Integrand size = 22, antiderivative size = 271 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=-\frac {5 b^2 c^2-4 a b c d+5 a^2 d^2}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {(b c+a d) \left (5 b^2 c^2-9 a b c d+5 a^2 d^2\right )}{2 a^3 c^3 (b c-a d)^2 x}+\frac {d (b c+a d)}{2 a c (b c-a d)^2 x^3 \left (c+d x^2\right )}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b^{7/2} (5 b c-9 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} (b c-a d)^3}+\frac {d^{7/2} (9 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^3} \]
1/6*(-5*a^2*d^2+4*a*b*c*d-5*b^2*c^2)/a^2/c^2/(-a*d+b*c)^2/x^3+1/2*(a*d+b*c )*(5*a^2*d^2-9*a*b*c*d+5*b^2*c^2)/a^3/c^3/(-a*d+b*c)^2/x+1/2*d*(a*d+b*c)/a /c/(-a*d+b*c)^2/x^3/(d*x^2+c)+1/2*b/a/(-a*d+b*c)/x^3/(b*x^2+a)/(d*x^2+c)+1 /2*b^(7/2)*(-9*a*d+5*b*c)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/(-a*d+b*c)^3+1 /2*d^(7/2)*(-5*a*d+9*b*c)*arctan(x*d^(1/2)/c^(1/2))/c^(7/2)/(-a*d+b*c)^3
Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {1}{6} \left (-\frac {2}{a^2 c^2 x^3}+\frac {12 (b c+a d)}{a^3 c^3 x}+\frac {3 b^4 x}{a^3 (b c-a d)^2 \left (a+b x^2\right )}+\frac {3 d^4 x}{c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {3 b^{7/2} (-5 b c+9 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} (-b c+a d)^3}+\frac {3 d^{7/2} (9 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)^3}\right ) \]
(-2/(a^2*c^2*x^3) + (12*(b*c + a*d))/(a^3*c^3*x) + (3*b^4*x)/(a^3*(b*c - a *d)^2*(a + b*x^2)) + (3*d^4*x)/(c^3*(b*c - a*d)^2*(c + d*x^2)) + (3*b^(7/2 )*(-5*b*c + 9*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(7/2)*(-(b*c) + a*d)^3) + (3*d^(7/2)*(9*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(7/2)*(b*c - a*d)^3))/6
Time = 0.56 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {374, 25, 441, 27, 445, 27, 445, 397, 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.
| \(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\int -\frac {7 b d x^2+5 b c-2 a d}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {7 b d x^2+5 b c-2 a d}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (5 b^2 c^2-4 a b d c+5 a^2 d^2+5 b d (b c+a d) x^2\right )}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {5 b^2 c^2-4 a b d c+5 a^2 d^2+5 b d (b c+a d) x^2}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )}dx}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {3 \left (b d \left (5 b^2 c^2-4 a b d c+5 a^2 d^2\right ) x^2+(b c+a d) \left (5 b^2 c^2-9 a b d c+5 a^2 d^2\right )\right )}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )}dx}{3 a c}-\frac {\frac {5 b^2 c}{a}+\frac {5 a d^2}{c}-4 b d}{3 x^3}}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {b d \left (5 b^2 c^2-4 a b d c+5 a^2 d^2\right ) x^2+(b c+a d) \left (5 b^2 c^2-9 a b d c+5 a^2 d^2\right )}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )}dx}{a c}-\frac {\frac {5 b^2 c}{a}+\frac {5 a d^2}{c}-4 b d}{3 x^3}}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {5 b^4 c^4-4 a b^3 d c^3-4 a^2 b^2 d^2 c^2-4 a^3 b d^3 c+5 a^4 d^4+b d (b c+a d) \left (5 b^2 c^2-9 a b d c+5 a^2 d^2\right ) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{a c}-\frac {(a d+b c) \left (5 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{a c x}}{a c}-\frac {\frac {5 b^2 c}{a}+\frac {5 a d^2}{c}-4 b d}{3 x^3}}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {\frac {a^3 d^4 (9 b c-5 a d) \int \frac {1}{d x^2+c}dx}{b c-a d}+\frac {b^4 c^3 (5 b c-9 a d) \int \frac {1}{b x^2+a}dx}{b c-a d}}{a c}-\frac {(a d+b c) \left (5 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{a c x}}{a c}-\frac {\frac {5 b^2 c}{a}+\frac {5 a d^2}{c}-4 b d}{3 x^3}}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {\frac {a^3 d^{7/2} (9 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}+\frac {b^{7/2} c^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (5 b c-9 a d)}{\sqrt {a} (b c-a d)}}{a c}-\frac {(a d+b c) \left (5 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{a c x}}{a c}-\frac {\frac {5 b^2 c}{a}+\frac {5 a d^2}{c}-4 b d}{3 x^3}}{c (b c-a d)}+\frac {d (a d+b c)}{c x^3 \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\) |
