Integrand size = 25, antiderivative size = 389 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=-\frac {3 A b^2-a b B-10 a A c}{2 a^2 \left (b^2-4 a c\right ) x}-\frac {a b B-A \left (b^2-2 a c\right )-(A b-2 a B) c x^2}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac {\sqrt {c} \left (a B \left (b^2-12 a c+b \sqrt {b^2-4 a c}\right )-A \left (3 b^3-16 a b c+3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (3 A b^2-a b B-10 a A c+\frac {a B \left (b^2-12 a c\right )-A \left (3 b^3-16 a b c\right )}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}} \]
1/2*(10*A*a*c-3*A*b^2+B*a*b)/a^2/(-4*a*c+b^2)/x+1/2*(-a*b*B+A*(-2*a*c+b^2) +(A*b-2*B*a)*c*x^2)/a/(-4*a*c+b^2)/x/(c*x^4+b*x^2+a)+1/4*arctan(x*2^(1/2)* c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(a*B*(b^2-12*a*c+b*(-4*a*c+b ^2)^(1/2))-A*(3*b^3-16*a*b*c+3*b^2*(-4*a*c+b^2)^(1/2)-10*a*c*(-4*a*c+b^2)^ (1/2)))/a^2/(-4*a*c+b^2)^(3/2)*2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-1/4*ar ctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(3*A*b^2-a*b* B-10*A*a*c+(a*B*(-12*a*c+b^2)-A*(-16*a*b*c+3*b^3))/(-4*a*c+b^2)^(1/2))/a^2 /(-4*a*c+b^2)*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.70 (sec) , antiderivative size = 382, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\frac {-\frac {4 A}{x}+\frac {2 x \left (a B \left (b^2-2 a c+b c x^2\right )-A \left (b^3-3 a b c+b^2 c x^2-2 a c^2 x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\sqrt {2} \sqrt {c} \left (a B \left (b^2-12 a c+b \sqrt {b^2-4 a c}\right )+A \left (-3 b^3+16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \sqrt {c} \left (a B \left (-b^2+12 a c+b \sqrt {b^2-4 a c}\right )+A \left (3 b^3-16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{4 a^2} \]
((-4*A)/x + (2*x*(a*B*(b^2 - 2*a*c + b*c*x^2) - A*(b^3 - 3*a*b*c + b^2*c*x ^2 - 2*a*c^2*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[2]*Sqrt[c] *(a*B*(b^2 - 12*a*c + b*Sqrt[b^2 - 4*a*c]) + A*(-3*b^3 + 16*a*b*c - 3*b^2* Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/ Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a *c]]) + (Sqrt[2]*Sqrt[c]*(a*B*(-b^2 + 12*a*c + b*Sqrt[b^2 - 4*a*c]) + A*(3 *b^3 - 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c]))*Arc Tan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2) *Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(4*a^2)
Time = 0.70 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1600, 25, 1604, 25, 1480, 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.
| \(\displaystyle \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1600 |
| \(\displaystyle -\frac {\int -\frac {3 A b^2-a B b+3 (A b-2 a B) c x^2-10 a A c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 A b^2-a B b+3 (A b-2 a B) c x^2-10 a A c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 1604 |
| \(\displaystyle \frac {-\frac {\int -\frac {-c \left (3 A b^2-a B b-10 a A c\right ) x^2+a B \left (b^2-6 a c\right )-A \left (3 b^3-13 a b c\right )}{c x^4+b x^2+a}dx}{a}-\frac {-10 a A c-a b B+3 A b^2}{a x}}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {-c \left (3 A b^2-a B b-10 a A c\right ) x^2+a B \left (b^2-6 a c\right )-A \left (3 b^3-13 a b c\right )}{c x^4+b x^2+a}dx}{a}-\frac {-10 a A c-a b B+3 A b^2}{a x}}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {c \left (a B \left (b \sqrt {b^2-4 a c}-12 a c+b^2\right )-A \left (3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx}{2 \sqrt {b^2-4 a c}}-\frac {1}{2} c \left (\frac {a B \left (b^2-12 a c\right )-A \left (3 b^3-16 a b c\right )}{\sqrt {b^2-4 a c}}-10 a A c-a b B+3 A b^2\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{a}-\frac {-10 a A c-a b B+3 A b^2}{a x}}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\sqrt {c} \left (a B \left (b \sqrt {b^2-4 a c}-12 a c+b^2\right )-A \left (3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (\frac {a B \left (b^2-12 a c\right )-A \left (3 b^3-16 a b c\right )}{\sqrt {b^2-4 a c}}-10 a A c-a b B+3 A b^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {-10 a A c-a b B+3 A b^2}{a x}}{2 a \left (b^2-4 a c\right )}-\frac {-A \left (b^2-2 a c\right )-\left (c x^2 (A b-2 a B)\right )+a b B}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
