Integrand size = 26, antiderivative size = 317 \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {B x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {A b-2 a C+(2 A c-b C) x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {B \sqrt {c} \left (2 b-\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {B \sqrt {c} \left (2 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}+\frac {(2 A c-b C) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]
-1/2*B*x*(2*c*x^2+b)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)+1/2*(-A*b+2*C*a-(2*A*c-C *b)*x^2)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)+(2*A*c-C*b)*arctanh((2*c*x^2+b)/(-4* a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)+1/2*B*arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a *c+b^2)^(1/2))^(1/2))*c^(1/2)*(2*b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)* 2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-1/2*B*arctan(x*2^(1/2)*c^(1/2)/(b+(-4 *a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(2*b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2 )*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.79 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.06 \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {1}{2} \left (\frac {2 a C-A \left (b+2 c x^2\right )+x \left (-b B+b C x-2 B c x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\sqrt {2} B \sqrt {c} \left (-2 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} B \sqrt {c} \left (2 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}+\frac {(-2 A c+b C) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac {(2 A c-b C) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}\right ) \]
((2*a*C - A*(b + 2*c*x^2) + x*(-(b*B) + b*C*x - 2*B*c*x^2))/((b^2 - 4*a*c) *(a + b*x^2 + c*x^4)) - (Sqrt[2]*B*Sqrt[c]*(-2*b + Sqrt[b^2 - 4*a*c])*ArcT an[(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]*B*Sqrt[c]*(2*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]]) + ((-2*A*c + b*C)*Log[-b + Sqrt[b^2 - 4 *a*c] - 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((2*A*c - b*C)*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/2
Time = 0.59 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {2193, 27, 1439, 1480, 218, 1576, 1159, 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.
| \(\displaystyle \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\int \frac {B x^2}{\left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+B \int \frac {x^2}{\left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 1439 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+B \left (\frac {\int \frac {b-2 c x^2}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+B \left (\frac {-c \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx-c \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+B \left (\frac {-\frac {\sqrt {2} \sqrt {c} \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \sqrt {c} \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {1}{2} \int \frac {C x^2+A}{\left (c x^4+b x^2+a\right )^2}dx^2+B \left (\frac {-\frac {\sqrt {2} \sqrt {c} \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \sqrt {c} \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(2 A c-b C) \int \frac {1}{c x^4+b x^2+a}dx^2}{b^2-4 a c}-\frac {-2 a C+x^2 (2 A c-b C)+A b}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\sqrt {2} \sqrt {c} \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \sqrt {c} \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 (2 A c-b C) \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{b^2-4 a c}-\frac {-2 a C+x^2 (2 A c-b C)+A b}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\sqrt {2} \sqrt {c} \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \sqrt {c} \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 (2 A c-b C) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {-2 a C+x^2 (2 A c-b C)+A b}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\sqrt {2} \sqrt {c} \left (1-\frac {2 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \sqrt {c} \left (\frac {2 b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
B*(-1/2*(x*(b + 2*c*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (-((Sqrt[2 ]*Sqrt[c]*(1 - (2*b)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (Sqrt[2]*Sqrt[c]*(1 + (2*b)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/Sqrt[b + Sqrt[b^2 - 4*a*c]])/(2*(b^2 - 4*a*c))) + (-((A*b - 2*a* C + (2*A*c - b*C)*x^2)/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4))) + (2*(2*A*c - b*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2))/2
3.1.32.3.1 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)^(-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[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & & LtQ[p, -1] && NeQ[p, -3/2]
