Integrand size = 26, antiderivative size = 119 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 b^3}-\frac {2 a \left (a+b x^3\right )^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 b^3}+\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^3} \] Output:
1/12*a^2*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/b^3-2/15*a*(b*x^3+a)^4*((b*x^3+a) ^2)^(1/2)/b^3+1/18*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/b^3
Time = 0.89 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x^9 \left (20 a^3+45 a^2 b x^3+36 a b^2 x^6+10 b^3 x^9\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{180 \left (-a^2-a b x^3+\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \] Input:
Integrate[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
(x^9*(20*a^3 + 45*a^2*b*x^3 + 36*a*b^2*x^6 + 10*b^3*x^9)*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])))/(180*(-a^2 - a*b*x^3 + Sqrt[a^2]*S qrt[(a + b*x^3)^2]))
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 798, 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.
\(\displaystyle \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int b^3 x^8 \left (b x^3+a\right )^3dx}{b^3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^8 \left (b x^3+a\right )^3dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^6 \left (b x^3+a\right )^3dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^3 x^{15}+3 a b^2 x^{12}+3 a^2 b x^9+a^3 x^6\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (\frac {a^3 x^9}{3}+\frac {3}{4} a^2 b x^{12}+\frac {3}{5} a b^2 x^{15}+\frac {b^3 x^{18}}{6}\right )}{3 \left (a+b x^3\right )}\) |
Input:
Int[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*((a^3*x^9)/3 + (3*a^2*b*x^12)/4 + (3*a*b^ 2*x^15)/5 + (b^3*x^18)/6))/(3*(a + b*x^3))
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)])
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{4} \left (10 b^{2} x^{6}-4 a \,x^{3} b +a^{2}\right )}{180 b^{3}}\) | \(42\) |
gosper | \(\frac {x^{9} \left (10 x^{9} b^{3}+36 a \,x^{6} b^{2}+45 a^{2} x^{3} b +20 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
default | \(\frac {x^{9} \left (10 x^{9} b^{3}+36 a \,x^{6} b^{2}+45 a^{2} x^{3} b +20 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
orering | \(\frac {x^{9} \left (10 x^{9} b^{3}+36 a \,x^{6} b^{2}+45 a^{2} x^{3} b +20 a^{3}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {3}{2}}}{180 \left (b \,x^{3}+a \right )^{3}}\) | \(67\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} x^{9}}{9 b \,x^{3}+9 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \,a^{2} x^{12}}{4 b \,x^{3}+4 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} a \,x^{15}}{5 b \,x^{3}+5 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} x^{18}}{18 b \,x^{3}+18 a}\) | \(116\) |
Input:
int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/180*csgn(b*x^3+a)*(b*x^3+a)^4*(10*b^2*x^6-4*a*b*x^3+a^2)/b^3
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{18} \, b^{3} x^{18} + \frac {1}{5} \, a b^{2} x^{15} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{9} \, a^{3} x^{9} \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")
Output:
1/18*b^3*x^18 + 1/5*a*b^2*x^15 + 1/4*a^2*b*x^12 + 1/9*a^3*x^9
\[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^{8} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**8*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)
Output:
Integral(x**8*((a + b*x**3)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{3}}{12 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} x^{3}}{18 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{12 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a}{90 \, b^{3}} \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")
Output:
1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^2*x^3/b^2 + 1/18*(b^2*x^6 + 2*a*b *x^3 + a^2)^(5/2)*x^3/b^2 + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^3/b^3 - 7/90*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a/b^3
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.56 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{18} \, b^{3} x^{18} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a b^{2} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{4} \, a^{2} b x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{9} \, a^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")
Output:
1/18*b^3*x^18*sgn(b*x^3 + a) + 1/5*a*b^2*x^15*sgn(b*x^3 + a) + 1/4*a^2*b*x ^12*sgn(b*x^3 + a) + 1/9*a^3*x^9*sgn(b*x^3 + a)
Timed out. \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^8\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \] Input:
int(x^8*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)
Output:
int(x^8*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.31 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x^{9} \left (10 b^{3} x^{9}+36 a \,b^{2} x^{6}+45 a^{2} b \,x^{3}+20 a^{3}\right )}{180} \] Input:
int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)
Output:
(x**9*(20*a**3 + 45*a**2*b*x**3 + 36*a*b**2*x**6 + 10*b**3*x**9))/180