Integrand size = 24, antiderivative size = 104 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {105 a x}{16 b^5}+\frac {35 x^3}{16 b^4}-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}+\frac {105 a^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{11/2}} \]
-105/16*a*x/b^5+35/16*x^3/b^4-1/6*x^9/b/(b*x^2+a)^3-3/8*x^7/b^2/(b*x^2+a)^ 2-21/16*x^5/b^3/(b*x^2+a)+105/16*a^(3/2)*arctan(x*b^(1/2)/a^(1/2))/b^(11/2 )
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {\sqrt {b} x \left (-315 a^4-840 a^3 b x^2-693 a^2 b^2 x^4-144 a b^3 x^6+16 b^4 x^8\right )}{\left (a+b x^2\right )^3}+315 a^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{48 b^{11/2}} \]
((Sqrt[b]*x*(-315*a^4 - 840*a^3*b*x^2 - 693*a^2*b^2*x^4 - 144*a*b^3*x^6 + 16*b^4*x^8))/(a + b*x^2)^3 + 315*a^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(48* b^(11/2))
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1380, 27, 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.
\(\displaystyle \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^{10}}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{10}}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {3 \int \frac {x^8}{\left (b x^2+a\right )^3}dx}{2 b}-\frac {x^9}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {3 \left (\frac {7 \int \frac {x^6}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^7}{4 b \left (a+b x^2\right )^2}\right )}{2 b}-\frac {x^9}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \int \frac {x^4}{b x^2+a}dx}{2 b}-\frac {x^5}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^7}{4 b \left (a+b x^2\right )^2}\right )}{2 b}-\frac {x^9}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \int \left (\frac {a^2}{b^2 \left (b x^2+a\right )}-\frac {a}{b^2}+\frac {x^2}{b}\right )dx}{2 b}-\frac {x^5}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^7}{4 b \left (a+b x^2\right )^2}\right )}{2 b}-\frac {x^9}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}-\frac {a x}{b^2}+\frac {x^3}{3 b}\right )}{2 b}-\frac {x^5}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^7}{4 b \left (a+b x^2\right )^2}\right )}{2 b}-\frac {x^9}{6 b \left (a+b x^2\right )^3}\) |
-1/6*x^9/(b*(a + b*x^2)^3) + (3*(-1/4*x^7/(b*(a + b*x^2)^2) + (7*(-1/2*x^5 /(b*(a + b*x^2)) + (5*(-((a*x)/b^2) + x^3/(3*b) + (a^(3/2)*ArcTan[(Sqrt[b] *x)/Sqrt[a]])/b^(5/2)))/(2*b)))/(4*b)))/(2*b)
3.6.3.3.1 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 = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {-\frac {1}{3} b \,x^{3}+4 a x}{b^{5}}+\frac {a^{2} \left (\frac {-\frac {55}{16} b^{2} x^{5}-\frac {35}{6} a b \,x^{3}-\frac {41}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {105 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{5}}\) | \(74\) |
risch | \(\frac {x^{3}}{3 b^{4}}-\frac {4 a x}{b^{5}}+\frac {-\frac {55}{16} a^{2} b^{2} x^{5}-\frac {35}{6} a^{3} b \,x^{3}-\frac {41}{16} a^{4} x}{b^{5} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {105 \sqrt {-a b}\, a \ln \left (-\sqrt {-a b}\, x +a \right )}{32 b^{6}}-\frac {105 \sqrt {-a b}\, a \ln \left (\sqrt {-a b}\, x +a \right )}{32 b^{6}}\) | \(124\) |
