Integrand size = 24, antiderivative size = 155 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {9009 a^2 x}{256 b^8}-\frac {3003 a x^3}{256 b^7}+\frac {9009 x^5}{1280 b^6}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}-\frac {3 x^{13}}{16 b^2 \left (a+b x^2\right )^4}-\frac {13 x^{11}}{32 b^3 \left (a+b x^2\right )^3}-\frac {143 x^9}{128 b^4 \left (a+b x^2\right )^2}-\frac {1287 x^7}{256 b^5 \left (a+b x^2\right )}-\frac {9009 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 b^{17/2}} \]
9009/256*a^2*x/b^8-3003/256*a*x^3/b^7+9009/1280*x^5/b^6-1/10*x^15/b/(b*x^2 +a)^5-3/16*x^13/b^2/(b*x^2+a)^4-13/32*x^11/b^3/(b*x^2+a)^3-143/128*x^9/b^4 /(b*x^2+a)^2-1287/256*x^7/b^5/(b*x^2+a)-9009/256*a^(5/2)*arctan(x*b^(1/2)/ a^(1/2))/b^(17/2)
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {\sqrt {b} x \left (45045 a^7+210210 a^6 b x^2+384384 a^5 b^2 x^4+338910 a^4 b^3 x^6+137995 a^3 b^4 x^8+16640 a^2 b^5 x^{10}-1280 a b^6 x^{12}+256 b^7 x^{14}\right )}{\left (a+b x^2\right )^5}-45045 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{1280 b^{17/2}} \]
((Sqrt[b]*x*(45045*a^7 + 210210*a^6*b*x^2 + 384384*a^5*b^2*x^4 + 338910*a^ 4*b^3*x^6 + 137995*a^3*b^4*x^8 + 16640*a^2*b^5*x^10 - 1280*a*b^6*x^12 + 25 6*b^7*x^14))/(a + b*x^2)^5 - 45045*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(1 280*b^(17/2))
Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1380, 27, 252, 252, 252, 252, 252, 254, 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.
| \(\displaystyle \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle b^6 \int \frac {x^{16}}{b^6 \left (b x^2+a\right )^6}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^{16}}{\left (a+b x^2\right )^6}dx\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {3 \int \frac {x^{14}}{\left (b x^2+a\right )^5}dx}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {3 \left (\frac {13 \int \frac {x^{12}}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {3 \left (\frac {13 \left (\frac {11 \int \frac {x^{10}}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {3 \left (\frac {13 \left (\frac {11 \left (\frac {9 \int \frac {x^8}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {3 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {7 \int \frac {x^6}{b x^2+a}dx}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {3 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {7 \int \left (\frac {x^4}{b}-\frac {a x^2}{b^2}-\frac {a^3}{b^3 \left (b x^2+a\right )}+\frac {a^2}{b^3}\right )dx}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \left (\frac {13 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (-\frac {a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 x}{b^3}-\frac {a x^3}{3 b^2}+\frac {x^5}{5 b}\right )}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^{13}}{8 b \left (a+b x^2\right )^4}\right )}{2 b}-\frac {x^{15}}{10 b \left (a+b x^2\right )^5}\) |
-1/10*x^15/(b*(a + b*x^2)^5) + (3*(-1/8*x^13/(b*(a + b*x^2)^4) + (13*(-1/6 *x^11/(b*(a + b*x^2)^3) + (11*(-1/4*x^9/(b*(a + b*x^2)^2) + (9*(-1/2*x^7/( b*(a + b*x^2)) + (7*((a^2*x)/b^3 - (a*x^3)/(3*b^2) + x^5/(5*b) - (a^(5/2)* ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(7/2)))/(2*b)))/(4*b)))/(6*b)))/(8*b)))/(2* b)
