3.49 \(\int \frac {x^3 (d+e x^2+f x^4)}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=144 \[ \frac {\log \left (a+b x^2+c x^4\right ) \left (-c (a f+b e)+b^2 f+c^2 d\right )}{4 c^3}-\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-b c (c d-3 a f)-2 a c^2 e+b^3 (-f)+b^2 c e\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {x^2 (c e-b f)}{2 c^2}+\frac {f x^4}{4 c} \]

[Out]

1/2*(-b*f+c*e)*x^2/c^2+1/4*f*x^4/c+1/4*(c^2*d+b^2*f-c*(a*f+b*e))*ln(c*x^4+b*x^2+a)/c^3-1/2*(b^2*c*e-2*a*c^2*e-
b^3*f-b*c*(-3*a*f+c*d))*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1663, 1628, 634, 618, 206, 628} \[ \frac {\log \left (a+b x^2+c x^4\right ) \left (-c (a f+b e)+b^2 f+c^2 d\right )}{4 c^3}-\frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-b c (c d-3 a f)-2 a c^2 e+b^2 c e+b^3 (-f)\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {x^2 (c e-b f)}{2 c^2}+\frac {f x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x^2 + f*x^4))/(a + b*x^2 + c*x^4),x]

[Out]

((c*e - b*f)*x^2)/(2*c^2) + (f*x^4)/(4*c) - ((b^2*c*e - 2*a*c^2*e - b^3*f - b*c*(c*d - 3*a*f))*ArcTanh[(b + 2*
c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*Sqrt[b^2 - 4*a*c]) + ((c^2*d + b^2*f - c*(b*e + a*f))*Log[a + b*x^2 + c*x^4]
)/(4*c^3)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {x^3 \left (d+e x^2+f x^4\right )}{a+b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x \left (d+e x+f x^2\right )}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {c e-b f}{c^2}+\frac {f x}{c}-\frac {a (c e-b f)-\left (c^2 d-b c e+b^2 f-a c f\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {(c e-b f) x^2}{2 c^2}+\frac {f x^4}{4 c}-\frac {\operatorname {Subst}\left (\int \frac {a (c e-b f)-\left (c^2 d-b c e+b^2 f-a c f\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2}\\ &=\frac {(c e-b f) x^2}{2 c^2}+\frac {f x^4}{4 c}-\frac {\left (-c^2 d+b c e-b^2 f+a c f\right ) \operatorname {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}+\frac {\left (b^2 c e-2 a c^2 e-b^3 f-b c (c d-3 a f)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}\\ &=\frac {(c e-b f) x^2}{2 c^2}+\frac {f x^4}{4 c}+\frac {\left (c^2 d-b c e+b^2 f-a c f\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}-\frac {\left (b^2 c e-2 a c^2 e-b^3 f-b c (c d-3 a f)\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3}\\ &=\frac {(c e-b f) x^2}{2 c^2}+\frac {f x^4}{4 c}-\frac {\left (b^2 c e-2 a c^2 e-b^3 f-b c (c d-3 a f)\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (c^2 d-b c e+b^2 f-a c f\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 136, normalized size = 0.94 \[ \frac {\log \left (a+b x^2+c x^4\right ) \left (-c (a f+b e)+b^2 f+c^2 d\right )-\frac {2 \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right ) \left (b c (c d-3 a f)+2 a c^2 e+b^3 f-b^2 c e\right )}{\sqrt {4 a c-b^2}}+2 c x^2 (c e-b f)+c^2 f x^4}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x^2 + f*x^4))/(a + b*x^2 + c*x^4),x]

[Out]

(2*c*(c*e - b*f)*x^2 + c^2*f*x^4 - (2*(-(b^2*c*e) + 2*a*c^2*e + b^3*f + b*c*(c*d - 3*a*f))*ArcTan[(b + 2*c*x^2
)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (c^2*d + b^2*f - c*(b*e + a*f))*Log[a + b*x^2 + c*x^4])/(4*c^3)

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fricas [A]  time = 1.50, size = 473, normalized size = 3.28 \[ \left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f x^{4} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} f\right )} x^{2} - {\left (b c^{2} d - {\left (b^{2} c - 2 \, a c^{2}\right )} e + {\left (b^{3} - 3 \, a b c\right )} f\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f x^{4} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} f\right )} x^{2} + 2 \, {\left (b c^{2} d - {\left (b^{2} c - 2 \, a c^{2}\right )} e + {\left (b^{3} - 3 \, a b c\right )} f\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

[1/4*((b^2*c^2 - 4*a*c^3)*f*x^4 + 2*((b^2*c^2 - 4*a*c^3)*e - (b^3*c - 4*a*b*c^2)*f)*x^2 - (b*c^2*d - (b^2*c -
2*a*c^2)*e + (b^3 - 3*a*b*c)*f)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqr
t(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + ((b^2*c^2 - 4*a*c^3)*d - (b^3*c - 4*a*b*c^2)*e + (b^4 - 5*a*b^2*c + 4*a
^2*c^2)*f)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4), 1/4*((b^2*c^2 - 4*a*c^3)*f*x^4 + 2*((b^2*c^2 - 4*a*c^3
)*e - (b^3*c - 4*a*b*c^2)*f)*x^2 + 2*(b*c^2*d - (b^2*c - 2*a*c^2)*e + (b^3 - 3*a*b*c)*f)*sqrt(-b^2 + 4*a*c)*ar
ctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + ((b^2*c^2 - 4*a*c^3)*d - (b^3*c - 4*a*b*c^2)*e + (b^4
- 5*a*b^2*c + 4*a^2*c^2)*f)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4)]

