Integrand size = 25, antiderivative size = 170 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {\left (b^2 B-3 A b c+2 a B c\right ) \left (b+2 c x^2\right )}{4 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {\left (b^2 B-3 A b c+2 a B c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]
1/4*(-a*(-2*A*c+B*b)-(-A*b*c-2*B*a*c+B*b^2)*x^2)/c/(-4*a*c+b^2)/(c*x^4+b*x ^2+a)^2+1/4*(-3*A*b*c+2*B*a*c+B*b^2)*(2*c*x^2+b)/c/(-4*a*c+b^2)^2/(c*x^4+b *x^2+a)-(-3*A*b*c+2*B*a*c+B*b^2)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/( -4*a*c+b^2)^(5/2)
Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {1}{4} \left (\frac {\left (b^2 B-3 A b c+2 a B c\right ) \left (b+2 c x^2\right )}{c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {a b B+b (b B-A c) x^2-2 a c \left (A+B x^2\right )}{c \left (-b^2+4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {4 \left (b^2 B-3 A b c+2 a B c\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{5/2}}\right ) \]
(((b^2*B - 3*A*b*c + 2*a*B*c)*(b + 2*c*x^2))/(c*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (a*b*B + b*(b*B - A*c)*x^2 - 2*a*c*(A + B*x^2))/(c*(-b^2 + 4* a*c)*(a + b*x^2 + c*x^4)^2) + (4*(b^2*B - 3*A*b*c + 2*a*B*c)*ArcTan[(b + 2 *c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/4
Time = 0.33 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1578, 1224, 1086, 1083, 219}
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.
| \(\displaystyle \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2 \left (B x^2+A\right )}{\left (c x^4+b x^2+a\right )^3}dx^2\) |
| \(\Big \downarrow \) 1224 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (2 a B c-3 A b c+b^2 B\right ) \int \frac {1}{\left (c x^4+b x^2+a\right )^2}dx^2}{2 c \left (b^2-4 a c\right )}-\frac {x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 1086 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (2 a B c-3 A b c+b^2 B\right ) \left (-\frac {2 c \int \frac {1}{c x^4+b x^2+a}dx^2}{b^2-4 a c}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{2 c \left (b^2-4 a c\right )}-\frac {x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (2 a B c-3 A b c+b^2 B\right ) \left (\frac {4 c \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{b^2-4 a c}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{2 c \left (b^2-4 a c\right )}-\frac {x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (2 a B c-3 A b c+b^2 B\right ) \left (\frac {4 c \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{2 c \left (b^2-4 a c\right )}-\frac {x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}\right )\) |
(-1/2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x^2)/(c*(b^2 - 4*a*c)*( a + b*x^2 + c*x^4)^2) - ((b^2*B - 3*A*b*c + 2*a*B*c)*(-((b + 2*c*x^2)/((b^ 2 - 4*a*c)*(a + b*x^2 + c*x^4))) + (4*c*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4 *a*c]])/(b^2 - 4*a*c)^(3/2)))/(2*c*(b^2 - 4*a*c)))/2
3.2.28.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && ILtQ[p, -1]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - ( b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x))*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c *(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(c*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, - 1] && !(IntegerQ[p] && NeQ[a, 0] && NiceSqrtQ[b^2 - 4*a*c])
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Time = 0.15 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.61
| method | result | size |
| default | \(\frac {-\frac {c \left (3 A b c -2 B a c -B \,b^{2}\right ) x^{6}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {3 b \left (3 A b c -2 B a c -B \,b^{2}\right ) x^{4}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 A a b c +A \,b^{3}+2 a^{2} B c -5 B a \,b^{2}\right ) x^{2}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {a \left (8 A a c +A \,b^{2}-6 a b B \right )}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{2 \left (c \,x^{4}+b \,x^{2}+a \right )^{2}}-\frac {\left (3 A b c -2 B a c -B \,b^{2}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}\) | \(273\) |
| risch | \(\frac {-\frac {c \left (3 A b c -2 B a c -B \,b^{2}\right ) x^{6}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {3 b \left (3 A b c -2 B a c -B \,b^{2}\right ) x^{4}}{4 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 A a b c +A \,b^{3}+2 a^{2} B c -5 B a \,b^{2}\right ) x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (8 A a c +A \,b^{2}-6 a b B \right )}{4 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) x^{2}-32 a^{3} c^{2}+16 a^{2} b^{2} c -2 b^{4} a \right ) A b c}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {\ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) x^{2}-32 a^{3} c^{2}+16 a^{2} b^{2} c -2 b^{4} a \right ) B a c}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {\ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) x^{2}-32 a^{3} c^{2}+16 a^{2} b^{2} c -2 b^{4} a \right ) B \,b^{2}}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) x^{2}+32 a^{3} c^{2}-16 a^{2} b^{2} c +2 b^{4} a \right ) A b c}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}-\frac {\ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) x^{2}+32 a^{3} c^{2}-16 a^{2} b^{2} c +2 b^{4} a \right ) B a c}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}-\frac {\ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) x^{2}+32 a^{3} c^{2}-16 a^{2} b^{2} c +2 b^{4} a \right ) B \,b^{2}}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\) | \(661\) |
1/2*(-c*(3*A*b*c-2*B*a*c-B*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^6-3/2*b*(3*A* b*c-2*B*a*c-B*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^4-(5*A*a*b*c+A*b^3+2*B*a^2 *c-5*B*a*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-1/2*a*(8*A*a*c+A*b^2-6*B*a*b) /(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^4+b*x^2+a)^2-(3*A*b*c-2*B*a*c-B*b^2)/(16 *a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^( 1/2))
Leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (162) = 324\).
