Integrand size = 24, antiderivative size = 93 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {35 x}{16 b^4}-\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac {35 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}} \]
35/16*x/b^4-1/6*x^7/b/(b*x^2+a)^3-7/24*x^5/b^2/(b*x^2+a)^2-35/48*x^3/b^3/( b*x^2+a)-35/16*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(9/2)
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.83 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {105 a^3 x+280 a^2 b x^3+231 a b^2 x^5+48 b^3 x^7}{48 b^4 \left (a+b x^2\right )^3}-\frac {35 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}} \]
(105*a^3*x + 280*a^2*b*x^3 + 231*a*b^2*x^5 + 48*b^3*x^7)/(48*b^4*(a + b*x^ 2)^3) - (35*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*b^(9/2))
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.20, 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, 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.
\(\displaystyle \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^8}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^8}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {7 \int \frac {x^6}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {7 \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \int \frac {x^2}{b x^2+a}dx}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^2+a}dx}{b}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\) |
-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)
3.6.4.3.1 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 = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67
method | result | size |
default | \(\frac {x}{b^{4}}-\frac {a \left (\frac {-\frac {29}{16} b^{2} x^{5}-\frac {17}{6} a b \,x^{3}-\frac {19}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {35 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{4}}\) | \(62\) |
risch | \(\frac {x}{b^{4}}+\frac {\frac {29}{16} b^{2} x^{5} a +\frac {17}{6} a^{2} b \,x^{3}+\frac {19}{16} a^{3} x}{b^{4} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {35 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{32 b^{5}}-\frac {35 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{32 b^{5}}\) | \(114\) |
x/b^4-1/b^4*a*((-29/16*b^2*x^5-17/6*a*b*x^3-19/16*a^2*x)/(b*x^2+a)^3+35/16 /(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.88 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {96 \, b^{3} x^{7} + 462 \, a b^{2} x^{5} + 560 \, a^{2} b x^{3} + 210 \, a^{3} x + 105 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{96 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {48 \, b^{3} x^{7} + 231 \, a b^{2} x^{5} + 280 \, a^{2} b x^{3} + 105 \, a^{3} x - 105 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{48 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
[1/96*(96*b^3*x^7 + 462*a*b^2*x^5 + 560*a^2*b*x^3 + 210*a^3*x + 105*(b^3*x ^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(- a/b) - a)/(b*x^2 + a)))/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4), 1/48*(48*b^3*x^7 + 231*a*b^2*x^5 + 280*a^2*b*x^3 + 105*a^3*x - 105*(b^3*x ^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/( b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)]
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.41 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {35 \sqrt {- \frac {a}{b^{9}}} \log {\left (- b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{32} - \frac {35 \sqrt {- \frac {a}{b^{9}}} \log {\left (b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{32} + \frac {57 a^{3} x + 136 a^{2} b x^{3} + 87 a b^{2} x^{5}}{48 a^{3} b^{4} + 144 a^{2} b^{5} x^{2} + 144 a b^{6} x^{4} + 48 b^{7} x^{6}} + \frac {x}{b^{4}} \]
35*sqrt(-a/b**9)*log(-b**4*sqrt(-a/b**9) + x)/32 - 35*sqrt(-a/b**9)*log(b* *4*sqrt(-a/b**9) + x)/32 + (57*a**3*x + 136*a**2*b*x**3 + 87*a*b**2*x**5)/ (48*a**3*b**4 + 144*a**2*b**5*x**2 + 144*a*b**6*x**4 + 48*b**7*x**6) + x/b **4
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {87 \, a b^{2} x^{5} + 136 \, a^{2} b x^{3} + 57 \, a^{3} x}{48 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} - \frac {35 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{4}} + \frac {x}{b^{4}} \]
1/48*(87*a*b^2*x^5 + 136*a^2*b*x^3 + 57*a^3*x)/(b^7*x^6 + 3*a*b^6*x^4 + 3* a^2*b^5*x^2 + a^3*b^4) - 35/16*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + x /b^4
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.70 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{4}} + \frac {x}{b^{4}} + \frac {87 \, a b^{2} x^{5} + 136 \, a^{2} b x^{3} + 57 \, a^{3} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{4}} \]
-35/16*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + x/b^4 + 1/48*(87*a*b^2*x^ 5 + 136*a^2*b*x^3 + 57*a^3*x)/((b*x^2 + a)^3*b^4)
Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {x}{b^4}+\frac {\frac {19\,a^3\,x}{16}+\frac {17\,a^2\,b\,x^3}{6}+\frac {29\,a\,b^2\,x^5}{16}}{a^3\,b^4+3\,a^2\,b^5\,x^2+3\,a\,b^6\,x^4+b^7\,x^6}-\frac {35\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,b^{9/2}} \]