Integrand size = 25, antiderivative size = 425 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {\left (3 b^2 B-A b c-10 a B c\right ) x}{2 c^2 \left (b^2-4 a c\right )}-\frac {(b B-2 A c) x^3}{2 c \left (b^2-4 a c\right )}-\frac {x^5 \left (A b-2 a B-(b B-2 A c) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\left (3 b^3 B-A b^2 c-13 a b B c+6 a A c^2-\frac {3 b^4 B-A b^3 c-19 a b^2 B c+8 a A b c^2+20 a^2 B c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (3 b^3 B-A b^2 c-13 a b B c+6 a A c^2+\frac {3 b^4 B-A b^3 c-19 a b^2 B c+8 a A b c^2+20 a^2 B c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}} \]
1/2*(-A*b*c-10*B*a*c+3*B*b^2)*x/c^2/(-4*a*c+b^2)-1/2*(-2*A*c+B*b)*x^3/c/(- 4*a*c+b^2)-1/2*x^5*(A*b-2*B*a-(-2*A*c+B*b)*x^2)/(-4*a*c+b^2)/(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))*(3*B*b^3-A*b ^2*c-13*B*a*b*c+6*A*a*c^2+(-8*A*a*b*c^2+A*b^3*c-20*B*a^2*c^2+19*B*a*b^2*c- 3*B*b^4)/(-4*a*c+b^2)^(1/2))/c^(5/2)/(-4*a*c+b^2)*2^(1/2)/(b-(-4*a*c+b^2)^ (1/2))^(1/2)-1/4*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*(3 *B*b^3-A*b^2*c-13*B*a*b*c+6*A*a*c^2+(8*A*a*b*c^2-A*b^3*c+20*B*a^2*c^2-19*B *a*b^2*c+3*B*b^4)/(-4*a*c+b^2)^(1/2))/c^(5/2)/(-4*a*c+b^2)*2^(1/2)/(b+(-4* a*c+b^2)^(1/2))^(1/2)
Time = 0.76 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.07 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {4 B \sqrt {c} x+\frac {2 \sqrt {c} x \left (-2 a^2 B c+b^2 (b B-A c) x^2+a \left (b^2 B+2 A c^2 x^2-b c \left (A+3 B x^2\right )\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\sqrt {2} \left (-3 b^4 B+b^2 c \left (19 a B-A \sqrt {b^2-4 a c}\right )+2 a c^2 \left (-10 a B+3 A \sqrt {b^2-4 a c}\right )+b^3 \left (A c+3 B \sqrt {b^2-4 a c}\right )-a b c \left (8 A c+13 B \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \left (3 b^4 B-b^2 c \left (19 a B+A \sqrt {b^2-4 a c}\right )+2 a c^2 \left (10 a B+3 A \sqrt {b^2-4 a c}\right )+a b c \left (8 A c-13 B \sqrt {b^2-4 a c}\right )+b^3 \left (-A c+3 B \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{4 c^{5/2}} \]
(4*B*Sqrt[c]*x + (2*Sqrt[c]*x*(-2*a^2*B*c + b^2*(b*B - A*c)*x^2 + a*(b^2*B + 2*A*c^2*x^2 - b*c*(A + 3*B*x^2))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (Sqrt[2]*(-3*b^4*B + b^2*c*(19*a*B - A*Sqrt[b^2 - 4*a*c]) + 2*a*c^2*(-10 *a*B + 3*A*Sqrt[b^2 - 4*a*c]) + b^3*(A*c + 3*B*Sqrt[b^2 - 4*a*c]) - a*b*c* (8*A*c + 13*B*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]*(3*b^4*B - b^2*c*(19*a*B + A*Sqrt[b^2 - 4*a*c]) + 2*a*c^2*(10*a*B + 3* A*Sqrt[b^2 - 4*a*c]) + a*b*c*(8*A*c - 13*B*Sqrt[b^2 - 4*a*c]) + b^3*(-(A*c ) + 3*B*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]]))/(4*c^(5/2))
Time = 0.76 (sec) , antiderivative size = 400, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1598, 1602, 27, 1602, 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.
