Integrand size = 34, antiderivative size = 356 \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\frac {B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\left (2 A c-b C-\frac {4 A b c-\left (b^2+4 a c\right ) C}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} \sqrt {c} \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (2 A c-b C+\frac {4 A b c-\left (b^2+4 a c\right ) C}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} \sqrt {c} \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {b B \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]
1/2*B*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-1/2*x*(A*b-2*C*a+(2*A*c-C*b )*x^2)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-b*B*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^( 1/2))/(-4*a*c+b^2)^(3/2)-1/4*arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2)^(1/2 ))^(1/2))*(2*A*c-C*b+(-4*A*b*c+(4*a*c+b^2)*C)/(-4*a*c+b^2)^(1/2))/(-4*a*c+ b^2)*2^(1/2)/c^(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))*(2*A*c-C*b+(4*A*b*c-(4*a*c+b^2)*C)/(-4* a*c+b^2)^(1/2))/(-4*a*c+b^2)*2^(1/2)/c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.10 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.06 \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\frac {1}{4} \left (\frac {4 a (B+C x)+2 x \left (b x (B+C x)-A \left (b+2 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\sqrt {2} \left (-2 A c \left (-2 b+\sqrt {b^2-4 a c}\right )+\left (-b^2-4 a c+b \sqrt {b^2-4 a c}\right ) C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {c} \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \left (-2 A c \left (2 b+\sqrt {b^2-4 a c}\right )+\left (b^2+4 a c+b \sqrt {b^2-4 a c}\right ) C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {c} \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}+\frac {2 b B \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {2 b B \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}\right ) \]
((4*a*(B + C*x) + 2*x*(b*x*(B + C*x) - A*(b + 2*c*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[2]*(-2*A*c*(-2*b + Sqrt[b^2 - 4*a*c]) + (-b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[ b^2 - 4*a*c]]])/(Sqrt[c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*(-2*A*c*(2*b + Sqrt[b^2 - 4*a*c]) + (b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt [c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*b*B*Log[-b + Sqr t[b^2 - 4*a*c] - 2*c*x^2])/(b^2 - 4*a*c)^(3/2) - (2*b*B*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/4
Time = 0.57 (sec) , antiderivative size = 348, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {9, 2193, 27, 1434, 1159, 1083, 219, 1598, 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.
| \(\displaystyle \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2}dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\int \frac {B x^3}{\left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+B \int \frac {x^3}{\left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 1434 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\frac {1}{2} B \int \frac {x^2}{\left (c x^4+b x^2+a\right )^2}dx^2\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\frac {1}{2} B \left (\frac {b \int \frac {1}{c x^4+b x^2+a}dx^2}{b^2-4 a c}+\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\frac {1}{2} B \left (\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {2 b \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{b^2-4 a c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \int \frac {x^2 \left (C x^2+A\right )}{\left (c x^4+b x^2+a\right )^2}dx+\frac {1}{2} B \left (\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {2 b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 1598 |
| \(\displaystyle \frac {\int \frac {-\left ((2 A c-b C) x^2\right )+A b-2 a C}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}-\frac {x \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {1}{2} B \left (\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {2 b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {-\frac {1}{2} \left (-\frac {4 A b c-C \left (4 a c+b^2\right )}{\sqrt {b^2-4 a c}}+2 A c-b C\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx-\frac {1}{2} \left (\frac {4 A b c-C \left (4 a c+b^2\right )}{\sqrt {b^2-4 a c}}+2 A c-b C\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{2 \left (b^2-4 a c\right )}-\frac {x \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {1}{2} B \left (\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {2 b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {\left (-\frac {4 A b c-C \left (4 a c+b^2\right )}{\sqrt {b^2-4 a c}}+2 A c-b C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (\frac {4 A b c-C \left (4 a c+b^2\right )}{\sqrt {b^2-4 a c}}+2 A c-b C\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{2 \left (b^2-4 a c\right )}-\frac {x \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {1}{2} B \left (\frac {2 a+b x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {2 b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\right )\) |
