Integrand size = 19, antiderivative size = 236 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\frac {d (b c-4 a d) (3 b c+a d) x}{8 a^2 c (b c-a d)^3 \left (c+d x^2\right )}+\frac {b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {3 b (b c-3 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b^{3/2} \left (3 b^2 c^2-14 a b c d+35 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} (b c-a d)^4}-\frac {d^{5/2} (7 b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{3/2} (b c-a d)^4} \]
1/8*d*(-4*a*d+b*c)*(a*d+3*b*c)*x/a^2/c/(-a*d+b*c)^3/(d*x^2+c)+1/4*b*x/a/(- a*d+b*c)/(b*x^2+a)^2/(d*x^2+c)+3/8*b*(-3*a*d+b*c)*x/a^2/(-a*d+b*c)^2/(b*x^ 2+a)/(d*x^2+c)+1/8*b^(3/2)*(35*a^2*d^2-14*a*b*c*d+3*b^2*c^2)*arctan(x*b^(1 /2)/a^(1/2))/a^(5/2)/(-a*d+b*c)^4-1/2*d^(5/2)*(-a*d+7*b*c)*arctan(x*d^(1/2 )/c^(1/2))/c^(3/2)/(-a*d+b*c)^4
Time = 0.26 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\frac {1}{8} \left (\frac {2 b^2 x}{a (b c-a d)^2 \left (a+b x^2\right )^2}+\frac {b^2 (-3 b c+11 a d) x}{a^2 (-b c+a d)^3 \left (a+b x^2\right )}-\frac {4 d^3 x}{c (b c-a d)^3 \left (c+d x^2\right )}+\frac {b^{3/2} \left (3 b^2 c^2-14 a b c d+35 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^4}+\frac {4 d^{5/2} (-7 b c+a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2} (b c-a d)^4}\right ) \]
((2*b^2*x)/(a*(b*c - a*d)^2*(a + b*x^2)^2) + (b^2*(-3*b*c + 11*a*d)*x)/(a^ 2*(-(b*c) + a*d)^3*(a + b*x^2)) - (4*d^3*x)/(c*(b*c - a*d)^3*(c + d*x^2)) + (b^(3/2)*(3*b^2*c^2 - 14*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a ]])/(a^(5/2)*(b*c - a*d)^4) + (4*d^(5/2)*(-7*b*c + a*d)*ArcTan[(Sqrt[d]*x) /Sqrt[c]])/(c^(3/2)*(b*c - a*d)^4))/8
Time = 0.46 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {316, 25, 402, 25, 402, 27, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}-\frac {\int -\frac {5 b d x^2+3 b c-4 a d}{\left (b x^2+a\right )^2 \left (d x^2+c\right )^2}dx}{4 a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 b d x^2+3 b c-4 a d}{\left (b x^2+a\right )^2 \left (d x^2+c\right )^2}dx}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\int -\frac {3 b^2 c^2-5 a b d c+8 a^2 d^2+9 b d (b c-3 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 a (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b^2 c^2-5 a b d c+8 a^2 d^2+9 b d (b c-3 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 a (b c-a d)}+\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (3 b^3 c^3-11 a b^2 d c^2+24 a^2 b d^2 c-4 a^3 d^3+b d (b c-4 a d) (3 b c+a d) x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {d x (b c-4 a d) (a d+3 b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 b^3 c^3-11 a b^2 d c^2+24 a^2 b d^2 c-4 a^3 d^3+b d (b c-4 a d) (3 b c+a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{c (b c-a d)}+\frac {d x (b c-4 a d) (a d+3 b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 c \left (35 a^2 d^2-14 a b c d+3 b^2 c^2\right ) \int \frac {1}{b x^2+a}dx}{b c-a d}-\frac {4 a^2 d^3 (7 b c-a d) \int \frac {1}{d x^2+c}dx}{b c-a d}}{c (b c-a d)}+\frac {d x (b c-4 a d) (a d+3 b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^{3/2} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (35 a^2 d^2-14 a b c d+3 b^2 c^2\right )}{\sqrt {a} (b c-a d)}-\frac {4 a^2 d^{5/2} (7 b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}}{c (b c-a d)}+\frac {d x (b c-4 a d) (a d+3 b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}+\frac {3 b x (b c-3 a d)}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)}\) |