3.6.23.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(\frac {\frac {1}{5} b^{2} x^{5}-2 a b \,x^{3}+21 a^{2} x}{b^{8}}-\frac {a^{3} \left (\frac {-\frac {5327}{256} b^{4} x^{9}-\frac {9443}{128} a \,b^{3} x^{7}-\frac {1001}{10} a^{2} b^{2} x^{5}-\frac {7837}{128} a^{3} b \,x^{3}-\frac {3633}{256} a^{4} x}{\left (b \,x^{2}+a \right )^{5}}+\frac {9009 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \sqrt {a b}}\right )}{b^{8}}\) | \(107\) |
| risch | \(\frac {x^{5}}{5 b^{6}}-\frac {2 a \,x^{3}}{b^{7}}+\frac {21 a^{2} x}{b^{8}}+\frac {\frac {5327}{256} a^{3} b^{4} x^{9}+\frac {9443}{128} a^{4} b^{3} x^{7}+\frac {1001}{10} a^{5} b^{2} x^{5}+\frac {7837}{128} a^{6} b \,x^{3}+\frac {3633}{256} a^{7} x}{b^{8} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}+\frac {9009 \sqrt {-a b}\, a^{2} \ln \left (-\sqrt {-a b}\, x -a \right )}{512 b^{9}}-\frac {9009 \sqrt {-a b}\, a^{2} \ln \left (\sqrt {-a b}\, x -a \right )}{512 b^{9}}\) | \(165\) |
1/b^8*(1/5*b^2*x^5-2*a*b*x^3+21*a^2*x)-1/b^8*a^3*((-5327/256*b^4*x^9-9443/ 128*a*b^3*x^7-1001/10*a^2*b^2*x^5-7837/128*a^3*b*x^3-3633/256*a^4*x)/(b*x^ 2+a)^5+9009/256/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.93 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\left [\frac {512 \, b^{7} x^{15} - 2560 \, a b^{6} x^{13} + 33280 \, a^{2} b^{5} x^{11} + 275990 \, a^{3} b^{4} x^{9} + 677820 \, a^{4} b^{3} x^{7} + 768768 \, a^{5} b^{2} x^{5} + 420420 \, a^{6} b x^{3} + 90090 \, a^{7} x + 45045 \, {\left (a^{2} b^{5} x^{10} + 5 \, a^{3} b^{4} x^{8} + 10 \, a^{4} b^{3} x^{6} + 10 \, a^{5} b^{2} x^{4} + 5 \, a^{6} b x^{2} + a^{7}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{2560 \, {\left (b^{13} x^{10} + 5 \, a b^{12} x^{8} + 10 \, a^{2} b^{11} x^{6} + 10 \, a^{3} b^{10} x^{4} + 5 \, a^{4} b^{9} x^{2} + a^{5} b^{8}\right )}}, \frac {256 \, b^{7} x^{15} - 1280 \, a b^{6} x^{13} + 16640 \, a^{2} b^{5} x^{11} + 137995 \, a^{3} b^{4} x^{9} + 338910 \, a^{4} b^{3} x^{7} + 384384 \, a^{5} b^{2} x^{5} + 210210 \, a^{6} b x^{3} + 45045 \, a^{7} x - 45045 \, {\left (a^{2} b^{5} x^{10} + 5 \, a^{3} b^{4} x^{8} + 10 \, a^{4} b^{3} x^{6} + 10 \, a^{5} b^{2} x^{4} + 5 \, a^{6} b x^{2} + a^{7}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{1280 \, {\left (b^{13} x^{10} + 5 \, a b^{12} x^{8} + 10 \, a^{2} b^{11} x^{6} + 10 \, a^{3} b^{10} x^{4} + 5 \, a^{4} b^{9} x^{2} + a^{5} b^{8}\right )}}\right ] \]
[1/2560*(512*b^7*x^15 - 2560*a*b^6*x^13 + 33280*a^2*b^5*x^11 + 275990*a^3* b^4*x^9 + 677820*a^4*b^3*x^7 + 768768*a^5*b^2*x^5 + 420420*a^6*b*x^3 + 900 90*a^7*x + 45045*(a^2*b^5*x^10 + 5*a^3*b^4*x^8 + 10*a^4*b^3*x^6 + 10*a^5*b ^2*x^4 + 5*a^6*b*x^2 + a^7)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/ (b*x^2 + a)))/(b^13*x^10 + 5*a*b^12*x^8 + 10*a^2*b^11*x^6 + 10*a^3*b^10*x^ 4 + 5*a^4*b^9*x^2 + a^5*b^8), 1/1280*(256*b^7*x^15 - 1280*a*b^6*x^13 + 166 40*a^2*b^5*x^11 + 137995*a^3*b^4*x^9 + 338910*a^4*b^3*x^7 + 384384*a^5*b^2 *x^5 + 210210*a^6*b*x^3 + 45045*a^7*x - 45045*(a^2*b^5*x^10 + 5*a^3*b^4*x^ 8 + 10*a^4*b^3*x^6 + 10*a^5*b^2*x^4 + 5*a^6*b*x^2 + a^7)*sqrt(a/b)*arctan( b*x*sqrt(a/b)/a))/(b^13*x^10 + 5*a*b^12*x^8 + 10*a^2*b^11*x^6 + 10*a^3*b^1 0*x^4 + 5*a^4*b^9*x^2 + a^5*b^8)]
