Integrand size = 28, antiderivative size = 339 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {(A c-b C) x}{c^2}+\frac {B x^2}{2 c}+\frac {C x^3}{3 c}-\frac {\left (A b c-b^2 C+a c C-\frac {A c \left (b^2-2 a c\right )-b \left (b^2-3 a c\right ) C}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} c^{5/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (A b c-b^2 C+a c C+\frac {A c \left (b^2-2 a c\right )-b \left (b^2-3 a c\right ) C}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} c^{5/2} \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {B \left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \sqrt {b^2-4 a c}}-\frac {b B \log \left (a+b x^2+c x^4\right )}{4 c^2} \]
(A*c-C*b)*x/c^2+1/2*B*x^2/c+1/3*C*x^3/c-1/4*b*B*ln(c*x^4+b*x^2+a)/c^2-1/2* B*(-2*a*c+b^2)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^2)^(1 /2)-1/2*arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*(A*b*c-b^2* C+a*c*C+(-A*c*(-2*a*c+b^2)+b*(-3*a*c+b^2)*C)/(-4*a*c+b^2)^(1/2))/c^(5/2)*2 ^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-1/2*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a* c+b^2)^(1/2))^(1/2))*(A*b*c-b^2*C+a*c*C+(A*c*(-2*a*c+b^2)-b*(-3*a*c+b^2)*C )/(-4*a*c+b^2)^(1/2))/c^(5/2)*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.36 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.36 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {12 \sqrt {c} (A c-b C) x+6 B c^{3/2} x^2+4 c^{3/2} C x^3+\frac {6 \sqrt {2} \left (A c \left (b^2-2 a c-b \sqrt {b^2-4 a c}\right )+\left (-b^3+3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {6 \sqrt {2} \left (-A c \left (b^2-2 a c+b \sqrt {b^2-4 a c}\right )+\left (b^3-3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {3 B \sqrt {c} \left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )}{\sqrt {b^2-4 a c}}-\frac {3 B \sqrt {c} \left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{12 c^{5/2}} \]
(12*Sqrt[c]*(A*c - b*C)*x + 6*B*c^(3/2)*x^2 + 4*c^(3/2)*C*x^3 + (6*Sqrt[2] *(A*c*(b^2 - 2*a*c - b*Sqrt[b^2 - 4*a*c]) + (-b^3 + 3*a*b*c + b^2*Sqrt[b^2 - 4*a*c] - a*c*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (6* Sqrt[2]*(-(A*c*(b^2 - 2*a*c + b*Sqrt[b^2 - 4*a*c])) + (b^3 - 3*a*b*c + b^2 *Sqrt[b^2 - 4*a*c] - a*c*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/ Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c ]]) - (3*B*Sqrt[c]*(-b^2 + 2*a*c + b*Sqrt[b^2 - 4*a*c])*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/Sqrt[b^2 - 4*a*c] - (3*B*Sqrt[c]*(b^2 - 2*a*c + b*Sqr t[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/( 12*c^(5/2))
Time = 0.86 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {2193, 27, 1434, 1143, 1602, 27, 1602, 25, 1480, 218, 2009}
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.