-1/2*(a*b*B - A*(b^2 - 2*a*c) - (A*b - 2*a*B)*c*x^2)/(a*(b^2 - 4*a*c)*x*(a + b*x^2 + c*x^4)) + (-((3*A*b^2 - a*b*B - 10*a*A*c)/(a*x)) + ((Sqrt[c]*(a *B*(b^2 - 12*a*c + b*Sqrt[b^2 - 4*a*c]) - A*(3*b^3 - 16*a*b*c + 3*b^2*Sqrt [b^2 - 4*a*c] - 10*a*c*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt [b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4 *a*c]]) - (Sqrt[c]*(3*A*b^2 - a*b*B - 10*a*A*c + (a*B*(b^2 - 12*a*c) - A*( 3*b^3 - 16*a*b*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/a)/(2*a*(b^2 - 4*a*c))
3.2.22.3.1 Defintions of rubi rules used
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[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^2 + c*x^4)^(p + 1) *((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2)/(2*a*f*(p + 1)*(b^2 - 4*a *c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^m*(a + b*x^2 + c *x^4)^(p + 1)*Simp[d*(b^2*(m + 2*(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) - a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Int egerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) /(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Time = 0.16 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.97
| method | result | size |
| default | \(-\frac {A}{a^{2} x}-\frac {\frac {\frac {c \left (2 A a c -A \,b^{2}+a b B \right ) x^{3}}{8 a c -2 b^{2}}+\frac {\left (3 A a b c -A \,b^{3}-2 a^{2} B c +B a \,b^{2}\right ) x}{8 a c -2 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {2 c \left (\frac {\left (10 A \sqrt {-4 a c +b^{2}}\, a c -3 A \sqrt {-4 a c +b^{2}}\, b^{2}-16 A a b c +3 A \,b^{3}+a b B \sqrt {-4 a c +b^{2}}+12 a^{2} B c -B a \,b^{2}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\left (10 A \sqrt {-4 a c +b^{2}}\, a c -3 A \sqrt {-4 a c +b^{2}}\, b^{2}+16 A a b c -3 A \,b^{3}+a b B \sqrt {-4 a c +b^{2}}-12 a^{2} B c +B a \,b^{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{a^{2}}\) | \(379\) |
| risch | \(\text {Expression too large to display}\) | \(1520\) |
-A/a^2/x-1/a^2*((1/2*c*(2*A*a*c-A*b^2+B*a*b)/(4*a*c-b^2)*x^3+1/2*(3*A*a*b* c-A*b^3-2*B*a^2*c+B*a*b^2)/(4*a*c-b^2)*x)/(c*x^4+b*x^2+a)+2/(4*a*c-b^2)*c* (1/8*(10*A*(-4*a*c+b^2)^(1/2)*a*c-3*A*(-4*a*c+b^2)^(1/2)*b^2-16*A*a*b*c+3* A*b^3+a*b*B*(-4*a*c+b^2)^(1/2)+12*a^2*B*c-B*a*b^2)/(-4*a*c+b^2)^(1/2)*2^(1 /2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^( 1/2))*c)^(1/2))-1/8*(10*A*(-4*a*c+b^2)^(1/2)*a*c-3*A*(-4*a*c+b^2)^(1/2)*b^ 2+16*A*a*b*c-3*A*b^3+a*b*B*(-4*a*c+b^2)^(1/2)-12*a^2*B*c+B*a*b^2)/(-4*a*c+ b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x*2^(1/2)/( (-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 7583 vs. \(2 (338) = 676\).
Time = 8.89 (sec) , antiderivative size = 7583, normalized size of antiderivative = 19.49 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {B x^{2} + A}{{\left (c x^{4} + b x^{2} + a\right )}^{2} x^{2}} \,d x } \]
1/2*((10*A*a*c^2 + (B*a*b - 3*A*b^2)*c)*x^4 - 2*A*a*b^2 + 8*A*a^2*c + (B*a *b^2 - 3*A*b^3 - (2*B*a^2 - 11*A*a*b)*c)*x^2)/((a^2*b^2*c - 4*a^3*c^2)*x^5 + (a^2*b^3 - 4*a^3*b*c)*x^3 + (a^3*b^2 - 4*a^4*c)*x) + 1/2*integrate((B*a *b^2 - 3*A*b^3 + (10*A*a*c^2 + (B*a*b - 3*A*b^2)*c)*x^2 - (6*B*a^2 - 13*A* a*b)*c)/(c*x^4 + b*x^2 + a), x)/(a^2*b^2 - 4*a^3*c)
Leaf count of result is larger than twice the leaf count of optimal. 5408 vs. \(2 (338) = 676\).
