Integrand size = 24, antiderivative size = 131 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {693 x}{256 b^6}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}-\frac {11 x^9}{80 b^2 \left (a+b x^2\right )^4}-\frac {33 x^7}{160 b^3 \left (a+b x^2\right )^3}-\frac {231 x^5}{640 b^4 \left (a+b x^2\right )^2}-\frac {231 x^3}{256 b^5 \left (a+b x^2\right )}-\frac {693 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 b^{13/2}} \]
693/256*x/b^6-1/10*x^11/b/(b*x^2+a)^5-11/80*x^9/b^2/(b*x^2+a)^4-33/160*x^7 /b^3/(b*x^2+a)^3-231/640*x^5/b^4/(b*x^2+a)^2-231/256*x^3/b^5/(b*x^2+a)-693 /256*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(13/2)
Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {\sqrt {b} x \left (3465 a^5+16170 a^4 b x^2+29568 a^3 b^2 x^4+26070 a^2 b^3 x^6+10615 a b^4 x^8+1280 b^5 x^{10}\right )}{\left (a+b x^2\right )^5}-3465 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{1280 b^{13/2}} \]
((Sqrt[b]*x*(3465*a^5 + 16170*a^4*b*x^2 + 29568*a^3*b^2*x^4 + 26070*a^2*b^ 3*x^6 + 10615*a*b^4*x^8 + 1280*b^5*x^10))/(a + b*x^2)^5 - 3465*Sqrt[a]*Arc Tan[(Sqrt[b]*x)/Sqrt[a]])/(1280*b^(13/2))
Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.27, 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, 262, 218}
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.
| \(\displaystyle \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle b^6 \int \frac {x^{12}}{b^6 \left (b x^2+a\right )^6}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^{12}}{\left (a+b x^2\right )^6}dx\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {11 \int \frac {x^{10}}{\left (b x^2+a\right )^5}dx}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {11 \left (\frac {9 \int \frac {x^8}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \int \frac {x^6}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {x^2}{b x^2+a}dx}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^2+a}dx}{b}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^{11}}{10 b \left (a+b x^2\right )^5}\) |
-1/10*x^11/(b*(a + b*x^2)^5) + (11*(-1/8*x^9/(b*(a + b*x^2)^4) + (9*(-1/6* x^7/(b*(a + b*x^2)^3) + (7*(-1/4*x^5/(b*(a + b*x^2)^2) + (5*(-1/2*x^3/(b*( a + b*x^2)) + (3*(x/b - (Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2)))/(2 *b)))/(4*b)))/(6*b)))/(8*b)))/(10*b)
3.6.25.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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 = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64
| method | result | size |
| default | \(\frac {x}{b^{6}}-\frac {a \left (\frac {-\frac {843}{256} b^{4} x^{9}-\frac {1327}{128} a \,b^{3} x^{7}-\frac {131}{10} a^{2} b^{2} x^{5}-\frac {977}{128} a^{3} b \,x^{3}-\frac {437}{256} a^{4} x}{\left (b \,x^{2}+a \right )^{5}}+\frac {693 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \sqrt {a b}}\right )}{b^{6}}\) | \(84\) |
| risch | \(\frac {x}{b^{6}}+\frac {\frac {843}{256} b^{4} a \,x^{9}+\frac {1327}{128} a^{2} b^{3} x^{7}+\frac {131}{10} a^{3} b^{2} x^{5}+\frac {977}{128} a^{4} b \,x^{3}+\frac {437}{256} a^{5} x}{b^{6} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}+\frac {693 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{512 b^{7}}-\frac {693 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{512 b^{7}}\) | \(136\) |
