Integrand size = 24, antiderivative size = 106 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35}{16 a^4 x^3}+\frac {105 b}{16 a^5 x}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}+\frac {3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac {21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{11/2}} \]
-35/16/a^4/x^3+105/16*b/a^5/x+1/6/a/x^3/(b*x^2+a)^3+3/8/a^2/x^3/(b*x^2+a)^ 2+21/16/a^3/x^3/(b*x^2+a)+105/16*b^(3/2)*arctan(x*b^(1/2)/a^(1/2))/a^(11/2 )
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {\sqrt {a} \left (-16 a^4+144 a^3 b x^2+693 a^2 b^2 x^4+840 a b^3 x^6+315 b^4 x^8\right )}{x^3 \left (a+b x^2\right )^3}+315 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{48 a^{11/2}} \]
((Sqrt[a]*(-16*a^4 + 144*a^3*b*x^2 + 693*a^2*b^2*x^4 + 840*a*b^3*x^6 + 315 *b^4*x^8))/(x^3*(a + b*x^2)^3) + 315*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/ (48*a^(11/2))
Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1380, 27, 253, 253, 253, 264, 264, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {1}{b^4 x^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {3 \int \frac {1}{x^4 \left (b x^2+a\right )^3}dx}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {3 \left (\frac {7 \int \frac {1}{x^4 \left (b x^2+a\right )^2}dx}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \int \frac {1}{x^4 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {3 \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )}{2 a}+\frac {1}{6 a x^3 \left (a+b x^2\right )^3}\) |
1/(6*a*x^3*(a + b*x^2)^3) + (3*(1/(4*a*x^3*(a + b*x^2)^2) + (7*(1/(2*a*x^3 *(a + b*x^2)) + (5*(-1/3*1/(a*x^3) - (b*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqr t[b]*x)/Sqrt[a]])/a^(3/2)))/a))/(2*a)))/(4*a)))/(2*a)
3.6.10.3.1 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*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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 = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {1}{3 a^{4} x^{3}}+\frac {4 b}{a^{5} x}+\frac {b^{2} \left (\frac {\frac {41}{16} b^{2} x^{5}+\frac {35}{6} a b \,x^{3}+\frac {55}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {105 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{5}}\) | \(75\) |
risch | \(\frac {\frac {105 b^{4} x^{8}}{16 a^{5}}+\frac {35 b^{3} x^{6}}{2 a^{4}}+\frac {231 b^{2} x^{4}}{16 a^{3}}+\frac {3 b \,x^{2}}{a^{2}}-\frac {1}{3 a}}{x^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}+\frac {105 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{11} \textit {\_Z}^{2}+b^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{11}+2 b^{3}\right ) x -a^{6} b \textit {\_R} \right )\right )}{32}\) | \(127\) |