-1/2*(x*(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*A*c - b*C - (4*A*b*c - (b^2 + 4*a*c)*C)/Sqrt[b^2 - 4*a*c])*A rcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*S qrt[b - Sqrt[b^2 - 4*a*c]])) - ((2*A*c - b*C + (4*A*b*c - (b^2 + 4*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]]))/(2*(b^2 - 4*a*c)) + (B*( (2*a + b*x^2)/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (2*b*ArcTanh[(b + 2*c* x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)))/2
3.1.45.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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[(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[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[(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 = 0.09 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.57
| method | result | size |
| risch | \(\frac {\frac {\left (2 A c -C b \right ) x^{3}}{8 a c -2 b^{2}}-\frac {x^{2} B b}{2 \left (4 a c -b^{2}\right )}+\frac {\left (A b -2 C a \right ) x}{8 a c -2 b^{2}}-\frac {B a}{4 a c -b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\frac {\left (2 A c -C b \right ) \textit {\_R}^{2}}{4 a c -b^{2}}-\frac {2 \textit {\_R} B b}{4 a c -b^{2}}-\frac {A b -2 C a}{4 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,\textit {\_R}^{3}+\textit {\_R} b}\right )}{4}\) | \(204\) |
| default | \(\frac {\frac {\left (2 A c -C b \right ) x^{3}}{8 a c -2 b^{2}}-\frac {x^{2} B b}{2 \left (4 a c -b^{2}\right )}+\frac {\left (A b -2 C a \right ) x}{8 a c -2 b^{2}}-\frac {B a}{4 a c -b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {2 c \left (\frac {-B \sqrt {-4 a c +b^{2}}\, b \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )+\frac {\left (-4 A b c \sqrt {-4 a c +b^{2}}+8 A a \,c^{2}-2 A \,b^{2} c +4 C \sqrt {-4 a c +b^{2}}\, a c +C \sqrt {-4 a c +b^{2}}\, b^{2}-4 C a b c +C \,b^{3}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )}+\frac {B \sqrt {-4 a c +b^{2}}\, b \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )+\frac {\left (-4 A b c \sqrt {-4 a c +b^{2}}-8 A a \,c^{2}+2 A \,b^{2} c +4 C \sqrt {-4 a c +b^{2}}\, a c +C \sqrt {-4 a c +b^{2}}\, b^{2}+4 C a b c -C \,b^{3}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )}\right )}{4 a c -b^{2}}\) | \(456\) |
(1/2*(2*A*c-C*b)/(4*a*c-b^2)*x^3-1/2/(4*a*c-b^2)*x^2*B*b+1/2*(A*b-2*C*a)/( 4*a*c-b^2)*x-1/(4*a*c-b^2)*B*a)/(c*x^4+b*x^2+a)+1/4*sum(((2*A*c-C*b)/(4*a* c-b^2)*_R^2-2/(4*a*c-b^2)*_R*B*b-(A*b-2*C*a)/(4*a*c-b^2))/(2*_R^3*c+_R*b)* ln(x-_R),_R=RootOf(_Z^4*c+_Z^2*b+a))
Timed out. \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {C x^{5} + B x^{4} + A x^{3}}{{\left (c x^{4} + b x^{2} + a\right )}^{2} x} \,d x } \]
1/2*(B*b*x^2 + (C*b - 2*A*c)*x^3 + 2*B*a + (2*C*a - A*b)*x)/((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*b* x + (C*b - 2*A*c)*x^2 - 2*C*a + A*b)/(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. 4438 vs. \(2 (306) = 612\).
Time = 1.42 (sec) , antiderivative size = 4438, normalized size of antiderivative = 12.47 \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
1/2*(C*b*x^3 - 2*A*c*x^3 + B*b*x^2 + 2*C*a*x - A*b*x + 2*B*a)/((c*x^4 + b* x^2 + a)*(b^2 - 4*a*c)) - 1/16*(2*(2*b^2*c^3 - 8*a*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c + 4*sqrt(2)*sqrt(b^2 - 4*a* c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqr t(b*c + sqrt(b^2 - 4*a*c)*c)*b*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*c^3 - 2*(b^2 - 4*a*c)*c^3)*(b^2 - 4*a*c)^2*A - (2*b^3 *c^2 - 8*a*b*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)* c)*b^3 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c - sqr t(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b*c^2 - 2*(b^2 - 4* a*c)*b*c^2)*(b^2 - 4*a*c)^2*C - 2*(sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c) *b^5*c - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^2 - 2*sqrt(2)*s qrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^2 - 2*b^5*c^2 + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c )*a*b^2*c^3 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^3 + 16*a*b^3*c ^3 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^4 - 32*a^2*b*c^4 + 2* (b^2 - 4*a*c)*b^3*c^2 - 8*(b^2 - 4*a*c)*a*b*c^3)*A*abs(b^2 - 4*a*c) + 4*(s qrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c - 8*sqrt(2)*sqrt(b*c + sqrt (b^2 - 4*a*c)*c)*a^2*b^2*c^2 - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a *b^3*c^2 - 2*a*b^4*c^2 + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3...