| \(\displaystyle \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {x^4 \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\int \frac {B x^5}{c x^4+b x^2+a}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^4 \left (C x^2+A\right )}{c x^4+b x^2+a}dx+B \int \frac {x^5}{c x^4+b x^2+a}dx\) |
| \(\Big \downarrow \) 1434 |
| \(\displaystyle \int \frac {x^4 \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {1}{2} B \int \frac {x^4}{c x^4+b x^2+a}dx^2\) |
| \(\Big \downarrow \) 1143 |
| \(\displaystyle \int \frac {x^4 \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle -\frac {\int \frac {3 x^2 \left (a C-(A c-b C) x^2\right )}{c x^4+b x^2+a}dx}{3 c}+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {x^2 \left (a C-(A c-b C) x^2\right )}{c x^4+b x^2+a}dx}{c}+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle -\frac {-\frac {\int -\frac {\left (-C b^2+A c b+a c C\right ) x^2+a (A c-b C)}{c x^4+b x^2+a}dx}{c}-\frac {x (A c-b C)}{c}}{c}+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {\left (-C b^2+A c b+a c C\right ) x^2+a (A c-b C)}{c x^4+b x^2+a}dx}{c}-\frac {x (A c-b C)}{c}}{c}+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {\frac {\frac {1}{2} \left (-\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} \left (\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{c}-\frac {x (A c-b C)}{c}}{c}+\frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{2} B \int \left (\frac {1}{c}-\frac {b x^2+a}{c \left (c x^4+b x^2+a\right )}\right )dx^2-\frac {\frac {\frac {\left (-\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{c}-\frac {x (A c-b C)}{c}}{c}+\frac {C x^3}{3 c}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {\frac {\left (-\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {A c \left (b^2-2 a c\right )-b C \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}+a c C+A b c+b^2 (-C)\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{c}-\frac {x (A c-b C)}{c}}{c}+\frac {1}{2} B \left (-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^2+c x^4\right )}{2 c^2}+\frac {x^2}{c}\right )+\frac {C x^3}{3 c}\) |
(C*x^3)/(3*c) - (-(((A*c - b*C)*x)/c) + (((A*b*c - b^2*C + a*c*C - (A*c*(b ^2 - 2*a*c) - b*(b^2 - 3*a*c)*C)/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]]) + ((A*b*c - b^2*C + a*c*C + (A*c*(b^2 - 2*a*c) - b*(b^2 - 3*a*c)*C) /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]]))/c)/c + (B*(x^2/c - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c] ) - (b*Log[a + b*x^2 + c*x^4])/(2*c^2)))/2
3.1.21.3.1 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_))^(m_)/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
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[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])
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 = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {C \,x^{3}}{3 c}+\frac {B \,x^{2}}{2 c}+\frac {A x}{c}-\frac {C b x}{c^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (-b B c \,\textit {\_R}^{3}+\left (-A b c -a c C +b^{2} C \right ) \textit {\_R}^{2}-B a c \textit {\_R} -A a c +a b C \right ) \ln \left (x -\textit {\_R} \right )}{2 c \,\textit {\_R}^{3}+\textit {\_R} b}}{2 c^{2}}\) | \(118\) |
| default | \(\frac {\frac {1}{3} c C \,x^{3}+\frac {1}{2} B \,x^{2} c +A c x -C b x}{c^{2}}+\frac {-\frac {\left (2 a c \sqrt {-4 a c +b^{2}}-b^{2} \sqrt {-4 a c +b^{2}}+4 a b c -b^{3}\right ) \left (\frac {B \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{2}+\frac {\left (2 A c -C \sqrt {-4 a c +b^{2}}-C b \right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 c \left (4 a c -b^{2}\right )}+\frac {\left (2 a c \sqrt {-4 a c +b^{2}}-b^{2} \sqrt {-4 a c +b^{2}}-4 a b c +b^{3}\right ) \left (\frac {B \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{2}+\frac {\left (-2 A c -C \sqrt {-4 a c +b^{2}}+C b \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 c \left (4 a c -b^{2}\right )}}{c}\) | \(331\) |
1/3*C*x^3/c+1/2*B*x^2/c+1/c*A*x-1/c^2*C*b*x+1/2/c^2*sum((-b*B*c*_R^3+(-A*b *c-C*a*c+C*b^2)*_R^2-B*a*c*_R-A*a*c+a*b*C)/(2*_R^3*c+_R*b)*ln(x-_R),_R=Roo tOf(_Z^4*c+_Z^2*b+a))
Timed out. \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} x^{4}}{c x^{4} + b x^{2} + a} \,d x } \]
1/6*(2*C*c*x^3 + 3*B*c*x^2 - 6*(C*b - A*c)*x)/c^2 - integrate((B*b*c*x^3 + B*a*c*x - C*a*b + A*a*c - (C*b^2 - (C*a + A*b)*c)*x^2)/(c*x^4 + b*x^2 + a ), x)/c^2
Leaf count of result is larger than twice the leaf count of optimal. 5304 vs. \(2 (294) = 588\).
