Integrand size = 24, antiderivative size = 53 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^2 \left (a+b x^2\right )^7}{14 b^3}-\frac {a \left (a+b x^2\right )^8}{8 b^3}+\frac {\left (a+b x^2\right )^9}{18 b^3} \] Output:
1/14*a^2*(b*x^2+a)^7/b^3-1/8*a*(b*x^2+a)^8/b^3+1/18*(b*x^2+a)^9/b^3
Time = 0.00 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.55 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^6 x^6}{6}+\frac {3}{4} a^5 b x^8+\frac {3}{2} a^4 b^2 x^{10}+\frac {5}{3} a^3 b^3 x^{12}+\frac {15}{14} a^2 b^4 x^{14}+\frac {3}{8} a b^5 x^{16}+\frac {b^6 x^{18}}{18} \] Input:
Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
Output:
(a^6*x^6)/6 + (3*a^5*b*x^8)/4 + (3*a^4*b^2*x^10)/2 + (5*a^3*b^3*x^12)/3 + (15*a^2*b^4*x^14)/14 + (3*a*b^5*x^16)/8 + (b^6*x^18)/18
Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1380, 27, 243, 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.
\(\displaystyle \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {\int b^6 x^5 \left (b x^2+a\right )^6dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^5 \left (a+b x^2\right )^6dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^4 \left (b x^2+a\right )^6dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right )^8}{b^2}-\frac {2 a \left (b x^2+a\right )^7}{b^2}+\frac {a^2 \left (b x^2+a\right )^6}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^2 \left (a+b x^2\right )^7}{7 b^3}+\frac {\left (a+b x^2\right )^9}{9 b^3}-\frac {a \left (a+b x^2\right )^8}{4 b^3}\right )\) |
Input:
Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
Output:
((a^2*(a + b*x^2)^7)/(7*b^3) - (a*(a + b*x^2)^8)/(4*b^3) + (a + b*x^2)^9/( 9*b^3))/2
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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^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]
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.30
method | result | size |
default | \(\frac {1}{6} a^{6} x^{6}+\frac {3}{4} a^{5} b \,x^{8}+\frac {3}{2} a^{4} b^{2} x^{10}+\frac {5}{3} a^{3} b^{3} x^{12}+\frac {15}{14} a^{2} b^{4} x^{14}+\frac {3}{8} b^{5} a \,x^{16}+\frac {1}{18} b^{6} x^{18}\) | \(69\) |
norman | \(\frac {1}{6} a^{6} x^{6}+\frac {3}{4} a^{5} b \,x^{8}+\frac {3}{2} a^{4} b^{2} x^{10}+\frac {5}{3} a^{3} b^{3} x^{12}+\frac {15}{14} a^{2} b^{4} x^{14}+\frac {3}{8} b^{5} a \,x^{16}+\frac {1}{18} b^{6} x^{18}\) | \(69\) |
risch | \(\frac {1}{6} a^{6} x^{6}+\frac {3}{4} a^{5} b \,x^{8}+\frac {3}{2} a^{4} b^{2} x^{10}+\frac {5}{3} a^{3} b^{3} x^{12}+\frac {15}{14} a^{2} b^{4} x^{14}+\frac {3}{8} b^{5} a \,x^{16}+\frac {1}{18} b^{6} x^{18}\) | \(69\) |
parallelrisch | \(\frac {1}{6} a^{6} x^{6}+\frac {3}{4} a^{5} b \,x^{8}+\frac {3}{2} a^{4} b^{2} x^{10}+\frac {5}{3} a^{3} b^{3} x^{12}+\frac {15}{14} a^{2} b^{4} x^{14}+\frac {3}{8} b^{5} a \,x^{16}+\frac {1}{18} b^{6} x^{18}\) | \(69\) |
gosper | \(\frac {x^{6} \left (28 b^{6} x^{12}+189 a \,b^{5} x^{10}+540 a^{2} b^{4} x^{8}+840 a^{3} b^{3} x^{6}+756 a^{4} b^{2} x^{4}+378 a^{5} b \,x^{2}+84 a^{6}\right )}{504}\) | \(71\) |
