Integrand size = 19, antiderivative size = 230 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {d (2 b c+a d) x}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d (4 b c-a d) (b c+3 a d) x}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac {b^{5/2} (b c-7 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)^4}+\frac {d^{3/2} \left (35 b^2 c^2-14 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} (b c-a d)^4} \]
1/4*d*(a*d+2*b*c)*x/a/c/(-a*d+b*c)^2/(d*x^2+c)^2+1/2*b*x/a/(-a*d+b*c)/(b*x ^2+a)/(d*x^2+c)^2+1/8*d*(-a*d+4*b*c)*(3*a*d+b*c)*x/a/c^2/(-a*d+b*c)^3/(d*x ^2+c)+1/2*b^(5/2)*(-7*a*d+b*c)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/(-a*d+b*c )^4+1/8*d^(3/2)*(3*a^2*d^2-14*a*b*c*d+35*b^2*c^2)*arctan(x*d^(1/2)/c^(1/2) )/c^(5/2)/(-a*d+b*c)^4
Time = 0.31 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {1}{8} \left (-\frac {4 b^3 x}{a (-b c+a d)^3 \left (a+b x^2\right )}+\frac {2 d^2 x}{c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac {d^2 (11 b c-3 a d) x}{c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac {4 b^{5/2} (b c-7 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^4}+\frac {d^{3/2} \left (35 b^2 c^2-14 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)^4}\right ) \]
((-4*b^3*x)/(a*(-(b*c) + a*d)^3*(a + b*x^2)) + (2*d^2*x)/(c*(b*c - a*d)^2* (c + d*x^2)^2) + (d^2*(11*b*c - 3*a*d)*x)/(c^2*(b*c - a*d)^3*(c + d*x^2)) + (4*b^(5/2)*(b*c - 7*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*(b*c - a* d)^4) + (d^(3/2)*(35*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/ Sqrt[c]])/(c^(5/2)*(b*c - a*d)^4))/8
Time = 0.44 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {316, 25, 402, 27, 402, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\int -\frac {5 b d x^2+b c-2 a d}{\left (b x^2+a\right ) \left (d x^2+c\right )^3}dx}{2 a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 b d x^2+b c-2 a d}{\left (b x^2+a\right ) \left (d x^2+c\right )^3}dx}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (2 b^2 c^2-8 a b d c+3 a^2 d^2+3 b d (2 b c+a d) x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}+\frac {d x (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {2 b^2 c^2-8 a b d c+3 a^2 d^2+3 b d (2 b c+a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 c (b c-a d)}+\frac {d x (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {4 b^3 c^3-24 a b^2 d c^2+11 a^2 b d^2 c-3 a^3 d^3+b d (4 b c-a d) (b c+3 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {d x (4 b c-a d) (3 a d+b c)}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}+\frac {d x (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {a d^2 \left (3 a^2 d^2-14 a b c d+35 b^2 c^2\right ) \int \frac {1}{d x^2+c}dx}{b c-a d}+\frac {4 b^3 c^2 (b c-7 a d) \int \frac {1}{b x^2+a}dx}{b c-a d}}{2 c (b c-a d)}+\frac {d x (4 b c-a d) (3 a d+b c)}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}+\frac {d x (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {a d^{3/2} \left (3 a^2 d^2-14 a b c d+35 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}+\frac {4 b^{5/2} c^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (b c-7 a d)}{\sqrt {a} (b c-a d)}}{2 c (b c-a d)}+\frac {d x (4 b c-a d) (3 a d+b c)}{2 c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}+\frac {d x (a d+2 b c)}{2 c \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}\) |