| \(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1598 |
| \(\displaystyle \frac {\int \frac {x^4 \left (5 (A b-2 a B)-3 (b B-2 A c) x^2\right )}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 x^2 \left (\left (3 B b^2-A c b-10 a B c\right ) x^2+3 a (b B-2 A c)\right )}{c x^4+b x^2+a}dx}{3 c}-\frac {x^3 (b B-2 A c)}{c}}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {x^2 \left (\left (3 B b^2-A c b-10 a B c\right ) x^2+3 a (b B-2 A c)\right )}{c x^4+b x^2+a}dx}{c}-\frac {x^3 (b B-2 A c)}{c}}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {\frac {\frac {x \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {\int \frac {\left (3 B b^3-A c b^2-13 a B c b+6 a A c^2\right ) x^2+a \left (3 B b^2-A c b-10 a B c\right )}{c x^4+b x^2+a}dx}{c}}{c}-\frac {x^3 (b B-2 A c)}{c}}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {x \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {\frac {1}{2} \left (-\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} \left (\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{c}}{c}-\frac {x^3 (b B-2 A c)}{c}}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {x \left (-10 a B c-A b c+3 b^2 B\right )}{c}-\frac {\frac {\left (-\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {20 a^2 B c^2+8 a A b c^2-19 a b^2 B c-A b^3 c+3 b^4 B}{\sqrt {b^2-4 a c}}+6 a A c^2-13 a b B c-A b^2 c+3 b^3 B\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{c}}{c}-\frac {x^3 (b B-2 A c)}{c}}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (-2 a B-\left (x^2 (b B-2 A c)\right )+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
-1/2*(x^5*(A*b - 2*a*B - (b*B - 2*A*c)*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c *x^4)) + (-(((b*B - 2*A*c)*x^3)/c) + (((3*b^2*B - A*b*c - 10*a*B*c)*x)/c - (((3*b^3*B - A*b^2*c - 13*a*b*B*c + 6*a*A*c^2 - (3*b^4*B - A*b^3*c - 19*a *b^2*B*c + 8*a*A*b*c^2 + 20*a^2*B*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]* Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^ 2 - 4*a*c]]) + ((3*b^3*B - A*b^2*c - 13*a*b*B*c + 6*a*A*c^2 + (3*b^4*B - A *b^3*c - 19*a*b^2*B*c + 8*a*A*b*c^2 + 20*a^2*B*c^2)/Sqrt[b^2 - 4*a*c])*Arc Tan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqr t[b + Sqrt[b^2 - 4*a*c]]))/c)/c)/(2*(b^2 - 4*a*c))
3.2.18.3.1 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[((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*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1) *((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[f ^2/(2*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1 )*Simp[(m - 1)*(b*d - 2*a*e) - (4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[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[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.51
| method | result | size |
| risch | \(\frac {B x}{c^{2}}+\frac {-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{3}}{2 \left (4 a c -b^{2}\right )}+\frac {a \left (A b c +2 B a c -B \,b^{2}\right ) x}{8 a c -2 b^{2}}}{c^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\frac {\left (6 A a \,c^{2}-A \,b^{2} c -13 B a b c +3 B \,b^{3}\right ) \textit {\_R}^{2}}{4 a c -b^{2}}-\frac {a \left (A b c +10 B a c -3 B \,b^{2}\right )}{4 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,\textit {\_R}^{3}+\textit {\_R} b}}{4 c^{2}}\) | \(217\) |
| default | \(\frac {B x}{c^{2}}+\frac {\frac {-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{3}}{2 \left (4 a c -b^{2}\right )}+\frac {a \left (A b c +2 B a c -B \,b^{2}\right ) x}{8 a c -2 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {2 c \left (\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}+8 A a b \,c^{2}-A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}+20 B \,a^{2} c^{2}-19 B a \,b^{2} c +3 B \,b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\left (6 A a \,c^{2} \sqrt {-4 a c +b^{2}}-A \,b^{2} c \sqrt {-4 a c +b^{2}}-8 A a b \,c^{2}+A \,b^{3} c -13 B a b c \sqrt {-4 a c +b^{2}}+3 B \,b^{3} \sqrt {-4 a c +b^{2}}-20 B \,a^{2} c^{2}+19 B a \,b^{2} c -3 B \,b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{c^{2}}\) | \(447\) |
B*x/c^2+(-1/2*(2*A*a*c^2-A*b^2*c-3*B*a*b*c+B*b^3)/(4*a*c-b^2)*x^3+1/2*a*(A *b*c+2*B*a*c-B*b^2)/(4*a*c-b^2)*x)/c^2/(c*x^4+b*x^2+a)+1/4/c^2*sum(((6*A*a *c^2-A*b^2*c-13*B*a*b*c+3*B*b^3)/(4*a*c-b^2)*_R^2-a*(A*b*c+10*B*a*c-3*B*b^ 2)/(4*a*c-b^2))/(2*_R^3*c+_R*b)*ln(x-_R),_R=RootOf(_Z^4*c+_Z^2*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 7252 vs. \(2 (379) = 758\).
Time = 7.53 (sec) , antiderivative size = 7252, normalized size of antiderivative = 17.06 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{6}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
1/2*((B*b^3 + 2*A*a*c^2 - (3*B*a*b + A*b^2)*c)*x^3 + (B*a*b^2 - (2*B*a^2 + A*a*b)*c)*x)/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*b*c^3)*x^2) + B*x/c^2 - 1/2*integrate((3*B*a*b^2 + (3*B*b^3 + 6*A*a* c^2 - (13*B*a*b + A*b^2)*c)*x^2 - (10*B*a^2 + A*a*b)*c)/(c*x^4 + b*x^2 + a ), x)/(b^2*c^2 - 4*a*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 5675 vs. \(2 (379) = 758\).
