Integrand size = 30, antiderivative size = 238 \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {a^3 (d x)^{1+m} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{d (1+m) \left (a+b x^n\right )}+\frac {3 a^2 b^2 x^{1+n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+n) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{1+2 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+2 n) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{1+3 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+3 n) \left (a b+b^2 x^n\right )} \]
a^3*(d*x)^(1+m)*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/d/(1+m)/(a+b*x^n)+3*a^2* b^2*x^(1+n)*(d*x)^m*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(1+m+n)/(a*b+b^2*x^n )+3*a*b^3*x^(1+2*n)*(d*x)^m*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(1+m+2*n)/(a *b+b^2*x^n)+b^4*x^(1+3*n)*(d*x)^m*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/(1+m+3 *n)/(a*b+b^2*x^n)
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.38 \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {x (d x)^m \left (\left (a+b x^n\right )^2\right )^{3/2} \left (\frac {a^3}{1+m}+\frac {3 a^2 b x^n}{1+m+n}+\frac {3 a b^2 x^{2 n}}{1+m+2 n}+\frac {b^3 x^{3 n}}{1+m+3 n}\right )}{\left (a+b x^n\right )^3} \]
(x*(d*x)^m*((a + b*x^n)^2)^(3/2)*(a^3/(1 + m) + (3*a^2*b*x^n)/(1 + m + n) + (3*a*b^2*x^(2*n))/(1 + m + 2*n) + (b^3*x^(3*n))/(1 + m + 3*n)))/(a + b*x ^n)^3
Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.58, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int (d x)^m \left (b^2 x^n+a b\right )^3dx}{a b^3+b^4 x^n}\) |
| \(\Big \downarrow \) 802 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (3 a^2 b^4 x^n (d x)^m+3 a b^5 x^{2 n} (d x)^m+b^6 x^{3 n} (d x)^m+a^3 b^3 (d x)^m\right )dx}{a b^3+b^4 x^n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \left (\frac {a^3 b^3 (d x)^{m+1}}{d (m+1)}+\frac {3 a^2 b^4 x^{n+1} (d x)^m}{m+n+1}+\frac {3 a b^5 x^{2 n+1} (d x)^m}{m+2 n+1}+\frac {b^6 x^{3 n+1} (d x)^m}{m+3 n+1}\right )}{a b^3+b^4 x^n}\) |
(Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*((3*a^2*b^4*x^(1 + n)*(d*x)^m)/(1 + m + n) + (3*a*b^5*x^(1 + 2*n)*(d*x)^m)/(1 + m + 2*n) + (b^6*x^(1 + 3*n)*(d* x)^m)/(1 + m + 3*n) + (a^3*b^3*(d*x)^(1 + m))/(d*(1 + m))))/(a*b^3 + b^4*x ^n)
3.6.22.3.1 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)])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 499, normalized size of antiderivative = 2.10
| method | result | size |
| risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, x \left (a^{3}+3 m \,a^{3}+3 a \,b^{2} m^{3} x^{2 n}+6 b^{3} m n \,x^{3 n}+9 a \,b^{2} m^{2} x^{2 n}+9 a \,b^{2} n^{2} x^{2 n}+9 m \,b^{2} a \,x^{2 n}+3 b^{3} m^{2} n \,x^{3 n}+2 b^{3} m \,n^{2} x^{3 n}+12 b^{2} a \,x^{2 n} n +3 a^{2} b \,m^{3} x^{n}+9 m \,a^{2} b \,x^{n}+15 a^{2} b n \,x^{n}+9 a^{2} b \,m^{2} x^{n}+18 a^{2} b \,n^{2} x^{n}+a^{3} m^{3}+3 a^{3} m^{2}+9 a \,b^{2} m \,n^{2} x^{2 n}+24 a \,b^{2} m n \,x^{2 n}+12 a \,b^{2} m^{2} n \,x^{2 n}+15 a^{2} b \,m^{2} n \,x^{n}+18 a^{2} b m \,n^{2} x^{n}+30 a^{2} b m n \,x^{n}+11 a^{3} m \,n^{2}+12 a^{3} m n +x^{3 n} b^{3}+6 a^{3} m^{2} n +3 b^{3} m^{2} x^{3 n}+b^{3} m^{3} x^{3 n}+2 b^{3} n^{2} x^{3 n}+3 m \,b^{3} x^{3 n}+3 b^{3} x^{3 n} n +3 x^{2 n} a \,b^{2}+11 a^{3} n^{2}+6 a^{3} n +3 x^{n} a^{2} b +6 a^{3} n^{3}\right ) d^{m} x^{m} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i d x \right ) m \left (\operatorname {csgn}\left (i d x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i d x \right )+\operatorname {csgn}\left (i d \right )\right )}{2}}}{\left (a +b \,x^{n}\right ) \left (1+m \right ) \left (1+m +n \right ) \left (1+m +2 n \right ) \left (1+m +3 n \right )}\) | \(499\) |