(b*x)/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^2) + ((d*(2*b*c + a*d)*x)/( 2*c*(b*c - a*d)*(c + d*x^2)^2) + ((d*(4*b*c - a*d)*(b*c + 3*a*d)*x)/(2*c*( b*c - a*d)*(c + d*x^2)) + ((4*b^(5/2)*c^2*(b*c - 7*a*d)*ArcTan[(Sqrt[b]*x) /Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) + (a*d^(3/2)*(35*b^2*c^2 - 14*a*b*c*d + 3 *a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(2*c*(b*c - a*d)))/(2*c*(b*c - a*d)))/(2*a*(b*c - a*d))
3.1.34.3.1 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 = 2.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.86
| method | result | size |
| default | \(-\frac {b^{3} \left (\frac {\left (a d -b c \right ) x}{2 a \left (b \,x^{2}+a \right )}+\frac {\left (7 a d -b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a d -b c \right )^{4}}+\frac {d^{2} \left (\frac {\frac {d \left (3 a^{2} d^{2}-14 a b c d +11 b^{2} c^{2}\right ) x^{3}}{8 c^{2}}+\frac {\left (5 a^{2} d^{2}-18 a b c d +13 b^{2} c^{2}\right ) x}{8 c}}{\left (d \,x^{2}+c \right )^{2}}+\frac {\left (3 a^{2} d^{2}-14 a b c d +35 b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 c^{2} \sqrt {c d}}\right )}{\left (a d -b c \right )^{4}}\) | \(198\) |
| risch | \(\text {Expression too large to display}\) | \(3743\) |
-b^3/(a*d-b*c)^4*(1/2*(a*d-b*c)/a*x/(b*x^2+a)+1/2*(7*a*d-b*c)/a/(a*b)^(1/2 )*arctan(b*x/(a*b)^(1/2)))+d^2/(a*d-b*c)^4*((1/8*d*(3*a^2*d^2-14*a*b*c*d+1 1*b^2*c^2)/c^2*x^3+1/8*(5*a^2*d^2-18*a*b*c*d+13*b^2*c^2)/c*x)/(d*x^2+c)^2+ 1/8*(3*a^2*d^2-14*a*b*c*d+35*b^2*c^2)/c^2/(c*d)^(1/2)*arctan(d*x/(c*d)^(1/ 2)))
Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (204) = 408\).
Time = 2.34 (sec) , antiderivative size = 3239, normalized size of antiderivative = 14.08 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \]
[1/16*(2*(4*b^4*c^3*d^2 + 7*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5 )*x^5 + 2*(8*b^4*c^4*d + 5*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3 - 9*a^3*b*c*d ^4 + 3*a^4*d^5)*x^3 - 4*(a*b^3*c^5 - 7*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 7*a* b^3*c^2*d^3)*x^6 + (2*b^4*c^4*d - 13*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3)*x^ 4 + (b^4*c^5 - 5*a*b^3*c^4*d - 14*a^2*b^2*c^3*d^2)*x^2)*sqrt(-b/a)*log((b* x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + (35*a^2*b^2*c^4*d - 14*a^3*b*c^ 3*d^2 + 3*a^4*c^2*d^3 + (35*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5 )*x^6 + (70*a*b^3*c^3*d^2 + 7*a^2*b^2*c^2*d^3 - 8*a^3*b*c*d^4 + 3*a^4*d^5) *x^4 + (35*a*b^3*c^4*d + 56*a^2*b^2*c^3*d^2 - 25*a^3*b*c^2*d^3 + 6*a^4*c*d ^4)*x^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 2*(4 *b^4*c^5 - 4*a*b^3*c^4*d + 13*a^2*b^2*c^3*d^2 - 18*a^3*b*c^2*d^3 + 5*a^4*c *d^4)*x)/(a^2*b^4*c^8 - 4*a^3*b^3*c^7*d + 6*a^4*b^2*c^6*d^2 - 4*a^5*b*c^5* d^3 + a^6*c^4*d^4 + (a*b^5*c^6*d^2 - 4*a^2*b^4*c^5*d^3 + 6*a^3*b^3*c^4*d^4 - 4*a^4*b^2*c^3*d^5 + a^5*b*c^2*d^6)*x^6 + (2*a*b^5*c^7*d - 7*a^2*b^4*c^6 *d^2 + 8*a^3*b^3*c^5*d^3 - 2*a^4*b^2*c^4*d^4 - 2*a^5*b*c^3*d^5 + a^6*c^2*d ^6)*x^4 + (a*b^5*c^8 - 2*a^2*b^4*c^7*d - 2*a^3*b^3*c^6*d^2 + 8*a^4*b^2*c^5 *d^3 - 7*a^5*b*c^4*d^4 + 2*a^6*c^3*d^5)*x^2), 1/8*((4*b^4*c^3*d^2 + 7*a*b^ 3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^5 + (8*b^4*c^4*d + 5*a*b^3*c ^3*d^2 - 7*a^2*b^2*c^2*d^3 - 9*a^3*b*c*d^4 + 3*a^4*d^5)*x^3 + (35*a^2*b^2* c^4*d - 14*a^3*b*c^3*d^2 + 3*a^4*c^2*d^3 + (35*a*b^3*c^2*d^3 - 14*a^2*b...
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (204) = 408\).