b/(2*a*(b*c - a*d)*x^3*(a + b*x^2)*(c + d*x^2)) + ((d*(b*c + a*d))/(c*(b*c - a*d)*x^3*(c + d*x^2)) + (-1/3*((5*b^2*c)/a - 4*b*d + (5*a*d^2)/c)/x^3 - (-(((b*c + a*d)*(5*b^2*c^2 - 9*a*b*c*d + 5*a^2*d^2))/(a*c*x)) - ((b^(7/2) *c^3*(5*b*c - 9*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) + (a^3*d^(7/2)*(9*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(a*c))/(a*c))/(c*(b*c - a*d)))/(2*a*(b*c - a*d))
3.4.9.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Time = 2.87 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.59
| method | result | size |
| default | \(-\frac {1}{3 a^{2} c^{2} x^{3}}-\frac {-2 a d -2 b c}{x \,c^{3} a^{3}}+\frac {b^{4} \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (9 a d -5 b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3} \left (a d -b c \right )^{3}}+\frac {d^{4} \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{d \,x^{2}+c}+\frac {\left (5 a d -9 b c \right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \sqrt {c d}}\right )}{c^{3} \left (a d -b c \right )^{3}}\) | \(159\) |
| risch | \(\text {Expression too large to display}\) | \(2929\) |
-1/3/a^2/c^2/x^3-(-2*a*d-2*b*c)/x/c^3/a^3+b^4/a^3/(a*d-b*c)^3*((1/2*a*d-1/ 2*b*c)*x/(b*x^2+a)+1/2*(9*a*d-5*b*c)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))+ d^4/c^3/(a*d-b*c)^3*((1/2*a*d-1/2*b*c)*x/(d*x^2+c)+1/2*(5*a*d-9*b*c)/(c*d) ^(1/2)*arctan(d*x/(c*d)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 591 vs. \(2 (243) = 486\).
Time = 3.38 (sec) , antiderivative size = 2457, normalized size of antiderivative = 9.07 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
[-1/12*(4*a^2*b^3*c^5 - 12*a^3*b^2*c^4*d + 12*a^4*b*c^3*d^2 - 4*a^5*c^2*d^ 3 - 6*(5*b^5*c^4*d - 9*a*b^4*c^3*d^2 + 9*a^3*b^2*c*d^4 - 5*a^4*b*d^5)*x^6 - 2*(15*b^5*c^5 - 17*a*b^4*c^4*d - 18*a^2*b^3*c^3*d^2 + 18*a^3*b^2*c^2*d^3 + 17*a^4*b*c*d^4 - 15*a^5*d^5)*x^4 - 20*(a*b^4*c^5 - 2*a^2*b^3*c^4*d + 2* a^4*b*c^2*d^3 - a^5*c*d^4)*x^2 - 3*((5*b^5*c^4*d - 9*a*b^4*c^3*d^2)*x^7 + (5*b^5*c^5 - 4*a*b^4*c^4*d - 9*a^2*b^3*c^3*d^2)*x^5 + (5*a*b^4*c^5 - 9*a^2 *b^3*c^4*d)*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a) ) - 3*((9*a^3*b^2*c*d^4 - 5*a^4*b*d^5)*x^7 + (9*a^3*b^2*c^2*d^3 + 4*a^4*b* c*d^4 - 5*a^5*d^5)*x^5 + (9*a^4*b*c^2*d^3 - 5*a^5*c*d^4)*x^3)*sqrt(-d/c)*l og((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^3*b^4*c^6*d - 3*a^4*b^ 3*c^5*d^2 + 3*a^5*b^2*c^4*d^3 - a^6*b*c^3*d^4)*x^7 + (a^3*b^4*c^7 - 2*a^4* b^3*c^6*d + 2*a^6*b*c^4*d^3 - a^7*c^3*d^4)*x^5 + (a^4*b^3*c^7 - 3*a^5*b^2* c^6*d + 3*a^6*b*c^5*d^2 - a^7*c^4*d^3)*x^3), -1/12*(4*a^2*b^3*c^5 - 12*a^3 *b^2*c^4*d + 12*a^4*b*c^3*d^2 - 4*a^5*c^2*d^3 - 6*(5*b^5*c^4*d - 9*a*b^4*c ^3*d^2 + 9*a^3*b^2*c*d^4 - 5*a^4*b*d^5)*x^6 - 2*(15*b^5*c^5 - 17*a*b^4*c^4 *d - 18*a^2*b^3*c^3*d^2 + 18*a^3*b^2*c^2*d^3 + 17*a^4*b*c*d^4 - 15*a^5*d^5 )*x^4 - 20*(a*b^4*c^5 - 2*a^2*b^3*c^4*d + 2*a^4*b*c^2*d^3 - a^5*c*d^4)*x^2 - 6*((9*a^3*b^2*c*d^4 - 5*a^4*b*d^5)*x^7 + (9*a^3*b^2*c^2*d^3 + 4*a^4*b*c *d^4 - 5*a^5*d^5)*x^5 + (9*a^4*b*c^2*d^3 - 5*a^5*c*d^4)*x^3)*sqrt(d/c)*arc tan(x*sqrt(d/c)) - 3*((5*b^5*c^4*d - 9*a*b^4*c^3*d^2)*x^7 + (5*b^5*c^5 ...
