Integrand size = 24, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {\left (a+b x^2\right )^7}{20 a x^{20}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}-\frac {b^2 \left (a+b x^2\right )^7}{240 a^3 x^{16}}+\frac {b^3 \left (a+b x^2\right )^7}{1680 a^4 x^{14}} \]
-1/20*(b*x^2+a)^7/a/x^20+1/60*b*(b*x^2+a)^7/a^2/x^18-1/240*b^2*(b*x^2+a)^7 /a^3/x^16+1/1680*b^3*(b*x^2+a)^7/a^4/x^14
Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {a^6}{20 x^{20}}-\frac {a^5 b}{3 x^{18}}-\frac {15 a^4 b^2}{16 x^{16}}-\frac {10 a^3 b^3}{7 x^{14}}-\frac {5 a^2 b^4}{4 x^{12}}-\frac {3 a b^5}{5 x^{10}}-\frac {b^6}{8 x^8} \]
-1/20*a^6/x^20 - (a^5*b)/(3*x^18) - (15*a^4*b^2)/(16*x^16) - (10*a^3*b^3)/ (7*x^14) - (5*a^2*b^4)/(4*x^12) - (3*a*b^5)/(5*x^10) - b^6/(8*x^8)
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1380, 27, 243, 55, 55, 55, 48}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {\int \frac {b^6 \left (b x^2+a\right )^6}{x^{21}}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (a+b x^2\right )^6}{x^{21}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^6}{x^{22}}dx^2\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 b \int \frac {\left (b x^2+a\right )^6}{x^{20}}dx^2}{10 a}-\frac {\left (a+b x^2\right )^7}{10 a x^{20}}\right )\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 b \left (-\frac {2 b \int \frac {\left (b x^2+a\right )^6}{x^{18}}dx^2}{9 a}-\frac {\left (a+b x^2\right )^7}{9 a x^{18}}\right )}{10 a}-\frac {\left (a+b x^2\right )^7}{10 a x^{20}}\right )\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 b \left (-\frac {2 b \left (-\frac {b \int \frac {\left (b x^2+a\right )^6}{x^{16}}dx^2}{8 a}-\frac {\left (a+b x^2\right )^7}{8 a x^{16}}\right )}{9 a}-\frac {\left (a+b x^2\right )^7}{9 a x^{18}}\right )}{10 a}-\frac {\left (a+b x^2\right )^7}{10 a x^{20}}\right )\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 b \left (-\frac {2 b \left (\frac {b \left (a+b x^2\right )^7}{56 a^2 x^{14}}-\frac {\left (a+b x^2\right )^7}{8 a x^{16}}\right )}{9 a}-\frac {\left (a+b x^2\right )^7}{9 a x^{18}}\right )}{10 a}-\frac {\left (a+b x^2\right )^7}{10 a x^{20}}\right )\) |
(-1/10*(a + b*x^2)^7/(a*x^20) - (3*b*(-1/9*(a + b*x^2)^7/(a*x^18) - (2*b*( -1/8*(a + b*x^2)^7/(a*x^16) + (b*(a + b*x^2)^7)/(56*a^2*x^14)))/(9*a)))/(1 0*a))/2
3.5.73.3.1 Defintions of rubi rules used
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] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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 = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {5 a^{2} b^{4}}{4 x^{12}}-\frac {10 a^{3} b^{3}}{7 x^{14}}-\frac {b^{6}}{8 x^{8}}-\frac {a^{6}}{20 x^{20}}-\frac {a^{5} b}{3 x^{18}}-\frac {15 a^{4} b^{2}}{16 x^{16}}-\frac {3 b^{5} a}{5 x^{10}}\) | \(69\) |
norman | \(\frac {-\frac {1}{20} a^{6}-\frac {1}{3} a^{5} b \,x^{2}-\frac {15}{16} a^{4} b^{2} x^{4}-\frac {10}{7} a^{3} b^{3} x^{6}-\frac {5}{4} a^{2} b^{4} x^{8}-\frac {3}{5} a \,b^{5} x^{10}-\frac {1}{8} b^{6} x^{12}}{x^{20}}\) | \(70\) |
risch | \(\frac {-\frac {1}{20} a^{6}-\frac {1}{3} a^{5} b \,x^{2}-\frac {15}{16} a^{4} b^{2} x^{4}-\frac {10}{7} a^{3} b^{3} x^{6}-\frac {5}{4} a^{2} b^{4} x^{8}-\frac {3}{5} a \,b^{5} x^{10}-\frac {1}{8} b^{6} x^{12}}{x^{20}}\) | \(70\) |
gosper | \(-\frac {210 b^{6} x^{12}+1008 a \,b^{5} x^{10}+2100 a^{2} b^{4} x^{8}+2400 a^{3} b^{3} x^{6}+1575 a^{4} b^{2} x^{4}+560 a^{5} b \,x^{2}+84 a^{6}}{1680 x^{20}}\) | \(71\) |
parallelrisch | \(\frac {-210 b^{6} x^{12}-1008 a \,b^{5} x^{10}-2100 a^{2} b^{4} x^{8}-2400 a^{3} b^{3} x^{6}-1575 a^{4} b^{2} x^{4}-560 a^{5} b \,x^{2}-84 a^{6}}{1680 x^{20}}\) | \(71\) |
-5/4*a^2*b^4/x^12-10/7*a^3*b^3/x^14-1/8*b^6/x^8-1/20*a^6/x^20-1/3*a^5*b/x^ 18-15/16*a^4*b^2/x^16-3/5*b^5*a/x^10
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]
-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3* x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x^2 + 84*a^6)/x^20
Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=\frac {- 84 a^{6} - 560 a^{5} b x^{2} - 1575 a^{4} b^{2} x^{4} - 2400 a^{3} b^{3} x^{6} - 2100 a^{2} b^{4} x^{8} - 1008 a b^{5} x^{10} - 210 b^{6} x^{12}}{1680 x^{20}} \]
(-84*a**6 - 560*a**5*b*x**2 - 1575*a**4*b**2*x**4 - 2400*a**3*b**3*x**6 - 2100*a**2*b**4*x**8 - 1008*a*b**5*x**10 - 210*b**6*x**12)/(1680*x**20)
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]
-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3* x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x^2 + 84*a^6)/x^20
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]
-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3* x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x^2 + 84*a^6)/x^20
Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {\frac {a^6}{20}+\frac {a^5\,b\,x^2}{3}+\frac {15\,a^4\,b^2\,x^4}{16}+\frac {10\,a^3\,b^3\,x^6}{7}+\frac {5\,a^2\,b^4\,x^8}{4}+\frac {3\,a\,b^5\,x^{10}}{5}+\frac {b^6\,x^{12}}{8}}{x^{20}} \]