Time = 0.30 (sec) , antiderivative size = 529, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {{\left (b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{4} c^{4} - 4 \, a^{2} b^{3} c^{3} d + 6 \, a^{3} b^{2} c^{2} d^{2} - 4 \, a^{4} b c d^{3} + a^{5} d^{4}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 14 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{4} c^{6} - 4 \, a b^{3} c^{5} d + 6 \, a^{2} b^{2} c^{4} d^{2} - 4 \, a^{3} b c^{3} d^{3} + a^{4} c^{2} d^{4}\right )} \sqrt {c d}} + \frac {{\left (4 \, b^{3} c^{2} d^{2} + 11 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} x^{5} + {\left (8 \, b^{3} c^{3} d + 13 \, a b^{2} c^{2} d^{2} + 6 \, a^{2} b c d^{3} - 3 \, a^{3} d^{4}\right )} x^{3} + {\left (4 \, b^{3} c^{4} + 13 \, a^{2} b c^{2} d^{2} - 5 \, a^{3} c d^{3}\right )} x}{8 \, {\left (a^{2} b^{3} c^{7} - 3 \, a^{3} b^{2} c^{6} d + 3 \, a^{4} b c^{5} d^{2} - a^{5} c^{4} d^{3} + {\left (a b^{4} c^{5} d^{2} - 3 \, a^{2} b^{3} c^{4} d^{3} + 3 \, a^{3} b^{2} c^{3} d^{4} - a^{4} b c^{2} d^{5}\right )} x^{6} + {\left (2 \, a b^{4} c^{6} d - 5 \, a^{2} b^{3} c^{5} d^{2} + 3 \, a^{3} b^{2} c^{4} d^{3} + a^{4} b c^{3} d^{4} - a^{5} c^{2} d^{5}\right )} x^{4} + {\left (a b^{4} c^{7} - a^{2} b^{3} c^{6} d - 3 \, a^{3} b^{2} c^{5} d^{2} + 5 \, a^{4} b c^{4} d^{3} - 2 \, a^{5} c^{3} d^{4}\right )} x^{2}\right )}} \]
1/2*(b^4*c - 7*a*b^3*d)*arctan(b*x/sqrt(a*b))/((a*b^4*c^4 - 4*a^2*b^3*c^3* d + 6*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + a^5*d^4)*sqrt(a*b)) + 1/8*(35*b^2* c^2*d^2 - 14*a*b*c*d^3 + 3*a^2*d^4)*arctan(d*x/sqrt(c*d))/((b^4*c^6 - 4*a* b^3*c^5*d + 6*a^2*b^2*c^4*d^2 - 4*a^3*b*c^3*d^3 + a^4*c^2*d^4)*sqrt(c*d)) + 1/8*((4*b^3*c^2*d^2 + 11*a*b^2*c*d^3 - 3*a^2*b*d^4)*x^5 + (8*b^3*c^3*d + 13*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 - 3*a^3*d^4)*x^3 + (4*b^3*c^4 + 13*a^2*b *c^2*d^2 - 5*a^3*c*d^3)*x)/(a^2*b^3*c^7 - 3*a^3*b^2*c^6*d + 3*a^4*b*c^5*d^ 2 - a^5*c^4*d^3 + (a*b^4*c^5*d^2 - 3*a^2*b^3*c^4*d^3 + 3*a^3*b^2*c^3*d^4 - a^4*b*c^2*d^5)*x^6 + (2*a*b^4*c^6*d - 5*a^2*b^3*c^5*d^2 + 3*a^3*b^2*c^4*d ^3 + a^4*b*c^3*d^4 - a^5*c^2*d^5)*x^4 + (a*b^4*c^7 - a^2*b^3*c^6*d - 3*a^3 *b^2*c^5*d^2 + 5*a^4*b*c^4*d^3 - 2*a^5*c^3*d^4)*x^2)