x/b^6-1/b^6*a*((-843/256*b^4*x^9-1327/128*a*b^3*x^7-131/10*a^2*b^2*x^5-977 /128*a^3*b*x^3-437/256*a^4*x)/(b*x^2+a)^5+693/256/(a*b)^(1/2)*arctan(b*x/( a*b)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.05 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\left [\frac {2560 \, b^{5} x^{11} + 21230 \, a b^{4} x^{9} + 52140 \, a^{2} b^{3} x^{7} + 59136 \, a^{3} b^{2} x^{5} + 32340 \, a^{4} b x^{3} + 6930 \, a^{5} x + 3465 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{2560 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}}, \frac {1280 \, b^{5} x^{11} + 10615 \, a b^{4} x^{9} + 26070 \, a^{2} b^{3} x^{7} + 29568 \, a^{3} b^{2} x^{5} + 16170 \, a^{4} b x^{3} + 3465 \, a^{5} x - 3465 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{1280 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}}\right ] \]
[1/2560*(2560*b^5*x^11 + 21230*a*b^4*x^9 + 52140*a^2*b^3*x^7 + 59136*a^3*b ^2*x^5 + 32340*a^4*b*x^3 + 6930*a^5*x + 3465*(b^5*x^10 + 5*a*b^4*x^8 + 10* a^2*b^3*x^6 + 10*a^3*b^2*x^4 + 5*a^4*b*x^2 + a^5)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^11*x^10 + 5*a*b^10*x^8 + 10*a^2*b^9 *x^6 + 10*a^3*b^8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6), 1/1280*(1280*b^5*x^11 + 10615*a*b^4*x^9 + 26070*a^2*b^3*x^7 + 29568*a^3*b^2*x^5 + 16170*a^4*b*x^3 + 3465*a^5*x - 3465*(b^5*x^10 + 5*a*b^4*x^8 + 10*a^2*b^3*x^6 + 10*a^3*b^2* x^4 + 5*a^4*b*x^2 + a^5)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^11*x^10 + 5 *a*b^10*x^8 + 10*a^2*b^9*x^6 + 10*a^3*b^8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6)]
Time = 0.44 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.36 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {693 \sqrt {- \frac {a}{b^{13}}} \log {\left (- b^{6} \sqrt {- \frac {a}{b^{13}}} + x \right )}}{512} - \frac {693 \sqrt {- \frac {a}{b^{13}}} \log {\left (b^{6} \sqrt {- \frac {a}{b^{13}}} + x \right )}}{512} + \frac {2185 a^{5} x + 9770 a^{4} b x^{3} + 16768 a^{3} b^{2} x^{5} + 13270 a^{2} b^{3} x^{7} + 4215 a b^{4} x^{9}}{1280 a^{5} b^{6} + 6400 a^{4} b^{7} x^{2} + 12800 a^{3} b^{8} x^{4} + 12800 a^{2} b^{9} x^{6} + 6400 a b^{10} x^{8} + 1280 b^{11} x^{10}} + \frac {x}{b^{6}} \]
693*sqrt(-a/b**13)*log(-b**6*sqrt(-a/b**13) + x)/512 - 693*sqrt(-a/b**13)* log(b**6*sqrt(-a/b**13) + x)/512 + (2185*a**5*x + 9770*a**4*b*x**3 + 16768 *a**3*b**2*x**5 + 13270*a**2*b**3*x**7 + 4215*a*b**4*x**9)/(1280*a**5*b**6 + 6400*a**4*b**7*x**2 + 12800*a**3*b**8*x**4 + 12800*a**2*b**9*x**6 + 640 0*a*b**10*x**8 + 1280*b**11*x**10) + x/b**6
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {4215 \, a b^{4} x^{9} + 13270 \, a^{2} b^{3} x^{7} + 16768 \, a^{3} b^{2} x^{5} + 9770 \, a^{4} b x^{3} + 2185 \, a^{5} x}{1280 \, {\left (b^{11} x^{10} + 5 \, a b^{10} x^{8} + 10 \, a^{2} b^{9} x^{6} + 10 \, a^{3} b^{8} x^{4} + 5 \, a^{4} b^{7} x^{2} + a^{5} b^{6}\right )}} - \frac {693 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{6}} + \frac {x}{b^{6}} \]
1/1280*(4215*a*b^4*x^9 + 13270*a^2*b^3*x^7 + 16768*a^3*b^2*x^5 + 9770*a^4* b*x^3 + 2185*a^5*x)/(b^11*x^10 + 5*a*b^10*x^8 + 10*a^2*b^9*x^6 + 10*a^3*b^ 8*x^4 + 5*a^4*b^7*x^2 + a^5*b^6) - 693/256*a*arctan(b*x/sqrt(a*b))/(sqrt(a *b)*b^6) + x/b^6
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {693 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{6}} + \frac {x}{b^{6}} + \frac {4215 \, a b^{4} x^{9} + 13270 \, a^{2} b^{3} x^{7} + 16768 \, a^{3} b^{2} x^{5} + 9770 \, a^{4} b x^{3} + 2185 \, a^{5} x}{1280 \, {\left (b x^{2} + a\right )}^{5} b^{6}} \]
-693/256*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^6) + x/b^6 + 1/1280*(4215*a* b^4*x^9 + 13270*a^2*b^3*x^7 + 16768*a^3*b^2*x^5 + 9770*a^4*b*x^3 + 2185*a^ 5*x)/((b*x^2 + a)^5*b^6)
Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {437\,a^5\,x}{256}+\frac {977\,a^4\,b\,x^3}{128}+\frac {131\,a^3\,b^2\,x^5}{10}+\frac {1327\,a^2\,b^3\,x^7}{128}+\frac {843\,a\,b^4\,x^9}{256}}{a^5\,b^6+5\,a^4\,b^7\,x^2+10\,a^3\,b^8\,x^4+10\,a^2\,b^9\,x^6+5\,a\,b^{10}\,x^8+b^{11}\,x^{10}}+\frac {x}{b^6}-\frac {693\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,b^{13/2}} \]