Time = 0.31 (sec) , antiderivative size = 1226, normalized size of antiderivative = 7.21 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \]
[1/4*(2*(B*b^4*c - 4*(2*B*a^2 - 3*A*a*b)*c^3 - (2*B*a*b^2 + 3*A*b^3)*c^2)* x^6 + 6*B*a^2*b^3 - A*a*b^4 + 32*A*a^3*c^2 + 3*(B*b^5 - 4*(2*B*a^2*b - 3*A *a*b^2)*c^2 - (2*B*a*b^3 + 3*A*b^4)*c)*x^4 + 2*(5*B*a*b^4 - A*b^5 + 4*(2*B *a^3 + 5*A*a^2*b)*c^2 - (22*B*a^2*b^2 + A*a*b^3)*c)*x^2 - 2*((B*b^2*c^2 + (2*B*a - 3*A*b)*c^3)*x^8 + 2*(B*b^3*c + (2*B*a*b - 3*A*b^2)*c^2)*x^6 + B*a ^2*b^2 + (B*b^4 + 2*(2*B*a^2 - 3*A*a*b)*c^2 + (4*B*a*b^2 - 3*A*b^3)*c)*x^4 + 2*(B*a*b^3 + (2*B*a^2*b - 3*A*a*b^2)*c)*x^2 + (2*B*a^3 - 3*A*a^2*b)*c)* sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c + (2*c*x^2 + b) *sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) - 4*(6*B*a^3*b + A*a^2*b^2)*c)/(( b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a ^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2* b^3*c^3 - 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3* b^2*c^3 - 128*a^4*c^4)*x^4 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64 *a^4*b*c^3)*x^2), 1/4*(2*(B*b^4*c - 4*(2*B*a^2 - 3*A*a*b)*c^3 - (2*B*a*b^2 + 3*A*b^3)*c^2)*x^6 + 6*B*a^2*b^3 - A*a*b^4 + 32*A*a^3*c^2 + 3*(B*b^5 - 4 *(2*B*a^2*b - 3*A*a*b^2)*c^2 - (2*B*a*b^3 + 3*A*b^4)*c)*x^4 + 2*(5*B*a*b^4 - A*b^5 + 4*(2*B*a^3 + 5*A*a^2*b)*c^2 - (22*B*a^2*b^2 + A*a*b^3)*c)*x^2 - 4*((B*b^2*c^2 + (2*B*a - 3*A*b)*c^3)*x^8 + 2*(B*b^3*c + (2*B*a*b - 3*A*b^ 2)*c^2)*x^6 + B*a^2*b^2 + (B*b^4 + 2*(2*B*a^2 - 3*A*a*b)*c^2 + (4*B*a*b^2 - 3*A*b^3)*c)*x^4 + 2*(B*a*b^3 + (2*B*a^2*b - 3*A*a*b^2)*c)*x^2 + (2*B*...