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.70 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {{\left (5 \, b^{5} c - 9 \, a b^{4} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3}\right )} \sqrt {a b}} + \frac {{\left (9 \, b c d^{4} - 5 \, a d^{5}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} - \frac {2 \, a^{2} b^{2} c^{4} - 4 \, a^{3} b c^{3} d + 2 \, a^{4} c^{2} d^{2} - 3 \, {\left (5 \, b^{4} c^{3} d - 4 \, a b^{3} c^{2} d^{2} - 4 \, a^{2} b^{2} c d^{3} + 5 \, a^{3} b d^{4}\right )} x^{6} - {\left (15 \, b^{4} c^{4} - 2 \, a b^{3} c^{3} d - 20 \, a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} + 15 \, a^{4} d^{4}\right )} x^{4} - 10 \, {\left (a b^{3} c^{4} - a^{2} b^{2} c^{3} d - a^{3} b c^{2} d^{2} + a^{4} c d^{3}\right )} x^{2}}{6 \, {\left ({\left (a^{3} b^{3} c^{5} d - 2 \, a^{4} b^{2} c^{4} d^{2} + a^{5} b c^{3} d^{3}\right )} x^{7} + {\left (a^{3} b^{3} c^{6} - a^{4} b^{2} c^{5} d - a^{5} b c^{4} d^{2} + a^{6} c^{3} d^{3}\right )} x^{5} + {\left (a^{4} b^{2} c^{6} - 2 \, a^{5} b c^{5} d + a^{6} c^{4} d^{2}\right )} x^{3}\right )}} \]
1/2*(5*b^5*c - 9*a*b^4*d)*arctan(b*x/sqrt(a*b))/((a^3*b^3*c^3 - 3*a^4*b^2* c^2*d + 3*a^5*b*c*d^2 - a^6*d^3)*sqrt(a*b)) + 1/2*(9*b*c*d^4 - 5*a*d^5)*ar ctan(d*x/sqrt(c*d))/((b^3*c^6 - 3*a*b^2*c^5*d + 3*a^2*b*c^4*d^2 - a^3*c^3* d^3)*sqrt(c*d)) - 1/6*(2*a^2*b^2*c^4 - 4*a^3*b*c^3*d + 2*a^4*c^2*d^2 - 3*( 5*b^4*c^3*d - 4*a*b^3*c^2*d^2 - 4*a^2*b^2*c*d^3 + 5*a^3*b*d^4)*x^6 - (15*b ^4*c^4 - 2*a*b^3*c^3*d - 20*a^2*b^2*c^2*d^2 - 2*a^3*b*c*d^3 + 15*a^4*d^4)* x^4 - 10*(a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + a^4*c*d^3)*x^2)/((a^ 3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 + a^5*b*c^3*d^3)*x^7 + (a^3*b^3*c^6 - a^4* b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x^5 + (a^4*b^2*c^6 - 2*a^5*b*c^5* d + a^6*c^4*d^2)*x^3)
Time = 0.29 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {{\left (5 \, b^{5} c - 9 \, a b^{4} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3}\right )} \sqrt {a b}} + \frac {{\left (9 \, b c d^{4} - 5 \, a d^{5}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{3} c^{6} - 3 \, a b^{2} c^{5} d + 3 \, a^{2} b c^{4} d^{2} - a^{3} c^{3} d^{3}\right )} \sqrt {c d}} + \frac {b^{4} c^{3} d x^{3} + a^{3} b d^{4} x^{3} + b^{4} c^{4} x + a^{4} d^{4} x}{2 \, {\left (a^{3} b^{2} c^{5} - 2 \, a^{4} b c^{4} d + a^{5} c^{3} d^{2}\right )} {\left (b d x^{4} + b c x^{2} + a d x^{2} + a c\right )}} + \frac {6 \, b c x^{2} + 6 \, a d x^{2} - a c}{3 \, a^{3} c^{3} x^{3}} \]