Time = 1.42 (sec) , antiderivative size = 5304, normalized size of antiderivative = 15.65 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
-1/4*B*b*log(abs(c*x^4 + b*x^2 + a))/c^2 - 1/8*((2*b^5*c^3 - 16*a*b^3*c^4 + 32*a^2*b*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c) *b^5*c + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3 *c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^2 - 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 - 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^3 + 4*sqrt (2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^4 - 2*(b^2 - 4 *a*c)*b^3*c^3 + 8*(b^2 - 4*a*c)*a*b*c^4)*A*c^2 - (2*b^6*c^2 - 18*a*b^4*c^3 + 48*a^2*b^2*c^4 - 32*a^3*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt (b^2 - 4*a*c)*c)*b^6 + 9*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)*sqrt(b*c + sqrt(b^2 - 4*a*c )*c)*b^5*c - 24*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)* a^2*b^2*c^2 - 10*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 + sqrt(b^2 - 4*a*c)*c)*b^4 *c^2 + 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*c^ 3 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 + 5*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^3 - 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^4 - 2*(b ^2 - 4*a*c)*b^4*c^2 + 10*(b^2 - 4*a*c)*a*b^2*c^3 - 8*(b^2 - 4*a*c)*a^2*...
Time = 7.92 (sec) , antiderivative size = 2588, normalized size of antiderivative = 7.63 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
x*(A/c - (C*b)/c^2) + symsum(log((C^3*a^4*c - C^3*a^3*b^2 - A*B^2*a^3*c^2 + A*C^2*a^2*b^3 + A^2*C*a^3*c^2 + A^3*a^2*b*c^2 + A*B^2*a^2*b^2*c - 2*A^2* C*a^2*b^2*c - B^2*C*a^3*b*c)/c^3 - root(128*a*b^2*c^6*z^4 - 16*b^4*c^5*z^4 - 256*a^2*c^7*z^4 - 256*B*a^2*b*c^5*z^3 + 128*B*a*b^3*c^4*z^3 - 16*B*b^5* c^3*z^3 - 64*A*C*a*b^4*c^2*z^2 + 144*A*C*a^2*b^2*c^3*z^2 + 8*A*C*b^6*c*z^2 + 80*C^2*a^3*b*c^3*z^2 + 32*B^2*a*b^4*c^2*z^2 - 48*A^2*a^2*b*c^4*z^2 + 28 *A^2*a*b^3*c^3*z^2 + 36*C^2*a*b^5*c*z^2 - 64*A*C*a^3*c^4*z^2 - 100*C^2*a^2 *b^3*c^2*z^2 - 56*B^2*a^2*b^2*c^3*z^2 - 4*B^2*b^6*c*z^2 - 32*B^2*a^3*c^4*z ^2 - 4*A^2*b^5*c^2*z^2 - 4*C^2*b^7*z^2 + 32*A*B*C*a^3*b*c^2*z - 8*A*B*C*a^ 2*b^3*c*z - 20*B*C^2*a^3*b^2*c*z + 4*A^2*B*a^2*b^2*c^2*z - 16*B^3*a^3*b*c^ 2*z + 4*B^3*a^2*b^3*c*z + 16*B*C^2*a^4*c^2*z + 4*B*C^2*a^2*b^4*z - 16*A^2* B*a^3*c^3*z + 2*A^3*C*a^3*b*c + 4*A*B^2*C*a^4*c - 2*A^2*C^2*a^4*c + 2*A*C^ 3*a^4*b - A^2*B^2*a^3*b*c - B^2*C^2*a^4*b - A^2*C^2*a^3*b^2 - A^4*a^3*c^2 - B^4*a^4*c - C^4*a^5, z, k)*(root(128*a*b^2*c^6*z^4 - 16*b^4*c^5*z^4 - 25 6*a^2*c^7*z^4 - 256*B*a^2*b*c^5*z^3 + 128*B*a*b^3*c^4*z^3 - 16*B*b^5*c^3*z ^3 - 64*A*C*a*b^4*c^2*z^2 + 144*A*C*a^2*b^2*c^3*z^2 + 8*A*C*b^6*c*z^2 + 80 *C^2*a^3*b*c^3*z^2 + 32*B^2*a*b^4*c^2*z^2 - 48*A^2*a^2*b*c^4*z^2 + 28*A^2* a*b^3*c^3*z^2 + 36*C^2*a*b^5*c*z^2 - 64*A*C*a^3*c^4*z^2 - 100*C^2*a^2*b^3* c^2*z^2 - 56*B^2*a^2*b^2*c^3*z^2 - 4*B^2*b^6*c*z^2 - 32*B^2*a^3*c^4*z^2 - 4*A^2*b^5*c^2*z^2 - 4*C^2*b^7*z^2 + 32*A*B*C*a^3*b*c^2*z - 8*A*B*C*a^2*...