Time = 8.19 (sec) , antiderivative size = 3835, normalized size of antiderivative = 10.77 \[ \int \frac {A x^3+B x^4+C x^5}{x \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
symsum(log((8*A^3*a*c^4 + 6*A^3*b^2*c^3 + A*C^2*b^4*c - 3*C^3*a*b^3*c + 4* A*B^2*b^3*c^2 + 8*A*C^2*a^2*c^3 - 5*A^2*C*b^3*c^2 - 4*C^3*a^2*b*c^2 + 18*A *C^2*a*b^2*c^2 - 8*B^2*C*a*b^2*c^2 - 28*A^2*C*a*b*c^3)/(8*(b^6 - 64*a^3*c^ 3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) - root(256*a*b^12*c*z^4 - 1572864*a^6*b^ 2*c^6*z^4 + 983040*a^5*b^4*c^5*z^4 - 327680*a^4*b^6*c^4*z^4 + 61440*a^3*b^ 8*c^3*z^4 - 6144*a^2*b^10*c^2*z^4 + 1048576*a^7*c^7*z^4 - 192*A*C*a*b^8*c* z^2 - 6144*A*C*a^3*b^4*c^3*z^2 + 2048*A*C*a^2*b^6*c^2*z^2 - 12288*C^2*a^5* b*c^4*z^2 - 12288*A^2*a^4*b*c^5*z^2 - 128*B^2*a*b^8*c*z^2 + 16384*A*C*a^5* c^5*z^2 + 8192*C^2*a^4*b^3*c^3*z^2 - 1536*C^2*a^3*b^5*c^2*z^2 + 8192*B^2*a ^4*b^2*c^4*z^2 - 6144*B^2*a^3*b^4*c^3*z^2 + 1536*B^2*a^2*b^6*c^2*z^2 + 819 2*A^2*a^3*b^3*c^4*z^2 - 1536*A^2*a^2*b^5*c^3*z^2 + 16*C^2*a*b^9*z^2 + 16*A ^2*b^9*c*z^2 + 1024*B*C^2*a^4*b*c^3*z + 192*B*C^2*a^2*b^5*c*z - 1024*A^2*B *a^3*b*c^4*z - 192*A^2*B*a*b^5*c^2*z - 768*B*C^2*a^3*b^3*c^2*z + 768*A^2*B *a^2*b^3*c^3*z + 16*A^2*B*b^7*c*z - 16*B*C^2*a*b^7*z - 64*A*B^2*C*a^2*b^2* c^2 - 48*A*B^2*C*a*b^4*c + 192*A^2*C^2*a^2*b^2*c^2 + 48*B^2*C^2*a^2*b^3*c + 48*A^2*B^2*a*b^3*c^2 - 96*A^3*C*a^2*b*c^3 - 96*A*C^3*a^3*b*c^2 - 80*A^3* C*a*b^3*c^2 - 80*A*C^3*a^2*b^3*c + 42*A^2*C^2*a*b^4*c + 24*C^4*a^3*b^2*c + 24*A^4*a*b^2*c^3 + 4*B^2*C^2*a*b^5 + 4*A^2*B^2*b^5*c + 16*B^4*a*b^4*c - 6 *A^3*C*b^5*c - 6*A*C^3*a*b^5 + 32*A^2*C^2*a^3*c^3 + 16*C^4*a^4*c^2 + 9*C^4 *a^2*b^4 + 9*A^4*b^4*c^2 + 16*A^4*a^2*c^4 + A^2*C^2*b^6, z, k)*(root(25...