Time = 1.36 (sec) , antiderivative size = 5408, normalized size of antiderivative = 13.90 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
1/2*(B*a*b*c*x^4 - 3*A*b^2*c*x^4 + 10*A*a*c^2*x^4 + B*a*b^2*x^2 - 3*A*b^3* x^2 - 2*B*a^2*c*x^2 + 11*A*a*b*c*x^2 - 2*A*a*b^2 + 8*A*a^2*c)/((c*x^5 + b* x^3 + a*x)*(a^2*b^2 - 4*a^3*c)) - 1/16*((6*b^4*c^2 - 44*a*b^2*c^3 + 80*a^2 *c^4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4 + 2 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c + 6*sq rt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c - 40*sqrt(2) *sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^2 - 20*sqrt(2)*sq rt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 - 3*sqrt(2)*sqrt(b ^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^2 + 10*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^3 - 6*(b^2 - 4*a*c)*b^2*c^2 + 20*(b^2 - 4*a*c)*a*c^3)*(a^2*b^2 - 4*a^3*c)^2*A - (2*a*b^3*c^2 - 8*a^2*b* c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3 + 4* sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c + 2*sqrt (2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c - sqrt(2)*sq rt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 - 2*(b^2 - 4*a*c)* a*b*c^2)*(a^2*b^2 - 4*a^3*c)^2*B + 2*(3*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a* c)*c)*a^2*b^7 - 37*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^5*c - 6*s qrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^6*c - 6*a^2*b^7*c + 152*sqrt( 2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^3*c^2 + 50*sqrt(2)*sqrt(b*c + sqr t(b^2 - 4*a*c)*c)*a^3*b^4*c^2 + 3*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*...
Time = 11.09 (sec) , antiderivative size = 17591, normalized size of antiderivative = 45.22 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
- (A/a - (x^2*(3*A*b^3 - B*a*b^2 + 2*B*a^2*c - 11*A*a*b*c))/(2*a^2*(4*a*c - b^2)) + (c*x^4*(10*A*a*c - 3*A*b^2 + B*a*b))/(2*a^2*(4*a*c - b^2)))/(a*x + b*x^3 + c*x^5) - atan((((-(9*A^2*b^13 + B^2*a^2*b^11 + 9*A^2*b^4*(-(4*a *c - b^2)^9)^(1/2) - 6*A*B*a*b^12 + 2077*A^2*a^2*b^9*c^2 - 10656*A^2*a^3*b ^7*c^3 + 30240*A^2*a^4*b^5*c^4 - 44800*A^2*a^5*b^3*c^5 + 25*A^2*a^2*c^2*(- (4*a*c - b^2)^9)^(1/2) + B^2*a^2*b^2*(-(4*a*c - b^2)^9)^(1/2) + 288*B^2*a^ 4*b^7*c^2 - 1504*B^2*a^5*b^5*c^3 + 3840*B^2*a^6*b^3*c^4 - 15360*A*B*a^7*c^ 6 - 213*A^2*a*b^11*c + 26880*A^2*a^6*b*c^6 - 27*B^2*a^3*b^9*c - 3840*B^2*a ^7*b*c^5 - 9*B^2*a^3*c*(-(4*a*c - b^2)^9)^(1/2) - 1548*A*B*a^3*b^8*c^2 + 8 064*A*B*a^4*b^6*c^3 - 22400*A*B*a^5*b^4*c^4 + 30720*A*B*a^6*b^2*c^5 - 51*A ^2*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2) - 6*A*B*a*b^3*(-(4*a*c - b^2)^9)^(1/2) + 152*A*B*a^2*b^10*c + 44*A*B*a^2*b*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a^5* b^12 + 4096*a^11*c^6 - 24*a^6*b^10*c + 240*a^7*b^8*c^2 - 1280*a^8*b^6*c^3 + 3840*a^9*b^4*c^4 - 6144*a^10*b^2*c^5)))^(1/2)*(x*(-(9*A^2*b^13 + B^2*a^2 *b^11 + 9*A^2*b^4*(-(4*a*c - b^2)^9)^(1/2) - 6*A*B*a*b^12 + 2077*A^2*a^2*b ^9*c^2 - 10656*A^2*a^3*b^7*c^3 + 30240*A^2*a^4*b^5*c^4 - 44800*A^2*a^5*b^3 *c^5 + 25*A^2*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) + B^2*a^2*b^2*(-(4*a*c - b^ 2)^9)^(1/2) + 288*B^2*a^4*b^7*c^2 - 1504*B^2*a^5*b^5*c^3 + 3840*B^2*a^6*b^ 3*c^4 - 15360*A*B*a^7*c^6 - 213*A^2*a*b^11*c + 26880*A^2*a^6*b*c^6 - 27*B^ 2*a^3*b^9*c - 3840*B^2*a^7*b*c^5 - 9*B^2*a^3*c*(-(4*a*c - b^2)^9)^(1/2)...