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[d*(d*x)^(m - 1)*(b + 2*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[d^2/(2*(p + 1)*(b^2 - 4*a*c)) Int[(d*x)^(m - 2)*(b*(m - 1) + 2*c*(m + 4*p + 5)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x ] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
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[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(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[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_S ymbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(d*x)^m*(a + b*x^2 + c*x^4)^p, x] + Simp[1/d Int[Sum[Coe ff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q + 1)/2}]*(d*x)^(m + 1)*(a + b*x^2 + c *x^4)^p, x], x]] /; FreeQ[{a, b, c, d, m, p}, x] && PolyQ[Pq, x] && !PolyQ [Pq, x^2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.62
| method | result | size |
| risch | \(\frac {\frac {B c \,x^{3}}{4 a c -b^{2}}+\frac {\left (2 A c -C b \right ) x^{2}}{8 a c -2 b^{2}}+\frac {B b x}{8 a c -2 b^{2}}+\frac {A b -2 C a}{8 a c -2 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\frac {2 c \,\textit {\_R}^{2} B}{4 a c -b^{2}}+\frac {2 \left (2 A c -C b \right ) \textit {\_R}}{4 a c -b^{2}}-\frac {b B}{4 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,\textit {\_R}^{3}+\textit {\_R} b}\right )}{4}\) | \(198\) |
| default | \(16 c^{2} \left (-\frac {\frac {-\frac {B \left (4 a c -b^{2}\right ) x}{8 c}-\frac {8 A a \,c^{2}-2 A \,b^{2} c +4 C \sqrt {-4 a c +b^{2}}\, a c -C \sqrt {-4 a c +b^{2}}\, b^{2}-4 C a b c +C \,b^{3}}{16 c^{2}}}{x^{2}+\frac {b}{2 c}-\frac {\sqrt {-4 a c +b^{2}}}{2 c}}-\frac {\left (-4 A c \sqrt {-4 a c +b^{2}}+2 C \sqrt {-4 a c +b^{2}}\, b \right ) \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{16 c}+\frac {\left (2 B b \sqrt {-4 a c +b^{2}}+4 B a c -B \,b^{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )^{2}}+\frac {\frac {\frac {B \left (4 a c -b^{2}\right ) x}{8 c}+\frac {8 A a \,c^{2}-2 A \,b^{2} c -4 C \sqrt {-4 a c +b^{2}}\, a c +C \sqrt {-4 a c +b^{2}}\, b^{2}-4 C a b c +C \,b^{3}}{16 c^{2}}}{x^{2}+\frac {\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}+\frac {\left (4 A c \sqrt {-4 a c +b^{2}}-2 C \sqrt {-4 a c +b^{2}}\, b \right ) \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{16 c}+\frac {\left (-2 B b \sqrt {-4 a c +b^{2}}+4 B a c -B \,b^{2}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )^{2}}\right )\) | \(510\) |
(B*c/(4*a*c-b^2)*x^3+1/2*(2*A*c-C*b)/(4*a*c-b^2)*x^2+1/2/(4*a*c-b^2)*x*B*b +1/2*(A*b-2*C*a)/(4*a*c-b^2))/(c*x^4+b*x^2+a)+1/4*sum((2*c/(4*a*c-b^2)*_R^ 2*B+2*(2*A*c-C*b)/(4*a*c-b^2)*_R-1/(4*a*c-b^2)*b*B)/(2*_R^3*c+_R*b)*ln(x-_ R),_R=RootOf(_Z^4*c+_Z^2*b+a))
Timed out. \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} x}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
-1/2*(2*B*c*x^3 + B*b*x - (C*b - 2*A*c)*x^2 - 2*C*a + A*b)/((b^2*c - 4*a*c ^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2) - 1/2*integrate((2*B*c*x^ 2 - B*b - 2*(C*b - 2*A*c)*x)/(c*x^4 + b*x^2 + a), x)/(b^2 - 4*a*c)
Leaf count of result is larger than twice the leaf count of optimal. 3014 vs. \(2 (270) = 540\).
Time = 1.24 (sec) , antiderivative size = 3014, normalized size of antiderivative = 9.51 \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
-1/2*(2*B*c*x^3 - C*b*x^2 + 2*A*c*x^2 + B*b*x - 2*C*a + A*b)/((c*x^4 + b*x ^2 + a)*(b^2 - 4*a*c)) - 1/8*((2*b^2*c^2 - 8*a*c^3 - sqrt(2)*sqrt(b^2 - 4* a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqr t(b*c + sqrt(b^2 - 4*a*c)*c)*a*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b*c - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*c^2 - 2*(b^2 - 4*a*c)*c^2)*(b^2 - 4*a*c)^2*B - (sqrt(2)*sqrt(b* c + sqrt(b^2 - 4*a*c)*c)*b^5 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a *b^3*c - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 2*b^5*c + 16*sq rt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 + 8*sqrt(2)*sqrt(b*c + sqr t(b^2 - 4*a*c)*c)*a*b^2*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3* c^2 + 16*a*b^3*c^2 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 - 3 2*a^2*b*c^3 + 2*(b^2 - 4*a*c)*b^3*c - 8*(b^2 - 4*a*c)*a*b*c^2)*B*abs(b^2 - 4*a*c) - 2*(2*b^6*c^2 - 16*a*b^4*c^3 + 32*a^2*b^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^6 + 8*sqrt(2)*sqrt(b^2 - 4*a*c) *sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqr t(b*c + sqrt(b^2 - 4*a*c)*c)*b^5*c - 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c^2 - 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + s qrt(b^2 - 4*a*c)*c)*b^4*c^2 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt( b^2 - 4*a*c)*c)*a*b^2*c^3 - 2*(b^2 - 4*a*c)*b^4*c^2 + 8*(b^2 - 4*a*c)*a...