((a+b*x^n)^2)^(1/2)/(a+b*x^n)*x*(a^3+3*m*a^3+3*a*b^2*m^3*(x^n)^2+6*b^3*m*n *(x^n)^3+3*a^2*b*m^3*x^n+9*m*a^2*b*x^n+15*a^2*b*n*x^n+9*a*b^2*m^2*(x^n)^2+ 9*a*b^2*n^2*(x^n)^2+9*a^2*b*m^2*x^n+18*a^2*b*n^2*x^n+9*m*b^2*a*(x^n)^2+3*b ^3*m^2*n*(x^n)^3+2*b^3*m*n^2*(x^n)^3+12*b^2*a*(x^n)^2*n+a^3*m^3+3*a^3*m^2+ (x^n)^3*b^3+9*a*b^2*m*n^2*(x^n)^2+15*a^2*b*m^2*n*x^n+18*a^2*b*m*n^2*x^n+24 *a*b^2*m*n*(x^n)^2+30*a^2*b*m*n*x^n+11*a^3*m*n^2+12*a^3*m*n+6*a^3*m^2*n+3* b^3*m^2*(x^n)^3+b^3*m^3*(x^n)^3+2*b^3*n^2*(x^n)^3+3*m*b^3*(x^n)^3+3*b^3*(x ^n)^3*n+11*a^3*n^2+6*a^3*n+3*(x^n)^2*a*b^2+3*x^n*a^2*b+6*a^3*n^3+12*a*b^2* m^2*n*(x^n)^2)/(1+m)/(1+m+n)/(1+m+2*n)/(1+m+3*n)*d^m*x^m*exp(1/2*I*Pi*csgn (I*d*x)*m*(csgn(I*d*x)-csgn(I*x))*(-csgn(I*d*x)+csgn(I*d)))
Time = 0.27 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.64 \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (b^{3} m^{3} + 3 \, b^{3} m^{2} + 3 \, b^{3} m + b^{3} + 2 \, {\left (b^{3} m + b^{3}\right )} n^{2} + 3 \, {\left (b^{3} m^{2} + 2 \, b^{3} m + b^{3}\right )} n\right )} x x^{3 \, n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + 3 \, {\left (a b^{2} m^{3} + 3 \, a b^{2} m^{2} + 3 \, a b^{2} m + a b^{2} + 3 \, {\left (a b^{2} m + a b^{2}\right )} n^{2} + 4 \, {\left (a b^{2} m^{2} + 2 \, a b^{2} m + a b^{2}\right )} n\right )} x x^{2 \, n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + 3 \, {\left (a^{2} b m^{3} + 3 \, a^{2} b m^{2} + 3 \, a^{2} b m + a^{2} b + 6 \, {\left (a^{2} b m + a^{2} b\right )} n^{2} + 5 \, {\left (a^{2} b m^{2} + 2 \, a^{2} b m + a^{2} b\right )} n\right )} x x^{n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + {\left (a^{3} m^{3} + 6 \, a^{3} n^{3} + 3 \, a^{3} m^{2} + 3 \, a^{3} m + a^{3} + 11 \, {\left (a^{3} m + a^{3}\right )} n^{2} + 6 \, {\left (a^{3} m^{2} + 2 \, a^{3} m + a^{3}\right )} n\right )} x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )}}{m^{4} + 6 \, {\left (m + 1\right )} n^{3} + 4 \, m^{3} + 11 \, {\left (m^{2} + 2 \, m + 1\right )} n^{2} + 6 \, m^{2} + 6 \, {\left (m^{3} + 3 \, m^{2} + 3 \, m + 1\right )} n + 4 \, m + 1} \]
((b^3*m^3 + 3*b^3*m^2 + 3*b^3*m + b^3 + 2*(b^3*m + b^3)*n^2 + 3*(b^3*m^2 + 2*b^3*m + b^3)*n)*x*x^(3*n)*e^(m*log(d) + m*log(x)) + 3*(a*b^2*m^3 + 3*a* b^2*m^2 + 3*a*b^2*m + a*b^2 + 3*(a*b^2*m + a*b^2)*n^2 + 4*(a*b^2*m^2 + 2*a *b^2*m + a*b^2)*n)*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 3*(a^2*b*m^3 + 3*a^ 2*b*m^2 + 3*a^2*b*m + a^2*b + 6*(a^2*b*m + a^2*b)*n^2 + 5*(a^2*b*m^2 + 2*a ^2*b*m + a^2*b)*n)*x*x^n*e^(m*log(d) + m*log(x)) + (a^3*m^3 + 6*a^3*n^3 + 3*a^3*m^2 + 3*a^3*m + a^3 + 11*(a^3*m + a^3)*n^2 + 6*(a^3*m^2 + 2*a^3*m + a^3)*n)*x*e^(m*log(d) + m*log(x)))/(m^4 + 6*(m + 1)*n^3 + 4*m^3 + 11*(m^2 + 2*m + 1)*n^2 + 6*m^2 + 6*(m^3 + 3*m^2 + 3*m + 1)*n + 4*m + 1)
