Integrand size = 24, antiderivative size = 72 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=-\frac {a^3 \left (a+b x^2\right )^7}{14 b^4}+\frac {3 a^2 \left (a+b x^2\right )^8}{16 b^4}-\frac {a \left (a+b x^2\right )^9}{6 b^4}+\frac {\left (a+b x^2\right )^{10}}{20 b^4} \] Output:
-1/14*a^3*(b*x^2+a)^7/b^4+3/16*a^2*(b*x^2+a)^8/b^4-1/6*a*(b*x^2+a)^9/b^4+1 /20*(b*x^2+a)^10/b^4
Time = 0.00 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^6 x^8}{8}+\frac {3}{5} a^5 b x^{10}+\frac {5}{4} a^4 b^2 x^{12}+\frac {10}{7} a^3 b^3 x^{14}+\frac {15}{16} a^2 b^4 x^{16}+\frac {1}{3} a b^5 x^{18}+\frac {b^6 x^{20}}{20} \] Input:
Integrate[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
Output:
(a^6*x^8)/8 + (3*a^5*b*x^10)/5 + (5*a^4*b^2*x^12)/4 + (10*a^3*b^3*x^14)/7 + (15*a^2*b^4*x^16)/16 + (a*b^5*x^18)/3 + (b^6*x^20)/20
Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {\int b^6 x^7 \left (b x^2+a\right )^6dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^7 \left (a+b x^2\right )^6dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^6 \left (b x^2+a\right )^6dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b x^2+a\right )^9}{b^3}-\frac {3 a \left (b x^2+a\right )^8}{b^3}+\frac {3 a^2 \left (b x^2+a\right )^7}{b^3}-\frac {a^3 \left (b x^2+a\right )^6}{b^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^3 \left (a+b x^2\right )^7}{7 b^4}+\frac {3 a^2 \left (a+b x^2\right )^8}{8 b^4}+\frac {\left (a+b x^2\right )^{10}}{10 b^4}-\frac {a \left (a+b x^2\right )^9}{3 b^4}\right )\) |
Input:
Int[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
Output:
(-1/7*(a^3*(a + b*x^2)^7)/b^4 + (3*a^2*(a + b*x^2)^8)/(8*b^4) - (a*(a + b* x^2)^9)/(3*b^4) + (a + b*x^2)^10/(10*b^4))/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 = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {1}{8} a^{6} x^{8}+\frac {3}{5} a^{5} b \,x^{10}+\frac {5}{4} a^{4} b^{2} x^{12}+\frac {10}{7} a^{3} b^{3} x^{14}+\frac {15}{16} a^{2} b^{4} x^{16}+\frac {1}{3} b^{5} a \,x^{18}+\frac {1}{20} b^{6} x^{20}\) | \(69\) |
norman | \(\frac {1}{8} a^{6} x^{8}+\frac {3}{5} a^{5} b \,x^{10}+\frac {5}{4} a^{4} b^{2} x^{12}+\frac {10}{7} a^{3} b^{3} x^{14}+\frac {15}{16} a^{2} b^{4} x^{16}+\frac {1}{3} b^{5} a \,x^{18}+\frac {1}{20} b^{6} x^{20}\) | \(69\) |
risch | \(\frac {1}{8} a^{6} x^{8}+\frac {3}{5} a^{5} b \,x^{10}+\frac {5}{4} a^{4} b^{2} x^{12}+\frac {10}{7} a^{3} b^{3} x^{14}+\frac {15}{16} a^{2} b^{4} x^{16}+\frac {1}{3} b^{5} a \,x^{18}+\frac {1}{20} b^{6} x^{20}\) | \(69\) |
parallelrisch | \(\frac {1}{8} a^{6} x^{8}+\frac {3}{5} a^{5} b \,x^{10}+\frac {5}{4} a^{4} b^{2} x^{12}+\frac {10}{7} a^{3} b^{3} x^{14}+\frac {15}{16} a^{2} b^{4} x^{16}+\frac {1}{3} b^{5} a \,x^{18}+\frac {1}{20} b^{6} x^{20}\) | \(69\) |
gosper | \(\frac {x^{8} \left (84 b^{6} x^{12}+560 a \,b^{5} x^{10}+1575 a^{2} b^{4} x^{8}+2400 a^{3} b^{3} x^{6}+2100 a^{4} b^{2} x^{4}+1008 a^{5} b \,x^{2}+210 a^{6}\right )}{1680}\) | \(71\) |
orering | \(\frac {x^{8} \left (84 b^{6} x^{12}+560 a \,b^{5} x^{10}+1575 a^{2} b^{4} x^{8}+2400 a^{3} b^{3} x^{6}+2100 a^{4} b^{2} x^{4}+1008 a^{5} b \,x^{2}+210 a^{6}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{3}}{1680 \left (b \,x^{2}+a \right )^{6}}\) | \(100\) |
