a b^3 x^{2 (1+n)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 (1+n) \left (a b+b^2 x^n\right )}+\frac {3 a^2 b^2 x^{2+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2+n) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{2+3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2+3 n) \left (a b+b^2 x^n\right )} \]
1/2*a^3*x^2*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(a+b*x^n)+3/2*a*b^3*x^(2+2*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^(2+n)* (a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(2+n)/(a*b+b^2*x^n)+b^4*x^(2+3*n)*(a^2+2 *a*b*x^n+b^2*x^(2*n))^(1/2)/(2+3*n)/(a*b+b^2*x^n)
Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.59 \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x^2 \sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (4+12 n+11 n^2+3 n^3\right )+6 a^2 b \left (2+5 n+3 n^2\right ) x^n+3 a b^2 \left (4+8 n+3 n^2\right ) x^{2 n}+2 b^3 \left (2+3 n+n^2\right ) x^{3 n}\right )}{2 (1+n) (2+n) (2+3 n) \left (a+b x^n\right )} \]
(x^2*Sqrt[(a + b*x^n)^2]*(a^3*(4 + 12*n + 11*n^2 + 3*n^3) + 6*a^2*b*(2 + 5 *n + 3*n^2)*x^n + 3*a*b^2*(4 + 8*n + 3*n^2)*x^(2*n) + 2*b^3*(2 + 3*n + n^2 )*x^(3*n)))/(2*(1 + n)*(2 + n)*(2 + 3*n)*(a + b*x^n))
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.52, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, 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.
\(\displaystyle \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x \left (b^2 x^n+a b\right )^3dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (3 a^2 b^4 x^{n+1}+3 a b^5 x^{2 n+1}+b^6 x^{3 n+1}+a^3 b^3 x\right )dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \left (\frac {1}{2} a^3 b^3 x^2+\frac {3 a^2 b^4 x^{n+2}}{n+2}+\frac {3 a b^5 x^{2 (n+1)}}{2 (n+1)}+\frac {b^6 x^{3 n+2}}{3 n+2}\right )}{a b^3+b^4 x^n}\) |
(Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*((a^3*b^3*x^2)/2 + (3*a*b^5*x^(2*(1 + n)))/(2*(1 + n)) + (3*a^2*b^4*x^(2 + n))/(2 + n) + (b^6*x^(2 + 3*n))/(2 + 3*n)))/(a*b^3 + b^4*x^n)
3.6.24.3.1 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 = 145, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} x^{2}}{2 a +2 b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{2} x^{3 n}}{\left (a +b \,x^{n}\right ) \left (2+3 n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,b^{2} x^{2} x^{2 n}}{2 \left (a +b \,x^{n}\right ) \left (1+n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b \,x^{2} x^{n}}{\left (a +b \,x^{n}\right ) \left (2+n \right )}\) | \(145\) |
1/2*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^3*x^2+((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^ 3/(2+3*n)*x^2*(x^n)^3+3/2*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a*b^2*x^2/(1+n)*(x ^n)^2+3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^2*b/(2+n)*x^2*x^n
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.69 \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {2 \, {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{2} x^{3 \, n} + 3 \, {\left (3 \, a b^{2} n^{2} + 8 \, a b^{2} n + 4 \, a b^{2}\right )} x^{2} x^{2 \, n} + 6 \, {\left (3 \, a^{2} b n^{2} + 5 \, a^{2} b n + 2 \, a^{2} b\right )} x^{2} x^{n} + {\left (3 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 12 \, a^{3} n + 4 \, a^{3}\right )} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \]
1/2*(2*(b^3*n^2 + 3*b^3*n + 2*b^3)*x^2*x^(3*n) + 3*(3*a*b^2*n^2 + 8*a*b^2* n + 4*a*b^2)*x^2*x^(2*n) + 6*(3*a^2*b*n^2 + 5*a^2*b*n + 2*a^2*b)*x^2*x^n + (3*a^3*n^3 + 11*a^3*n^2 + 12*a^3*n + 4*a^3)*x^2)/(3*n^3 + 11*n^2 + 12*n + 4)
\[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int x \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.52 \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {2 \, {\left (n^{2} + 3 \, n + 2\right )} b^{3} x^{2} x^{3 \, n} + 3 \, {\left (3 \, n^{2} + 8 \, n + 4\right )} a b^{2} x^{2} x^{2 \, n} + 6 \, {\left (3 \, n^{2} + 5 \, n + 2\right )} a^{2} b x^{2} x^{n} + {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )} a^{3} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \]
1/2*(2*(n^2 + 3*n + 2)*b^3*x^2*x^(3*n) + 3*(3*n^2 + 8*n + 4)*a*b^2*x^2*x^( 2*n) + 6*(3*n^2 + 5*n + 2)*a^2*b*x^2*x^n + (3*n^3 + 11*n^2 + 12*n + 4)*a^3 *x^2)/(3*n^3 + 11*n^2 + 12*n + 4)
Time = 0.31 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.38 \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {2 \, b^{3} n^{2} x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, a^{3} n^{3} x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 6 \, b^{3} n x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 24 \, a b^{2} n x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 30 \, a^{2} b n x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 4 \, b^{3} x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a b^{2} x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a^{2} b x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a^{3} n x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 4 \, a^{3} x^{2} \mathrm {sgn}\left (b x^{n} + a\right )}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \]
1/2*(2*b^3*n^2*x^2*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2*n^2*x^2*x^(2*n)*sgn(b* x^n + a) + 18*a^2*b*n^2*x^2*x^n*sgn(b*x^n + a) + 3*a^3*n^3*x^2*sgn(b*x^n + a) + 6*b^3*n*x^2*x^(3*n)*sgn(b*x^n + a) + 24*a*b^2*n*x^2*x^(2*n)*sgn(b*x^ n + a) + 30*a^2*b*n*x^2*x^n*sgn(b*x^n + a) + 11*a^3*n^2*x^2*sgn(b*x^n + a) + 4*b^3*x^2*x^(3*n)*sgn(b*x^n + a) + 12*a*b^2*x^2*x^(2*n)*sgn(b*x^n + a) + 12*a^2*b*x^2*x^n*sgn(b*x^n + a) + 12*a^3*n*x^2*sgn(b*x^n + a) + 4*a^3*x^ 2*sgn(b*x^n + a))/(3*n^3 + 11*n^2 + 12*n + 4)
Timed out. \[ \int x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int x\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,d x \]