Time = 1.51 (sec) , antiderivative size = 5675, normalized size of antiderivative = 13.35 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
B*x/c^2 + 1/2*(B*b^3*x^3 - 3*B*a*b*c*x^3 - A*b^2*c*x^3 + 2*A*a*c^2*x^3 + B *a*b^2*x - 2*B*a^2*c*x - A*a*b*c*x)/((c*x^4 + b*x^2 + a)*(b^2*c^2 - 4*a*c^ 3)) + 1/16*((2*b^4*c^3 - 20*a*b^2*c^4 + 48*a^2*c^5 - sqrt(2)*sqrt(b^2 - 4* a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c + 10*sqrt(2)*sqrt(b^2 - 4*a*c)* sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sq rt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 - 24*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt( b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^3 - 12*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr t(b^2 - 4*a*c)*c)*b^2*c^3 + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^ 2 - 4*a*c)*c)*a*c^4 - 2*(b^2 - 4*a*c)*b^2*c^3 + 12*(b^2 - 4*a*c)*a*c^4)*(b ^2*c^2 - 4*a*c^3)^2*A - (6*b^5*c^2 - 50*a*b^3*c^3 + 104*a^2*b*c^4 - 3*sqrt (2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5 + 25*sqrt(2)*sqr t(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c + 6*sqrt(2)*sqrt(b^ 2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 52*sqrt(2)*sqrt(b^2 - 4 *a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 - 26*sqrt(2)*sqrt(b^2 - 4* a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 - 3*sqrt(2)*sqrt(b^2 - 4*a* c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 + 13*sqrt(2)*sqrt(b^2 - 4*a*c)* sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 - 6*(b^2 - 4*a*c)*b^3*c^2 + 26*(b^ 2 - 4*a*c)*a*b*c^3)*(b^2*c^2 - 4*a*c^3)^2*B + 2*(sqrt(2)*sqrt(b*c + sqrt(b ^2 - 4*a*c)*c)*a*b^5*c^4 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^...
Time = 10.14 (sec) , antiderivative size = 16604, normalized size of antiderivative = 39.07 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
(B*x)/c^2 - atan(((((10240*B*a^5*c^7 - 16*A*a*b^7*c^4 + 1024*A*a^4*b*c^7 + 48*B*a*b^8*c^3 + 192*A*a^2*b^5*c^5 - 768*A*a^3*b^3*c^6 - 736*B*a^2*b^6*c^ 4 + 4224*B*a^3*b^4*c^5 - 10752*B*a^4*b^2*c^6)/(8*(64*a^3*c^6 - b^6*c^3 + 1 2*a*b^4*c^4 - 48*a^2*b^2*c^5)) - (x*((9*B^2*b^4*(-(4*a*c - b^2)^9)^(1/2) - A^2*b^11*c^2 - 9*B^2*b^13 + 6*A*B*b^12*c - 288*A^2*a^2*b^7*c^4 + 1504*A^2 *a^3*b^5*c^5 - 3840*A^2*a^4*b^3*c^6 - 2077*B^2*a^2*b^9*c^2 + 10656*B^2*a^3 *b^7*c^3 - 30240*B^2*a^4*b^5*c^4 + 44800*B^2*a^5*b^3*c^5 + A^2*b^2*c^2*(-( 4*a*c - b^2)^9)^(1/2) + 25*B^2*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) + 15360*A* B*a^6*c^7 + 213*B^2*a*b^11*c + 27*A^2*a*b^9*c^3 + 3840*A^2*a^5*b*c^7 - 9*A ^2*a*c^3*(-(4*a*c - b^2)^9)^(1/2) - 26880*B^2*a^6*b*c^6 + 1548*A*B*a^2*b^8 *c^3 - 8064*A*B*a^3*b^6*c^4 + 22400*A*B*a^4*b^4*c^5 - 30720*A*B*a^5*b^2*c^ 6 - 51*B^2*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2) - 152*A*B*a*b^10*c^2 - 6*A*B*b ^3*c*(-(4*a*c - b^2)^9)^(1/2) + 44*A*B*a*b*c^2*(-(4*a*c - b^2)^9)^(1/2))/( 32*(4096*a^6*c^11 + b^12*c^5 - 24*a*b^10*c^6 + 240*a^2*b^8*c^7 - 1280*a^3* b^6*c^8 + 3840*a^4*b^4*c^9 - 6144*a^5*b^2*c^10)))^(1/2)*(16*b^7*c^5 - 192* a*b^5*c^6 - 1024*a^3*b*c^8 + 768*a^2*b^3*c^7))/(2*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))*((9*B^2*b^4*(-(4*a*c - b^2)^9)^(1/2) - A^2*b^11*c^2 - 9*B^2 *b^13 + 6*A*B*b^12*c - 288*A^2*a^2*b^7*c^4 + 1504*A^2*a^3*b^5*c^5 - 3840*A ^2*a^4*b^3*c^6 - 2077*B^2*a^2*b^9*c^2 + 10656*B^2*a^3*b^7*c^3 - 30240*B^2* a^4*b^5*c^4 + 44800*B^2*a^5*b^3*c^5 + A^2*b^2*c^2*(-(4*a*c - b^2)^9)^(1...