Input:
int(x^7*(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)
Output:
1/8*a^6*x^8+3/5*a^5*b*x^10+5/4*a^4*b^2*x^12+10/7*a^3*b^3*x^14+15/16*a^2*b^ 4*x^16+1/3*b^5*a*x^18+1/20*b^6*x^20
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{20} \, b^{6} x^{20} + \frac {1}{3} \, a b^{5} x^{18} + \frac {15}{16} \, a^{2} b^{4} x^{16} + \frac {10}{7} \, a^{3} b^{3} x^{14} + \frac {5}{4} \, a^{4} b^{2} x^{12} + \frac {3}{5} \, a^{5} b x^{10} + \frac {1}{8} \, a^{6} x^{8} \] Input:
integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")
Output:
1/20*b^6*x^20 + 1/3*a*b^5*x^18 + 15/16*a^2*b^4*x^16 + 10/7*a^3*b^3*x^14 + 5/4*a^4*b^2*x^12 + 3/5*a^5*b*x^10 + 1/8*a^6*x^8
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^{6} x^{8}}{8} + \frac {3 a^{5} b x^{10}}{5} + \frac {5 a^{4} b^{2} x^{12}}{4} + \frac {10 a^{3} b^{3} x^{14}}{7} + \frac {15 a^{2} b^{4} x^{16}}{16} + \frac {a b^{5} x^{18}}{3} + \frac {b^{6} x^{20}}{20} \] Input:
integrate(x**7*(b**2*x**4+2*a*b*x**2+a**2)**3,x)
Output:
a**6*x**8/8 + 3*a**5*b*x**10/5 + 5*a**4*b**2*x**12/4 + 10*a**3*b**3*x**14/ 7 + 15*a**2*b**4*x**16/16 + a*b**5*x**18/3 + b**6*x**20/20
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{20} \, b^{6} x^{20} + \frac {1}{3} \, a b^{5} x^{18} + \frac {15}{16} \, a^{2} b^{4} x^{16} + \frac {10}{7} \, a^{3} b^{3} x^{14} + \frac {5}{4} \, a^{4} b^{2} x^{12} + \frac {3}{5} \, a^{5} b x^{10} + \frac {1}{8} \, a^{6} x^{8} \] Input:
integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")
Output:
1/20*b^6*x^20 + 1/3*a*b^5*x^18 + 15/16*a^2*b^4*x^16 + 10/7*a^3*b^3*x^14 + 5/4*a^4*b^2*x^12 + 3/5*a^5*b*x^10 + 1/8*a^6*x^8
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {1}{20} \, b^{6} x^{20} + \frac {1}{3} \, a b^{5} x^{18} + \frac {15}{16} \, a^{2} b^{4} x^{16} + \frac {10}{7} \, a^{3} b^{3} x^{14} + \frac {5}{4} \, a^{4} b^{2} x^{12} + \frac {3}{5} \, a^{5} b x^{10} + \frac {1}{8} \, a^{6} x^{8} \] Input:
integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")
Output:
1/20*b^6*x^20 + 1/3*a*b^5*x^18 + 15/16*a^2*b^4*x^16 + 10/7*a^3*b^3*x^14 + 5/4*a^4*b^2*x^12 + 3/5*a^5*b*x^10 + 1/8*a^6*x^8
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {a^6\,x^8}{8}+\frac {3\,a^5\,b\,x^{10}}{5}+\frac {5\,a^4\,b^2\,x^{12}}{4}+\frac {10\,a^3\,b^3\,x^{14}}{7}+\frac {15\,a^2\,b^4\,x^{16}}{16}+\frac {a\,b^5\,x^{18}}{3}+\frac {b^6\,x^{20}}{20} \] Input:
int(x^7*(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)
Output:
(a^6*x^8)/8 + (b^6*x^20)/20 + (3*a^5*b*x^10)/5 + (a*b^5*x^18)/3 + (5*a^4*b ^2*x^12)/4 + (10*a^3*b^3*x^14)/7 + (15*a^2*b^4*x^16)/16
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {x^{8} \left (84 b^{6} x^{12}+560 a \,b^{5} x^{10}+1575 a^{2} b^{4} x^{8}+2400 a^{3} b^{3} x^{6}+2100 a^{4} b^{2} x^{4}+1008 a^{5} b \,x^{2}+210 a^{6}\right )}{1680} \] Input:
int(x^7*(b^2*x^4+2*a*b*x^2+a^2)^3,x)
Output:
(x**8*(210*a**6 + 1008*a**5*b*x**2 + 2100*a**4*b**2*x**4 + 2400*a**3*b**3* x**6 + 1575*a**2*b**4*x**8 + 560*a*b**5*x**10 + 84*b**6*x**12))/1680