-1/3/a^4/x^3+4*b/a^5/x+b^2/a^5*((41/16*b^2*x^5+35/6*a*b*x^3+55/16*a^2*x)/( b*x^2+a)^3+105/16/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.87 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {630 \, b^{4} x^{8} + 1680 \, a b^{3} x^{6} + 1386 \, a^{2} b^{2} x^{4} + 288 \, a^{3} b x^{2} - 32 \, a^{4} + 315 \, {\left (b^{4} x^{9} + 3 \, a b^{3} x^{7} + 3 \, a^{2} b^{2} x^{5} + a^{3} b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{96 \, {\left (a^{5} b^{3} x^{9} + 3 \, a^{6} b^{2} x^{7} + 3 \, a^{7} b x^{5} + a^{8} x^{3}\right )}}, \frac {315 \, b^{4} x^{8} + 840 \, a b^{3} x^{6} + 693 \, a^{2} b^{2} x^{4} + 144 \, a^{3} b x^{2} - 16 \, a^{4} + 315 \, {\left (b^{4} x^{9} + 3 \, a b^{3} x^{7} + 3 \, a^{2} b^{2} x^{5} + a^{3} b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{48 \, {\left (a^{5} b^{3} x^{9} + 3 \, a^{6} b^{2} x^{7} + 3 \, a^{7} b x^{5} + a^{8} x^{3}\right )}}\right ] \]
[1/96*(630*b^4*x^8 + 1680*a*b^3*x^6 + 1386*a^2*b^2*x^4 + 288*a^3*b*x^2 - 3 2*a^4 + 315*(b^4*x^9 + 3*a*b^3*x^7 + 3*a^2*b^2*x^5 + a^3*b*x^3)*sqrt(-b/a) *log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^5*b^3*x^9 + 3*a^6*b^2 *x^7 + 3*a^7*b*x^5 + a^8*x^3), 1/48*(315*b^4*x^8 + 840*a*b^3*x^6 + 693*a^2 *b^2*x^4 + 144*a^3*b*x^2 - 16*a^4 + 315*(b^4*x^9 + 3*a*b^3*x^7 + 3*a^2*b^2 *x^5 + a^3*b*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^5*b^3*x^9 + 3*a^6*b^2* x^7 + 3*a^7*b*x^5 + a^8*x^3)]
Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.53 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=- \frac {105 \sqrt {- \frac {b^{3}}{a^{11}}} \log {\left (- \frac {a^{6} \sqrt {- \frac {b^{3}}{a^{11}}}}{b^{2}} + x \right )}}{32} + \frac {105 \sqrt {- \frac {b^{3}}{a^{11}}} \log {\left (\frac {a^{6} \sqrt {- \frac {b^{3}}{a^{11}}}}{b^{2}} + x \right )}}{32} + \frac {- 16 a^{4} + 144 a^{3} b x^{2} + 693 a^{2} b^{2} x^{4} + 840 a b^{3} x^{6} + 315 b^{4} x^{8}}{48 a^{8} x^{3} + 144 a^{7} b x^{5} + 144 a^{6} b^{2} x^{7} + 48 a^{5} b^{3} x^{9}} \]
-105*sqrt(-b**3/a**11)*log(-a**6*sqrt(-b**3/a**11)/b**2 + x)/32 + 105*sqrt (-b**3/a**11)*log(a**6*sqrt(-b**3/a**11)/b**2 + x)/32 + (-16*a**4 + 144*a* *3*b*x**2 + 693*a**2*b**2*x**4 + 840*a*b**3*x**6 + 315*b**4*x**8)/(48*a**8 *x**3 + 144*a**7*b*x**5 + 144*a**6*b**2*x**7 + 48*a**5*b**3*x**9)
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {315 \, b^{4} x^{8} + 840 \, a b^{3} x^{6} + 693 \, a^{2} b^{2} x^{4} + 144 \, a^{3} b x^{2} - 16 \, a^{4}}{48 \, {\left (a^{5} b^{3} x^{9} + 3 \, a^{6} b^{2} x^{7} + 3 \, a^{7} b x^{5} + a^{8} x^{3}\right )}} + \frac {105 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{5}} \]
1/48*(315*b^4*x^8 + 840*a*b^3*x^6 + 693*a^2*b^2*x^4 + 144*a^3*b*x^2 - 16*a ^4)/(a^5*b^3*x^9 + 3*a^6*b^2*x^7 + 3*a^7*b*x^5 + a^8*x^3) + 105/16*b^2*arc tan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5)
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {105 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{5}} + \frac {315 \, b^{4} x^{8} + 840 \, a b^{3} x^{6} + 693 \, a^{2} b^{2} x^{4} + 144 \, a^{3} b x^{2} - 16 \, a^{4}}{48 \, {\left (b x^{3} + a x\right )}^{3} a^{5}} \]
105/16*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/48*(315*b^4*x^8 + 840 *a*b^3*x^6 + 693*a^2*b^2*x^4 + 144*a^3*b*x^2 - 16*a^4)/((b*x^3 + a*x)^3*a^ 5)
Time = 13.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {3\,b\,x^2}{a^2}-\frac {1}{3\,a}+\frac {231\,b^2\,x^4}{16\,a^3}+\frac {35\,b^3\,x^6}{2\,a^4}+\frac {105\,b^4\,x^8}{16\,a^5}}{a^3\,x^3+3\,a^2\,b\,x^5+3\,a\,b^2\,x^7+b^3\,x^9}+\frac {105\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{11/2}} \]