Time = 0.28 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {b^{3} x}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} {\left (b x^{2} + a\right )}} + \frac {{\left (b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{4} c^{4} - 4 \, a^{2} b^{3} c^{3} d + 6 \, a^{3} b^{2} c^{2} d^{2} - 4 \, a^{4} b c d^{3} + a^{5} d^{4}\right )} \sqrt {a b}} + \frac {{\left (35 \, b^{2} c^{2} d^{2} - 14 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{4} c^{6} - 4 \, a b^{3} c^{5} d + 6 \, a^{2} b^{2} c^{4} d^{2} - 4 \, a^{3} b c^{3} d^{3} + a^{4} c^{2} d^{4}\right )} \sqrt {c d}} + \frac {11 \, b c d^{3} x^{3} - 3 \, a d^{4} x^{3} + 13 \, b c^{2} d^{2} x - 5 \, a c d^{3} x}{8 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} {\left (d x^{2} + c\right )}^{2}} \]
1/2*b^3*x/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*(b*x^2 + a)) + 1/2*(b^4*c - 7*a*b^3*d)*arctan(b*x/sqrt(a*b))/((a*b^4*c^4 - 4*a^2* b^3*c^3*d + 6*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + a^5*d^4)*sqrt(a*b)) + 1/8* (35*b^2*c^2*d^2 - 14*a*b*c*d^3 + 3*a^2*d^4)*arctan(d*x/sqrt(c*d))/((b^4*c^ 6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*d^2 - 4*a^3*b*c^3*d^3 + a^4*c^2*d^4)*sqr t(c*d)) + 1/8*(11*b*c*d^3*x^3 - 3*a*d^4*x^3 + 13*b*c^2*d^2*x - 5*a*c*d^3*x )/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*(d*x^2 + c)^2 )
Time = 7.47 (sec) , antiderivative size = 8649, normalized size of antiderivative = 37.60 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \]
(atan(((((x*(9*a^6*b^3*d^9 + 16*b^9*c^6*d^3 - 224*a*b^8*c^5*d^4 - 84*a^5*b ^4*c*d^8 + 2009*a^2*b^7*c^4*d^5 - 980*a^3*b^6*c^3*d^6 + 406*a^4*b^5*c^2*d^ 7))/(32*(a^2*b^6*c^10 + a^8*c^4*d^6 - 6*a^3*b^5*c^9*d - 6*a^7*b*c^5*d^5 + 15*a^4*b^4*c^8*d^2 - 20*a^5*b^3*c^7*d^3 + 15*a^6*b^2*c^6*d^4)) - (((2*a*b^ 13*c^13*d^2 - 28*a^2*b^12*c^12*d^3 + (315*a^3*b^11*c^11*d^4)/2 - (987*a^4* b^10*c^10*d^5)/2 + 978*a^5*b^9*c^9*d^6 - 1302*a^6*b^8*c^8*d^7 + 1197*a^7*b ^7*c^7*d^8 - 765*a^8*b^6*c^6*d^9 + 336*a^9*b^5*c^5*d^10 - 98*a^10*b^4*c^4* d^11 + (35*a^11*b^3*c^3*d^12)/2 - (3*a^12*b^2*c^2*d^13)/2)/(a^2*b^9*c^13 - a^11*c^4*d^9 - 9*a^3*b^8*c^12*d + 9*a^10*b*c^5*d^8 + 36*a^4*b^7*c^11*d^2 - 84*a^5*b^6*c^10*d^3 + 126*a^6*b^5*c^9*d^4 - 126*a^7*b^4*c^8*d^5 + 84*a^8 *b^3*c^7*d^6 - 36*a^9*b^2*c^6*d^7) - (x*(-c^5*d^3)^(1/2)*(3*a^2*d^2 + 35*b ^2*c^2 - 14*a*b*c*d)*(256*a^2*b^11*c^13*d^2 - 1792*a^3*b^10*c^12*d^3 + 512 0*a^4*b^9*c^11*d^4 - 7168*a^5*b^8*c^10*d^5 + 3584*a^6*b^7*c^9*d^6 + 3584*a ^7*b^6*c^8*d^7 - 7168*a^8*b^5*c^7*d^8 + 5120*a^9*b^4*c^6*d^9 - 1792*a^10*b ^3*c^5*d^10 + 256*a^11*b^2*c^4*d^11))/(512*(b^4*c^9 + a^4*c^5*d^4 - 4*a^3* b*c^6*d^3 + 6*a^2*b^2*c^7*d^2 - 4*a*b^3*c^8*d)*(a^2*b^6*c^10 + a^8*c^4*d^6 - 6*a^3*b^5*c^9*d - 6*a^7*b*c^5*d^5 + 15*a^4*b^4*c^8*d^2 - 20*a^5*b^3*c^7 *d^3 + 15*a^6*b^2*c^6*d^4)))*(-c^5*d^3)^(1/2)*(3*a^2*d^2 + 35*b^2*c^2 - 14 *a*b*c*d))/(16*(b^4*c^9 + a^4*c^5*d^4 - 4*a^3*b*c^6*d^3 + 6*a^2*b^2*c^7*d^ 2 - 4*a*b^3*c^8*d)))*(-c^5*d^3)^(1/2)*(3*a^2*d^2 + 35*b^2*c^2 - 14*a*b*...