(b*x)/(4*a*(b*c - a*d)*(a + b*x^2)^2*(c + d*x^2)) + ((3*b*(b*c - 3*a*d)*x) /(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)) + ((d*(b*c - 4*a*d)*(3*b*c + a* d)*x)/(c*(b*c - a*d)*(c + d*x^2)) + ((b^(3/2)*c*(3*b^2*c^2 - 14*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) - (4*a^2*d^ (5/2)*(7*b*c - a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(c *(b*c - a*d)))/(2*a*(b*c - a*d)))/(4*a*(b*c - a*d))
3.1.41.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Time = 8.53 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.83
| method | result | size |
| default | \(\frac {b^{2} \left (\frac {\frac {b \left (11 a^{2} d^{2}-14 a b c d +3 b^{2} c^{2}\right ) x^{3}}{8 a^{2}}+\frac {\left (13 a^{2} d^{2}-18 a b c d +5 b^{2} c^{2}\right ) x}{8 a}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (35 a^{2} d^{2}-14 a b c d +3 b^{2} c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a d -b c \right )^{4}}+\frac {d^{3} \left (\frac {\left (a d -b c \right ) x}{2 c \left (d \,x^{2}+c \right )}+\frac {\left (a d -7 b c \right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 c \sqrt {c d}}\right )}{\left (a d -b c \right )^{4}}\) | \(196\) |
| risch | \(\text {Expression too large to display}\) | \(3743\) |
b^2/(a*d-b*c)^4*((1/8*b*(11*a^2*d^2-14*a*b*c*d+3*b^2*c^2)/a^2*x^3+1/8*(13* a^2*d^2-18*a*b*c*d+5*b^2*c^2)/a*x)/(b*x^2+a)^2+1/8*(35*a^2*d^2-14*a*b*c*d+ 3*b^2*c^2)/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))+d^3/(a*d-b*c)^4*(1/2*( a*d-b*c)/c*x/(d*x^2+c)+1/2*(a*d-7*b*c)/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. 788 vs. \(2 (210) = 420\).
Time = 2.41 (sec) , antiderivative size = 3251, normalized size of antiderivative = 13.78 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
[1/16*(2*(3*b^5*c^3*d - 14*a*b^4*c^2*d^2 + 7*a^2*b^3*c*d^3 + 4*a^3*b^2*d^4 )*x^5 + 2*(3*b^5*c^4 - 9*a*b^4*c^3*d - 7*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 + 8*a^4*b*d^4)*x^3 + (3*a^2*b^3*c^4 - 14*a^3*b^2*c^3*d + 35*a^4*b*c^2*d^2 + (3*b^5*c^3*d - 14*a*b^4*c^2*d^2 + 35*a^2*b^3*c*d^3)*x^6 + (3*b^5*c^4 - 8*a*b^4*c^3*d + 7*a^2*b^3*c^2*d^2 + 70*a^3*b^2*c*d^3)*x^4 + (6*a*b^4*c^4 - 25*a^2*b^3*c^3*d + 56*a^3*b^2*c^2*d^2 + 35*a^4*b*c*d^3)*x^2)*sqrt(-b/a)*l og((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 4*(7*a^4*b*c^2*d^2 - a^5* c*d^3 + (7*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^6 + (7*a^2*b^3*c^2*d^2 + 13*a^3* b^2*c*d^3 - 2*a^4*b*d^4)*x^4 + (14*a^3*b^2*c^2*d^2 + 5*a^4*b*c*d^3 - a^5*d ^4)*x^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 2*(5 *a*b^4*c^4 - 18*a^2*b^3*c^3*d + 13*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + 4*a^5 *d^4)*x)/(a^4*b^4*c^6 - 4*a^5*b^3*c^5*d + 6*a^6*b^2*c^4*d^2 - 4*a^7*b*c^3* d^3 + a^8*c^2*d^4 + (a^2*b^6*c^5*d - 4*a^3*b^5*c^4*d^2 + 6*a^4*b^4*c^3*d^3 - 4*a^5*b^3*c^2*d^4 + a^6*b^2*c*d^5)*x^6 + (a^2*b^6*c^6 - 2*a^3*b^5*c^5*d - 2*a^4*b^4*c^4*d^2 + 8*a^5*b^3*c^3*d^3 - 7*a^6*b^2*c^2*d^4 + 2*a^7*b*c*d ^5)*x^4 + (2*a^3*b^5*c^6 - 7*a^4*b^4*c^5*d + 8*a^5*b^3*c^4*d^2 - 2*a^6*b^2 *c^3*d^3 - 2*a^7*b*c^2*d^4 + a^8*c*d^5)*x^2), 1/16*(2*(3*b^5*c^3*d - 14*a* b^4*c^2*d^2 + 7*a^2*b^3*c*d^3 + 4*a^3*b^2*d^4)*x^5 + 2*(3*b^5*c^4 - 9*a*b^ 4*c^3*d - 7*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 + 8*a^4*b*d^4)*x^3 - 8*(7*a^ 4*b*c^2*d^2 - a^5*c*d^3 + (7*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^6 + (7*a^2*...