1/2*(5*b^5*c - 9*a*b^4*d)*arctan(b*x/sqrt(a*b))/((a^3*b^3*c^3 - 3*a^4*b^2* c^2*d + 3*a^5*b*c*d^2 - a^6*d^3)*sqrt(a*b)) + 1/2*(9*b*c*d^4 - 5*a*d^5)*ar ctan(d*x/sqrt(c*d))/((b^3*c^6 - 3*a*b^2*c^5*d + 3*a^2*b*c^4*d^2 - a^3*c^3* d^3)*sqrt(c*d)) + 1/2*(b^4*c^3*d*x^3 + a^3*b*d^4*x^3 + b^4*c^4*x + a^4*d^4 *x)/((a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2)*(b*d*x^4 + b*c*x^2 + a*d* x^2 + a*c)) + 1/3*(6*b*c*x^2 + 6*a*d*x^2 - a*c)/(a^3*c^3*x^3)
Time = 6.71 (sec) , antiderivative size = 3978, normalized size of antiderivative = 14.68 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
(atan((a^9*d^3*x*(-c^7*d^7)^(3/2)*25i + b^9*c^16*d*x*(-c^7*d^7)^(1/2)*25i + a^2*b^7*c^14*d^3*x*(-c^7*d^7)^(1/2)*81i - a^8*b*c*d^2*x*(-c^7*d^7)^(3/2) *90i + a^7*b^2*c^2*d*x*(-c^7*d^7)^(3/2)*81i - a*b^8*c^15*d^2*x*(-c^7*d^7)^ (1/2)*90i)/(25*a^9*c^11*d^13 - 25*b^9*c^20*d^4 + 90*a*b^8*c^19*d^5 - 90*a^ 8*b*c^12*d^12 - 81*a^2*b^7*c^18*d^6 + 81*a^7*b^2*c^13*d^11))*(5*a*d - 9*b* c)*(-c^7*d^7)^(1/2)*1i)/(2*(b^3*c^10 - a^3*c^7*d^3 + 3*a^2*b*c^8*d^2 - 3*a *b^2*c^9*d)) - (1/(3*a*c) - (5*x^2*(a*d + b*c))/(3*a^2*c^2) + (x^4*(20*a^2 *b^2*c^2*d^2 - 15*b^4*c^4 - 15*a^4*d^4 + 2*a*b^3*c^3*d + 2*a^3*b*c*d^3))/( 6*a^3*c^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (b*d*x^6*(5*a^3*d^3 + 5*b^3*c ^3 - 4*a*b^2*c^2*d - 4*a^2*b*c*d^2))/(2*a^3*c^3*(a^2*d^2 + b^2*c^2 - 2*a*b *c*d)))/(x^5*(a*d + b*c) + a*c*x^3 + b*d*x^7) + (atan((((x*(400*a^9*b^17*c ^23*d^3 - 3840*a^10*b^16*c^22*d^4 + 15936*a^11*b^15*c^21*d^5 - 37376*a^12* b^14*c^20*d^6 + 54240*a^13*b^13*c^19*d^7 - 49920*a^14*b^12*c^18*d^8 + 2977 6*a^15*b^11*c^17*d^9 - 18432*a^16*b^10*c^16*d^10 + 29776*a^17*b^9*c^15*d^1 1 - 49920*a^18*b^8*c^14*d^12 + 54240*a^19*b^7*c^13*d^13 - 37376*a^20*b^6*c ^12*d^14 + 15936*a^21*b^5*c^11*d^15 - 3840*a^22*b^4*c^10*d^16 + 400*a^23*b ^3*c^9*d^17) - ((9*a*d - 5*b*c)*(-a^7*b^7)^(1/2)*(320*a^12*b^16*c^26*d^2 - 3456*a^13*b^15*c^25*d^3 + 16704*a^14*b^14*c^24*d^4 - 47616*a^15*b^13*c^23 *d^5 + 89280*a^16*b^12*c^22*d^6 - 118400*a^17*b^11*c^21*d^7 + 123072*a^18* b^10*c^20*d^8 - 119808*a^19*b^9*c^19*d^9 + 123072*a^20*b^8*c^18*d^10 - ...