-1/b^5*(-1/3*b*x^3+4*a*x)+1/b^5*a^2*((-55/16*b^2*x^5-35/6*a*b*x^3-41/16*a^ 2*x)/(b*x^2+a)^3+105/16/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.85 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {32 \, b^{4} x^{9} - 288 \, a b^{3} x^{7} - 1386 \, a^{2} b^{2} x^{5} - 1680 \, a^{3} b x^{3} - 630 \, a^{4} x + 315 \, {\left (a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + 3 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{96 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}}, \frac {16 \, b^{4} x^{9} - 144 \, a b^{3} x^{7} - 693 \, a^{2} b^{2} x^{5} - 840 \, a^{3} b x^{3} - 315 \, a^{4} x + 315 \, {\left (a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + 3 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{48 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}}\right ] \]
[1/96*(32*b^4*x^9 - 288*a*b^3*x^7 - 1386*a^2*b^2*x^5 - 1680*a^3*b*x^3 - 63 0*a^4*x + 315*(a*b^3*x^6 + 3*a^2*b^2*x^4 + 3*a^3*b*x^2 + a^4)*sqrt(-a/b)*l og((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^8*x^6 + 3*a*b^7*x^4 + 3 *a^2*b^6*x^2 + a^3*b^5), 1/48*(16*b^4*x^9 - 144*a*b^3*x^7 - 693*a^2*b^2*x^ 5 - 840*a^3*b*x^3 - 315*a^4*x + 315*(a*b^3*x^6 + 3*a^2*b^2*x^4 + 3*a^3*b*x ^2 + a^4)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^8*x^6 + 3*a*b^7*x^4 + 3*a^ 2*b^6*x^2 + a^3*b^5)]
Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.50 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=- \frac {4 a x}{b^{5}} - \frac {105 \sqrt {- \frac {a^{3}}{b^{11}}} \log {\left (x - \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{11}}}}{a} \right )}}{32} + \frac {105 \sqrt {- \frac {a^{3}}{b^{11}}} \log {\left (x + \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{11}}}}{a} \right )}}{32} + \frac {- 123 a^{4} x - 280 a^{3} b x^{3} - 165 a^{2} b^{2} x^{5}}{48 a^{3} b^{5} + 144 a^{2} b^{6} x^{2} + 144 a b^{7} x^{4} + 48 b^{8} x^{6}} + \frac {x^{3}}{3 b^{4}} \]
-4*a*x/b**5 - 105*sqrt(-a**3/b**11)*log(x - b**5*sqrt(-a**3/b**11)/a)/32 + 105*sqrt(-a**3/b**11)*log(x + b**5*sqrt(-a**3/b**11)/a)/32 + (-123*a**4*x - 280*a**3*b*x**3 - 165*a**2*b**2*x**5)/(48*a**3*b**5 + 144*a**2*b**6*x** 2 + 144*a*b**7*x**4 + 48*b**8*x**6) + x**3/(3*b**4)
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {165 \, a^{2} b^{2} x^{5} + 280 \, a^{3} b x^{3} + 123 \, a^{4} x}{48 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}} + \frac {105 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{5}} + \frac {b x^{3} - 12 \, a x}{3 \, b^{5}} \]
-1/48*(165*a^2*b^2*x^5 + 280*a^3*b*x^3 + 123*a^4*x)/(b^8*x^6 + 3*a*b^7*x^4 + 3*a^2*b^6*x^2 + a^3*b^5) + 105/16*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)* b^5) + 1/3*(b*x^3 - 12*a*x)/b^5
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.81 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {105 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{5}} - \frac {165 \, a^{2} b^{2} x^{5} + 280 \, a^{3} b x^{3} + 123 \, a^{4} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{5}} + \frac {b^{8} x^{3} - 12 \, a b^{7} x}{3 \, b^{12}} \]
105/16*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/48*(165*a^2*b^2*x^5 + 280*a^3*b*x^3 + 123*a^4*x)/((b*x^2 + a)^3*b^5) + 1/3*(b^8*x^3 - 12*a*b^7* x)/b^12
Time = 13.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {x^3}{3\,b^4}-\frac {\frac {41\,a^4\,x}{16}+\frac {35\,a^3\,b\,x^3}{6}+\frac {55\,a^2\,b^2\,x^5}{16}}{a^3\,b^5+3\,a^2\,b^6\,x^2+3\,a\,b^7\,x^4+b^8\,x^6}+\frac {105\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,b^{11/2}}-\frac {4\,a\,x}{b^5} \]