orering | \(\frac {x^{6} \left (28 b^{6} x^{12}+189 a \,b^{5} x^{10}+540 a^{2} b^{4} x^{8}+840 a^{3} b^{3} x^{6}+756 a^{4} b^{2} x^{4}+378 a^{5} b \,x^{2}+84 a^{6}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{3}}{504 \left (b \,x^{2}+a \right )^{6}}\) | \(100\) |
Input:
int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)
Output:
1/6*a^6*x^6+3/4*a^5*b*x^8+3/2*a^4*b^2*x^10+5/3*a^3*b^3*x^12+15/14*a^2*b^4* x^14+3/8*b^5*a*x^16+1/18*b^6*x^18
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{18} \, b^{6} x^{18} + \frac {3}{8} \, a b^{5} x^{16} + \frac {15}{14} \, a^{2} b^{4} x^{14} + \frac {5}{3} \, a^{3} b^{3} x^{12} + \frac {3}{2} \, a^{4} b^{2} x^{10} + \frac {3}{4} \, a^{5} b x^{8} + \frac {1}{6} \, a^{6} x^{6} \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")
Output:
1/18*b^6*x^18 + 3/8*a*b^5*x^16 + 15/14*a^2*b^4*x^14 + 5/3*a^3*b^3*x^12 + 3 /2*a^4*b^2*x^10 + 3/4*a^5*b*x^8 + 1/6*a^6*x^6
Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^{6} x^{6}}{6} + \frac {3 a^{5} b x^{8}}{4} + \frac {3 a^{4} b^{2} x^{10}}{2} + \frac {5 a^{3} b^{3} x^{12}}{3} + \frac {15 a^{2} b^{4} x^{14}}{14} + \frac {3 a b^{5} x^{16}}{8} + \frac {b^{6} x^{18}}{18} \] Input:
integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**3,x)
Output:
a**6*x**6/6 + 3*a**5*b*x**8/4 + 3*a**4*b**2*x**10/2 + 5*a**3*b**3*x**12/3 + 15*a**2*b**4*x**14/14 + 3*a*b**5*x**16/8 + b**6*x**18/18
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{18} \, b^{6} x^{18} + \frac {3}{8} \, a b^{5} x^{16} + \frac {15}{14} \, a^{2} b^{4} x^{14} + \frac {5}{3} \, a^{3} b^{3} x^{12} + \frac {3}{2} \, a^{4} b^{2} x^{10} + \frac {3}{4} \, a^{5} b x^{8} + \frac {1}{6} \, a^{6} x^{6} \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")
Output:
1/18*b^6*x^18 + 3/8*a*b^5*x^16 + 15/14*a^2*b^4*x^14 + 5/3*a^3*b^3*x^12 + 3 /2*a^4*b^2*x^10 + 3/4*a^5*b*x^8 + 1/6*a^6*x^6
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{18} \, b^{6} x^{18} + \frac {3}{8} \, a b^{5} x^{16} + \frac {15}{14} \, a^{2} b^{4} x^{14} + \frac {5}{3} \, a^{3} b^{3} x^{12} + \frac {3}{2} \, a^{4} b^{2} x^{10} + \frac {3}{4} \, a^{5} b x^{8} + \frac {1}{6} \, a^{6} x^{6} \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")
Output:
1/18*b^6*x^18 + 3/8*a*b^5*x^16 + 15/14*a^2*b^4*x^14 + 5/3*a^3*b^3*x^12 + 3 /2*a^4*b^2*x^10 + 3/4*a^5*b*x^8 + 1/6*a^6*x^6
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^6\,x^6}{6}+\frac {3\,a^5\,b\,x^8}{4}+\frac {3\,a^4\,b^2\,x^{10}}{2}+\frac {5\,a^3\,b^3\,x^{12}}{3}+\frac {15\,a^2\,b^4\,x^{14}}{14}+\frac {3\,a\,b^5\,x^{16}}{8}+\frac {b^6\,x^{18}}{18} \] Input:
int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)
Output:
(a^6*x^6)/6 + (b^6*x^18)/18 + (3*a^5*b*x^8)/4 + (3*a*b^5*x^16)/8 + (3*a^4* b^2*x^10)/2 + (5*a^3*b^3*x^12)/3 + (15*a^2*b^4*x^14)/14
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {x^{6} \left (28 b^{6} x^{12}+189 a \,b^{5} x^{10}+540 a^{2} b^{4} x^{8}+840 a^{3} b^{3} x^{6}+756 a^{4} b^{2} x^{4}+378 a^{5} b \,x^{2}+84 a^{6}\right )}{504} \] Input:
int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^3,x)
Output:
(x**6*(84*a**6 + 378*a**5*b*x**2 + 756*a**4*b**2*x**4 + 840*a**3*b**3*x**6 + 540*a**2*b**4*x**8 + 189*a*b**5*x**10 + 28*b**6*x**12))/504