Time = 0.48 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.41 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {21 a^{2} x}{b^{8}} - \frac {2 a x^{3}}{b^{7}} + \frac {9009 \sqrt {- \frac {a^{5}}{b^{17}}} \log {\left (x - \frac {b^{8} \sqrt {- \frac {a^{5}}{b^{17}}}}{a^{2}} \right )}}{512} - \frac {9009 \sqrt {- \frac {a^{5}}{b^{17}}} \log {\left (x + \frac {b^{8} \sqrt {- \frac {a^{5}}{b^{17}}}}{a^{2}} \right )}}{512} + \frac {18165 a^{7} x + 78370 a^{6} b x^{3} + 128128 a^{5} b^{2} x^{5} + 94430 a^{4} b^{3} x^{7} + 26635 a^{3} b^{4} x^{9}}{1280 a^{5} b^{8} + 6400 a^{4} b^{9} x^{2} + 12800 a^{3} b^{10} x^{4} + 12800 a^{2} b^{11} x^{6} + 6400 a b^{12} x^{8} + 1280 b^{13} x^{10}} + \frac {x^{5}}{5 b^{6}} \]
21*a**2*x/b**8 - 2*a*x**3/b**7 + 9009*sqrt(-a**5/b**17)*log(x - b**8*sqrt( -a**5/b**17)/a**2)/512 - 9009*sqrt(-a**5/b**17)*log(x + b**8*sqrt(-a**5/b* *17)/a**2)/512 + (18165*a**7*x + 78370*a**6*b*x**3 + 128128*a**5*b**2*x**5 + 94430*a**4*b**3*x**7 + 26635*a**3*b**4*x**9)/(1280*a**5*b**8 + 6400*a** 4*b**9*x**2 + 12800*a**3*b**10*x**4 + 12800*a**2*b**11*x**6 + 6400*a*b**12 *x**8 + 1280*b**13*x**10) + x**5/(5*b**6)
Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {26635 \, a^{3} b^{4} x^{9} + 94430 \, a^{4} b^{3} x^{7} + 128128 \, a^{5} b^{2} x^{5} + 78370 \, a^{6} b x^{3} + 18165 \, a^{7} x}{1280 \, {\left (b^{13} x^{10} + 5 \, a b^{12} x^{8} + 10 \, a^{2} b^{11} x^{6} + 10 \, a^{3} b^{10} x^{4} + 5 \, a^{4} b^{9} x^{2} + a^{5} b^{8}\right )}} - \frac {9009 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{8}} + \frac {b^{2} x^{5} - 10 \, a b x^{3} + 105 \, a^{2} x}{5 \, b^{8}} \]
1/1280*(26635*a^3*b^4*x^9 + 94430*a^4*b^3*x^7 + 128128*a^5*b^2*x^5 + 78370 *a^6*b*x^3 + 18165*a^7*x)/(b^13*x^10 + 5*a*b^12*x^8 + 10*a^2*b^11*x^6 + 10 *a^3*b^10*x^4 + 5*a^4*b^9*x^2 + a^5*b^8) - 9009/256*a^3*arctan(b*x/sqrt(a* b))/(sqrt(a*b)*b^8) + 1/5*(b^2*x^5 - 10*a*b*x^3 + 105*a^2*x)/b^8
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {9009 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{8}} + \frac {26635 \, a^{3} b^{4} x^{9} + 94430 \, a^{4} b^{3} x^{7} + 128128 \, a^{5} b^{2} x^{5} + 78370 \, a^{6} b x^{3} + 18165 \, a^{7} x}{1280 \, {\left (b x^{2} + a\right )}^{5} b^{8}} + \frac {b^{24} x^{5} - 10 \, a b^{23} x^{3} + 105 \, a^{2} b^{22} x}{5 \, b^{30}} \]
-9009/256*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^8) + 1/1280*(26635*a^3*b^ 4*x^9 + 94430*a^4*b^3*x^7 + 128128*a^5*b^2*x^5 + 78370*a^6*b*x^3 + 18165*a ^7*x)/((b*x^2 + a)^5*b^8) + 1/5*(b^24*x^5 - 10*a*b^23*x^3 + 105*a^2*b^22*x )/b^30
Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.99 \[ \int \frac {x^{16}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {3633\,a^7\,x}{256}+\frac {7837\,a^6\,b\,x^3}{128}+\frac {1001\,a^5\,b^2\,x^5}{10}+\frac {9443\,a^4\,b^3\,x^7}{128}+\frac {5327\,a^3\,b^4\,x^9}{256}}{a^5\,b^8+5\,a^4\,b^9\,x^2+10\,a^3\,b^{10}\,x^4+10\,a^2\,b^{11}\,x^6+5\,a\,b^{12}\,x^8+b^{13}\,x^{10}}+\frac {x^5}{5\,b^6}-\frac {2\,a\,x^3}{b^7}+\frac {21\,a^2\,x}{b^8}-\frac {9009\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,b^{17/2}} \]
((3633*a^7*x)/256 + (7837*a^6*b*x^3)/128 + (1001*a^5*b^2*x^5)/10 + (9443*a ^4*b^3*x^7)/128 + (5327*a^3*b^4*x^9)/256)/(a^5*b^8 + b^13*x^10 + 5*a*b^12* x^8 + 5*a^4*b^9*x^2 + 10*a^3*b^10*x^4 + 10*a^2*b^11*x^6) + x^5/(5*b^6) - ( 2*a*x^3)/b^7 + (21*a^2*x)/b^8 - (9009*a^(5/2)*atan((b^(1/2)*x)/a^(1/2)))/( 256*b^(17/2))