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (210) = 420\).
Time = 0.30 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.25 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\frac {{\left (3 \, b^{4} c^{2} - 14 \, a b^{3} c d + 35 \, a^{2} b^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )} \sqrt {a b}} - \frac {{\left (7 \, b c d^{3} - a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{4} c^{5} - 4 \, a b^{3} c^{4} d + 6 \, a^{2} b^{2} c^{3} d^{2} - 4 \, a^{3} b c^{2} d^{3} + a^{4} c d^{4}\right )} \sqrt {c d}} + \frac {{\left (3 \, b^{4} c^{2} d - 11 \, a b^{3} c d^{2} - 4 \, a^{2} b^{2} d^{3}\right )} x^{5} + {\left (3 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d - 13 \, a^{2} b^{2} c d^{2} - 8 \, a^{3} b d^{3}\right )} x^{3} + {\left (5 \, a b^{3} c^{3} - 13 \, a^{2} b^{2} c^{2} d - 4 \, a^{4} d^{3}\right )} x}{8 \, {\left (a^{4} b^{3} c^{5} - 3 \, a^{5} b^{2} c^{4} d + 3 \, a^{6} b c^{3} d^{2} - a^{7} c^{2} d^{3} + {\left (a^{2} b^{5} c^{4} d - 3 \, a^{3} b^{4} c^{3} d^{2} + 3 \, a^{4} b^{3} c^{2} d^{3} - a^{5} b^{2} c d^{4}\right )} x^{6} + {\left (a^{2} b^{5} c^{5} - a^{3} b^{4} c^{4} d - 3 \, a^{4} b^{3} c^{3} d^{2} + 5 \, a^{5} b^{2} c^{2} d^{3} - 2 \, a^{6} b c d^{4}\right )} x^{4} + {\left (2 \, a^{3} b^{4} c^{5} - 5 \, a^{4} b^{3} c^{4} d + 3 \, a^{5} b^{2} c^{3} d^{2} + a^{6} b c^{2} d^{3} - a^{7} c d^{4}\right )} x^{2}\right )}} \]
1/8*(3*b^4*c^2 - 14*a*b^3*c*d + 35*a^2*b^2*d^2)*arctan(b*x/sqrt(a*b))/((a^ 2*b^4*c^4 - 4*a^3*b^3*c^3*d + 6*a^4*b^2*c^2*d^2 - 4*a^5*b*c*d^3 + a^6*d^4) *sqrt(a*b)) - 1/2*(7*b*c*d^3 - a*d^4)*arctan(d*x/sqrt(c*d))/((b^4*c^5 - 4* a*b^3*c^4*d + 6*a^2*b^2*c^3*d^2 - 4*a^3*b*c^2*d^3 + a^4*c*d^4)*sqrt(c*d)) + 1/8*((3*b^4*c^2*d - 11*a*b^3*c*d^2 - 4*a^2*b^2*d^3)*x^5 + (3*b^4*c^3 - 6 *a*b^3*c^2*d - 13*a^2*b^2*c*d^2 - 8*a^3*b*d^3)*x^3 + (5*a*b^3*c^3 - 13*a^2 *b^2*c^2*d - 4*a^4*d^3)*x)/(a^4*b^3*c^5 - 3*a^5*b^2*c^4*d + 3*a^6*b*c^3*d^ 2 - a^7*c^2*d^3 + (a^2*b^5*c^4*d - 3*a^3*b^4*c^3*d^2 + 3*a^4*b^3*c^2*d^3 - a^5*b^2*c*d^4)*x^6 + (a^2*b^5*c^5 - a^3*b^4*c^4*d - 3*a^4*b^3*c^3*d^2 + 5 *a^5*b^2*c^2*d^3 - 2*a^6*b*c*d^4)*x^4 + (2*a^3*b^4*c^5 - 5*a^4*b^3*c^4*d + 3*a^5*b^2*c^3*d^2 + a^6*b*c^2*d^3 - a^7*c*d^4)*x^2)
Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=-\frac {d^{3} x}{2 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} {\left (d x^{2} + c\right )}} + \frac {{\left (3 \, b^{4} c^{2} - 14 \, a b^{3} c d + 35 \, a^{2} b^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )} \sqrt {a b}} - \frac {{\left (7 \, b c d^{3} - a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{4} c^{5} - 4 \, a b^{3} c^{4} d + 6 \, a^{2} b^{2} c^{3} d^{2} - 4 \, a^{3} b c^{2} d^{3} + a^{4} c d^{4}\right )} \sqrt {c d}} + \frac {3 \, b^{4} c x^{3} - 11 \, a b^{3} d x^{3} + 5 \, a b^{3} c x - 13 \, a^{2} b^{2} d x}{8 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} {\left (b x^{2} + a\right )}^{2}} \]
-1/2*d^3*x/((b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*(d*x^2 + c)) + 1/8*(3*b^4*c^2 - 14*a*b^3*c*d + 35*a^2*b^2*d^2)*arctan(b*x/sqrt(a *b))/((a^2*b^4*c^4 - 4*a^3*b^3*c^3*d + 6*a^4*b^2*c^2*d^2 - 4*a^5*b*c*d^3 + a^6*d^4)*sqrt(a*b)) - 1/2*(7*b*c*d^3 - a*d^4)*arctan(d*x/sqrt(c*d))/((b^4 *c^5 - 4*a*b^3*c^4*d + 6*a^2*b^2*c^3*d^2 - 4*a^3*b*c^2*d^3 + a^4*c*d^4)*sq rt(c*d)) + 1/8*(3*b^4*c*x^3 - 11*a*b^3*d*x^3 + 5*a*b^3*c*x - 13*a^2*b^2*d* x)/((a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - a^5*d^3)*(b*x^2 + a)^ 2)
Time = 7.58 (sec) , antiderivative size = 8635, normalized size of antiderivative = 36.59 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
((x^5*(4*a^2*b^2*d^3 - 3*b^4*c^2*d + 11*a*b^3*c*d^2))/(8*a^2*c*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)) + (x*(4*a^3*d^3 - 5*b^3*c^3 + 13 *a*b^2*c^2*d))/(8*a*c*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (b*x^ 3*(8*a^3*d^3 - 3*b^3*c^3 + 6*a*b^2*c^2*d + 13*a^2*b*c*d^2))/(8*a^2*c*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(a^2*c + x^2*(a^2*d + 2*a*b*c) + x^4*(b^2*c + 2*a*b*d) + b^2*d*x^6) - (atan(((((x*(16*a^6*b^3*d^9 + 9*b^9*c ^6*d^3 - 84*a*b^8*c^5*d^4 - 224*a^5*b^4*c*d^8 + 406*a^2*b^7*c^4*d^5 - 980* a^3*b^6*c^3*d^6 + 2009*a^4*b^5*c^2*d^7))/(32*(a^4*b^6*c^8 + a^10*c^2*d^6 - 6*a^5*b^5*c^7*d - 6*a^9*b*c^3*d^5 + 15*a^6*b^4*c^6*d^2 - 20*a^7*b^3*c^5*d ^3 + 15*a^8*b^2*c^4*d^4)) - (((2*a^13*b^2*c*d^13 - (3*a^2*b^13*c^12*d^2)/2 + (35*a^3*b^12*c^11*d^3)/2 - 98*a^4*b^11*c^10*d^4 + 336*a^5*b^10*c^9*d^5 - 765*a^6*b^9*c^8*d^6 + 1197*a^7*b^8*c^7*d^7 - 1302*a^8*b^7*c^6*d^8 + 978* a^9*b^6*c^5*d^9 - (987*a^10*b^5*c^4*d^10)/2 + (315*a^11*b^4*c^3*d^11)/2 - 28*a^12*b^3*c^2*d^12)/(a^4*b^9*c^11 - a^13*c^2*d^9 - 9*a^5*b^8*c^10*d + 9* a^12*b*c^3*d^8 + 36*a^6*b^7*c^9*d^2 - 84*a^7*b^6*c^8*d^3 + 126*a^8*b^5*c^7 *d^4 - 126*a^9*b^4*c^6*d^5 + 84*a^10*b^3*c^5*d^6 - 36*a^11*b^2*c^4*d^7) - (x*(-a^5*b^3)^(1/2)*(35*a^2*d^2 + 3*b^2*c^2 - 14*a*b*c*d)*(256*a^4*b^11*c^ 11*d^2 - 1792*a^5*b^10*c^10*d^3 + 5120*a^6*b^9*c^9*d^4 - 7168*a^7*b^8*c^8* d^5 + 3584*a^8*b^7*c^7*d^6 + 3584*a^9*b^6*c^6*d^7 - 7168*a^10*b^5*c^5*d^8 + 5120*a^11*b^4*c^4*d^9 - 1792*a^12*b^3*c^3*d^10 + 256*a^13*b^2*c^2*d^1...