Integrand size = 24, antiderivative size = 206 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {a^3 x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{\left (a+b x^n\right )^3}+\frac {3 a^2 b^4 x^{1+n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+n) \left (a b+b^2 x^n\right )^3}+\frac {3 a b^5 x^{1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+2 n) \left (a b+b^2 x^n\right )^3}+\frac {b^6 x^{1+3 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+3 n) \left (a b+b^2 x^n\right )^3} \]
a^3*x*(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2)/(a+b*x^n)^3+3*a^2*b^4*x^(1+n)*(a^2 +2*a*b*x^n+b^2*x^(2*n))^(3/2)/(1+n)/(a*b+b^2*x^n)^3+3*a*b^5*x^(1+2*n)*(a^2 +2*a*b*x^n+b^2*x^(2*n))^(3/2)/(1+2*n)/(a*b+b^2*x^n)^3+b^6*x^(1+3*n)*(a^2+2 *a*b*x^n+b^2*x^(2*n))^(3/2)/(1+3*n)/(a*b+b^2*x^n)^3
Time = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.59 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x \sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (1+6 n+11 n^2+6 n^3\right )+3 a^2 b \left (1+5 n+6 n^2\right ) x^n+3 a b^2 \left (1+4 n+3 n^2\right ) x^{2 n}+b^3 \left (1+3 n+2 n^2\right ) x^{3 n}\right )}{(1+n) (1+2 n) (1+3 n) \left (a+b x^n\right )} \]
(x*Sqrt[(a + b*x^n)^2]*(a^3*(1 + 6*n + 11*n^2 + 6*n^3) + 3*a^2*b*(1 + 5*n + 6*n^2)*x^n + 3*a*b^2*(1 + 4*n + 3*n^2)*x^(2*n) + b^3*(1 + 3*n + 2*n^2)*x ^(3*n)))/((1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b*x^n))
Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1384, 775, 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.
\(\displaystyle \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (b^2 x^n+a b\right )^3dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 775 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (3 a^2 b^4 x^n+3 a b^5 x^{2 n}+b^6 x^{3 n}+a^3 b^3\right )dx}{a b^3+b^4 x^n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \left (a^3 b^3 x+\frac {3 a^2 b^4 x^{n+1}}{n+1}+\frac {3 a b^5 x^{2 n+1}}{2 n+1}+\frac {b^6 x^{3 n+1}}{3 n+1}\right )}{a b^3+b^4 x^n}\) |
(Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*(a^3*b^3*x + (3*a^2*b^4*x^(1 + n))/(1 + n) + (3*a*b^5*x^(1 + 2*n))/(1 + 2*n) + (b^6*x^(1 + 3*n))/(1 + 3*n)))/(a *b^3 + b^4*x^n)
3.6.25.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b* x^n)^p, x], x] /; FreeQ[{a, b, 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 = 138, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} x}{a +b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x \,x^{3 n}}{\left (a +b \,x^{n}\right ) \left (1+3 n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} a x \,x^{2 n}}{\left (a +b \,x^{n}\right ) \left (1+2 n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b x \,x^{n}}{\left (a +b \,x^{n}\right ) \left (1+n \right )}\) | \(138\) |
((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^3*x+((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^3/(1+3 *n)*x*(x^n)^3+3*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^2*a/(1+2*n)*x*(x^n)^2+3*(( a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^2*b/(1+n)*x*x^n
Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.63 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (2 \, b^{3} n^{2} + 3 \, b^{3} n + b^{3}\right )} x x^{3 \, n} + 3 \, {\left (3 \, a b^{2} n^{2} + 4 \, a b^{2} n + a b^{2}\right )} x x^{2 \, n} + 3 \, {\left (6 \, a^{2} b n^{2} + 5 \, a^{2} b n + a^{2} b\right )} x x^{n} + {\left (6 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 6 \, a^{3} n + a^{3}\right )} x}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \]
((2*b^3*n^2 + 3*b^3*n + b^3)*x*x^(3*n) + 3*(3*a*b^2*n^2 + 4*a*b^2*n + a*b^ 2)*x*x^(2*n) + 3*(6*a^2*b*n^2 + 5*a^2*b*n + a^2*b)*x*x^n + (6*a^3*n^3 + 11 *a^3*n^2 + 6*a^3*n + a^3)*x)/(6*n^3 + 11*n^2 + 6*n + 1)
\[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int \left (a^{2} + 2 a b x^{n} + b^{2} x^{2 n}\right )^{\frac {3}{2}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.49 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (2 \, n^{2} + 3 \, n + 1\right )} b^{3} x x^{3 \, n} + 3 \, {\left (3 \, n^{2} + 4 \, n + 1\right )} a b^{2} x x^{2 \, n} + 3 \, {\left (6 \, n^{2} + 5 \, n + 1\right )} a^{2} b x x^{n} + {\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a^{3} x}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \]
((2*n^2 + 3*n + 1)*b^3*x*x^(3*n) + 3*(3*n^2 + 4*n + 1)*a*b^2*x*x^(2*n) + 3 *(6*n^2 + 5*n + 1)*a^2*b*x*x^n + (6*n^3 + 11*n^2 + 6*n + 1)*a^3*x)/(6*n^3 + 11*n^2 + 6*n + 1)
Time = 0.30 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.28 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {6 \, a^{3} n^{3} x \mathrm {sgn}\left (b x^{n} + a\right ) + 2 \, b^{3} n^{2} x x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, b^{3} n x x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a b^{2} n x x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 15 \, a^{2} b n x x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 6 \, a^{3} n x \mathrm {sgn}\left (b x^{n} + a\right ) + b^{3} x x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, a b^{2} x x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, a^{2} b x x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{n} + a\right )}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \]
(6*a^3*n^3*x*sgn(b*x^n + a) + 2*b^3*n^2*x*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2 *n^2*x*x^(2*n)*sgn(b*x^n + a) + 18*a^2*b*n^2*x*x^n*sgn(b*x^n + a) + 11*a^3 *n^2*x*sgn(b*x^n + a) + 3*b^3*n*x*x^(3*n)*sgn(b*x^n + a) + 12*a*b^2*n*x*x^ (2*n)*sgn(b*x^n + a) + 15*a^2*b*n*x*x^n*sgn(b*x^n + a) + 6*a^3*n*x*sgn(b*x ^n + a) + b^3*x*x^(3*n)*sgn(b*x^n + a) + 3*a*b^2*x*x^(2*n)*sgn(b*x^n + a) + 3*a^2*b*x*x^n*sgn(b*x^n + a) + a^3*x*sgn(b*x^n + a))/(6*n^3 + 11*n^2 + 6 *n + 1)
Timed out. \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int {\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,d x \]