Timed out. \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 1.55 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {{\left (B b^{2} + 2 \, B a c - 3 \, A b c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {2 \, B b^{2} c x^{6} + 4 \, B a c^{2} x^{6} - 6 \, A b c^{2} x^{6} + 3 \, B b^{3} x^{4} + 6 \, B a b c x^{4} - 9 \, A b^{2} c x^{4} + 10 \, B a b^{2} x^{2} - 2 \, A b^{3} x^{2} - 4 \, B a^{2} c x^{2} - 10 \, A a b c x^{2} + 6 \, B a^{2} b - A a b^{2} - 8 \, A a^{2} c}{4 \, {\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} \]
(B*b^2 + 2*B*a*c - 3*A*b*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) + 1/4*(2*B*b^2*c*x^6 + 4*B* a*c^2*x^6 - 6*A*b*c^2*x^6 + 3*B*b^3*x^4 + 6*B*a*b*c*x^4 - 9*A*b^2*c*x^4 + 10*B*a*b^2*x^2 - 2*A*b^3*x^2 - 4*B*a^2*c*x^2 - 10*A*a*b*c*x^2 + 6*B*a^2*b - A*a*b^2 - 8*A*a^2*c)/((c*x^4 + b*x^2 + a)^2*(b^4 - 8*a*b^2*c + 16*a^2*c^ 2))
Time = 7.82 (sec) , antiderivative size = 587, normalized size of antiderivative = 3.45 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {\left (B\,b^2\,c^2-3\,A\,b\,c^3+2\,B\,a\,c^3\right )\,\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}{a\,{\left (4\,a\,c-b^2\right )}^{9/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {b\,{\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}^2\,\left (32\,a^2\,b\,c^4-16\,a\,b^3\,c^3+2\,b^5\,c^2\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{15/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )+\frac {2\,b\,c^2\,{\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}^2}{{\left (4\,a\,c-b^2\right )}^{15/2}}\right )\,\left (b^4\,{\left (4\,a\,c-b^2\right )}^5+16\,a^2\,c^2\,{\left (4\,a\,c-b^2\right )}^5-8\,a\,b^2\,c\,{\left (4\,a\,c-b^2\right )}^5\right )}{18\,A^2\,b^2\,c^4-24\,A\,B\,a\,b\,c^4-12\,A\,B\,b^3\,c^3+8\,B^2\,a^2\,c^4+8\,B^2\,a\,b^2\,c^3+2\,B^2\,b^4\,c^2}\right )\,\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {-6\,B\,a^2\,b+8\,A\,c\,a^2+A\,a\,b^2}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (2\,B\,c\,a^2-5\,B\,a\,b^2+5\,A\,c\,a\,b+A\,b^3\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {3\,b\,x^4\,\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {c\,x^6\,\left (B\,b^2-3\,A\,c\,b+2\,B\,a\,c\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^4\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^8+2\,a\,b\,x^2+2\,b\,c\,x^6} \]
(atan(((x^2*(((B*b^2*c^2 - 3*A*b*c^3 + 2*B*a*c^3)*(B*b^2 - 3*A*b*c + 2*B*a *c))/(a*(4*a*c - b^2)^(9/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b*(B*b^2 - 3*A*b*c + 2*B*a*c)^2*(2*b^5*c^2 - 16*a*b^3*c^3 + 32*a^2*b*c^4))/(2*a*(4*a* c - b^2)^(15/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))) + (2*b*c^2*(B*b^2 - 3*A*b *c + 2*B*a*c)^2)/(4*a*c - b^2)^(15/2))*(b^4*(4*a*c - b^2)^5 + 16*a^2*c^2*( 4*a*c - b^2)^5 - 8*a*b^2*c*(4*a*c - b^2)^5))/(18*A^2*b^2*c^4 + 8*B^2*a^2*c ^4 + 2*B^2*b^4*c^2 - 12*A*B*b^3*c^3 + 8*B^2*a*b^2*c^3 - 24*A*B*a*b*c^4))*( B*b^2 - 3*A*b*c + 2*B*a*c))/(4*a*c - b^2)^(5/2) - ((A*a*b^2 + 8*A*a^2*c - 6*B*a^2*b)/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(A*b^3 - 5*B*a*b^2 + 2*B*a^2*c + 5*A*a*b*c))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (3*b*x^4*(B*b ^2 - 3*A*b*c + 2*B*a*c))/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (c*x^6*(B*b^ 2 - 3*A*b*c + 2*B*a*c))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^4*(2*a*c + b^2) + a^2 + c^2*x^8 + 2*a*b*x^2 + 2*b*c*x^6)