\[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int \left (d x\right )^{m} \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.16 \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\frac {{\left (m^{3} + 3 \, m^{2} {\left (2 \, n + 1\right )} + 6 \, n^{3} + {\left (11 \, n^{2} + 12 \, n + 3\right )} m + 11 \, n^{2} + 6 \, n + 1\right )} a^{3} d^{m} x x^{m} + {\left (m^{3} + 3 \, m^{2} {\left (n + 1\right )} + {\left (2 \, n^{2} + 6 \, n + 3\right )} m + 2 \, n^{2} + 3 \, n + 1\right )} b^{3} d^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )} + 3 \, {\left (m^{3} + m^{2} {\left (4 \, n + 3\right )} + {\left (3 \, n^{2} + 8 \, n + 3\right )} m + 3 \, n^{2} + 4 \, n + 1\right )} a b^{2} d^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )} + 3 \, {\left (m^{3} + m^{2} {\left (5 \, n + 3\right )} + {\left (6 \, n^{2} + 10 \, n + 3\right )} m + 6 \, n^{2} + 5 \, n + 1\right )} a^{2} b d^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m^{4} + 2 \, m^{3} {\left (3 \, n + 2\right )} + {\left (11 \, n^{2} + 18 \, n + 6\right )} m^{2} + 6 \, n^{3} + 2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 9 \, n + 2\right )} m + 11 \, n^{2} + 6 \, n + 1} \]
((m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1 )*a^3*d^m*x*x^m + (m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*b^3*d^m*x*e^(m*log(x) + 3*n*log(x)) + 3*(m^3 + m^2*(4*n + 3) + (3*n^ 2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*a*b^2*d^m*x*e^(m*log(x) + 2*n*log(x)) + 3*(m^3 + m^2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*a^2*b*d^m *x*e^(m*log(x) + n*log(x)))/(m^4 + 2*m^3*(3*n + 2) + (11*n^2 + 18*n + 6)*m ^2 + 6*n^3 + 2*(3*n^3 + 11*n^2 + 9*n + 2)*m + 11*n^2 + 6*n + 1)
Leaf count of result is larger than twice the leaf count of optimal. 2719 vs. \(2 (230) = 460\).
Time = 0.41 (sec) , antiderivative size = 2719, normalized size of antiderivative = 11.42 \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\text {Too large to display} \]
(b^3*m^3*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x* x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*x^(3*n)*e^( m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a*b^2*m^3*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b^3*m^3*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b *x^n + a) + 12*a*b^2*m^2*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a ) + 3*b^3*m^2*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2 *m*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*x^ (2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a^2*b*m^3*x*x^n*e^(m*log( d) + m*log(x))*sgn(b*x^n + a) + 3*a*b^2*m^3*x*x^n*e^(m*log(d) + m*log(x))* sgn(b*x^n + a) + b^3*m^3*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 15 *a^2*b*m^2*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a*b^2*m^2*n *x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x*x^n*e^(m*log (d) + m*log(x))*sgn(b*x^n + a) + 18*a^2*b*m*n^2*x*x^n*e^(m*log(d) + m*log( x))*sgn(b*x^n + a) + 9*a*b^2*m*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + a^3*m^3 *x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a^2*b*m^3*x*e^(m*log(d) + m* log(x))*sgn(b*x^n + a) + 3*a*b^2*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b^3*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 6*a^3*m^2*n*x*e^( m*log(d) + m*log(x))*sgn(b*x^n + a) + 15*a^2*b*m^2*n*x*e^(m*log(d) + m*log (x))*sgn(b*x^n + a) + 12*a*b^2*m^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^...
Timed out. \[ \int (d x)^m \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx=\int {\left (d\,x\right )}^m\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,d x \]