Time = 8.31 (sec) , antiderivative size = 3198, normalized size of antiderivative = 10.09 \[ \int \frac {x \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
symsum(log((4*B^3*a*c^4 + 3*B^3*b^2*c^3 + 8*A^2*B*b*c^4 + 2*B*C^2*b^3*c^2 - 8*A*B*C*b^2*c^3)/(4*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) - root(1572864*a^6*b^2*c^5*z^4 - 983040*a^5*b^4*c^4*z^4 + 327680*a^4*b^6*c^3 *z^4 - 61440*a^3*b^8*c^2*z^4 + 6144*a^2*b^10*c*z^4 - 1048576*a^7*c^6*z^4 - 256*a*b^12*z^4 + 32768*A*C*a^4*b*c^4*z^2 - 512*A*C*a*b^7*c*z^2 - 24576*A* C*a^3*b^3*c^3*z^2 + 6144*A*C*a^2*b^5*c^2*z^2 - 1536*C^2*a^2*b^6*c*z^2 + 12 288*B^2*a^4*b*c^4*z^2 + 512*A^2*a*b^6*c^2*z^2 - 8192*C^2*a^4*b^2*c^3*z^2 + 6144*C^2*a^3*b^4*c^2*z^2 - 8192*B^2*a^3*b^3*c^3*z^2 + 1536*B^2*a^2*b^5*c^ 2*z^2 + 24576*A^2*a^3*b^2*c^4*z^2 - 6144*A^2*a^2*b^4*c^3*z^2 + 128*C^2*a*b ^8*z^2 - 32768*A^2*a^4*c^5*z^2 - 16*B^2*b^9*z^2 + 1024*B^2*C*a^3*b*c^3*z - 384*A*B^2*a*b^4*c^2*z + 192*B^2*C*a*b^5*c*z - 768*B^2*C*a^2*b^3*c^2*z + 1 536*A*B^2*a^2*b^2*c^3*z + 32*A*B^2*b^6*c*z - 2048*A*B^2*a^3*c^4*z - 16*B^2 *C*b^7*z + 192*A*B^2*C*a*b^2*c^2 + 512*A^3*C*a*b*c^3 + 128*A*C^3*a*b^3*c + 16*A*B^2*C*b^4*c - 384*A^2*C^2*a*b^2*c^2 - 48*B^2*C^2*a*b^3*c - 192*A^2*B ^2*a*b*c^3 - 24*B^4*a*b^2*c^2 - 16*A^2*B^2*b^3*c^2 - 16*B^4*a^2*c^3 - 4*B^ 2*C^2*b^5 - 9*B^4*b^4*c - 16*C^4*a*b^4 - 256*A^4*a*c^4, z, k)*(root(157286 4*a^6*b^2*c^5*z^4 - 983040*a^5*b^4*c^4*z^4 + 327680*a^4*b^6*c^3*z^4 - 6144 0*a^3*b^8*c^2*z^4 + 6144*a^2*b^10*c*z^4 - 1048576*a^7*c^6*z^4 - 256*a*b^12 *z^4 + 32768*A*C*a^4*b*c^4*z^2 - 512*A*C*a*b^7*c*z^2 - 24576*A*C*a^3*b^3*c ^3*z^2 + 6144*A*C*a^2*b^5*c^2*z^2 - 1536*C^2*a^2*b^6*c*z^2 + 12288*B^2*...