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giac [A]  time = 1.99, size = 141, normalized size = 0.98 \[ \frac {c f x^{4} - 2 \, b f x^{2} + 2 \, c x^{2} e}{4 \, c^{2}} + \frac {{\left (c^{2} d + b^{2} f - a c f - b c e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac {{\left (b c^{2} d + b^{3} f - 3 \, a b c f - b^{2} c e + 2 \, a c^{2} e\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

1/4*(c*f*x^4 - 2*b*f*x^2 + 2*c*x^2*e)/c^2 + 1/4*(c^2*d + b^2*f - a*c*f - b*c*e)*log(c*x^4 + b*x^2 + a)/c^3 - 1
/2*(b*c^2*d + b^3*f - 3*a*b*c*f - b^2*c*e + 2*a*c^2*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4
*a*c)*c^3)

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maple [B]  time = 0.01, size = 321, normalized size = 2.23 \[ \frac {f \,x^{4}}{4 c}+\frac {3 a b f \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}-\frac {a e \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {b^{3} f \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{3}}+\frac {b^{2} e \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}-\frac {b d \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c}-\frac {b f \,x^{2}}{2 c^{2}}+\frac {e \,x^{2}}{2 c}-\frac {a f \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}}+\frac {b^{2} f \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{3}}-\frac {b e \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}}+\frac {d \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x)

[Out]

1/4*f*x^4/c-1/2/c^2*x^2*b*f+1/2/c*x^2*e-1/4/c^2*ln(c*x^4+b*x^2+a)*a*f+1/4/c^3*ln(c*x^4+b*x^2+a)*b^2*f-1/4/c^2*
ln(c*x^4+b*x^2+a)*b*e+1/4/c*ln(c*x^4+b*x^2+a)*d+3/2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)
)*a*b*f-1/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*e-1/2/c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*
x^2+b)/(4*a*c-b^2)^(1/2))*b^3*f+1/2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^2*e-1/2/c/(4
*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.30, size = 1689, normalized size = 11.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x^2 + f*x^4))/(a + b*x^2 + c*x^4),x)

[Out]

x^2*(e/(2*c) - (b*f)/(2*c^2)) + (f*x^4)/(4*c) - (log(a + b*x^2 + c*x^4)*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f -
8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(2*(16*a*c^4 - 4*b^2*c^3)) - (atan((2*c^4*(4*a*c - b^2)*
(x^2*(((((6*b^3*c^3*f - 6*b^2*c^4*e + 4*a*c^5*e + 6*b*c^5*d - 10*a*b*c^4*f)/c^4 + (4*b*c^2*(2*b^4*f + 2*b^2*c^
2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(16*a*c^4 - 4*b^2*c^3))*(b^3*f + 2*a*
c^2*e + b*c^2*d - b^2*c*e - 3*a*b*c*f))/(8*c^3*(4*a*c - b^2)^(1/2)) + (b*(b^3*f + 2*a*c^2*e + b*c^2*d - b^2*c*
e - 3*a*b*c*f)*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(2*
c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a - (b*((b^5*f^2 + b*c^4*d^2 + b^3*c^2*e^2 + 2*a^2*b*c^2*f^2 +
a*c^4*d*e - 2*b^4*c*e*f - a*b*c^3*e^2 - 3*a*b^3*c*f^2 - 2*b^2*c^3*d*e - a^2*c^3*e*f + 2*b^3*c^2*d*f + 4*a*b^2*
c^2*e*f - 3*a*b*c^3*d*f)/c^4 + (((6*b^3*c^3*f - 6*b^2*c^4*e + 4*a*c^5*e + 6*b*c^5*d - 10*a*b*c^4*f)/c^4 + (4*b
*c^2*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(16*a*c^4 - 4
*b^2*c^3))*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(2*(16*
a*c^4 - 4*b^2*c^3)) - (b*(b^3*f + 2*a*c^2*e + b*c^2*d - b^2*c*e - 3*a*b*c*f)^2)/(2*c^4*(4*a*c - b^2))))/(2*a*(
4*a*c - b^2)^(1/2))) - ((((8*a^2*c^4*f - 8*a*c^5*d + 8*a*b*c^4*e - 8*a*b^2*c^3*f)/c^4 - (8*a*c^2*(2*b^4*f + 2*
b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f))/(16*a*c^4 - 4*b^2*c^3))*(b^3*f
+ 2*a*c^2*e + b*c^2*d - b^2*c*e - 3*a*b*c*f))/(8*c^3*(4*a*c - b^2)^(1/2)) - (a*(b^3*f + 2*a*c^2*e + b*c^2*d -
b^2*c*e - 3*a*b*c*f)*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*f
))/(c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a + (b*((((8*a^2*c^4*f - 8*a*c^5*d + 8*a*b*c^4*e - 8*a*b^2*
c^3*f)/c^4 - (8*a*c^2*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*b^2*c*
f))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4*f + 2*b^2*c^2*d + 8*a^2*c^2*f - 8*a*c^3*d - 2*b^3*c*e + 8*a*b*c^2*e - 10*a*
b^2*c*f))/(2*(16*a*c^4 - 4*b^2*c^3)) - (a*c^4*d^2 + a*b^4*f^2 + a^3*c^2*f^2 + a*b^2*c^2*e^2 - 2*a^2*b^2*c*f^2
- 2*a^2*c^3*d*f + 2*a*b^2*c^2*d*f + 2*a^2*b*c^2*e*f - 2*a*b*c^3*d*e - 2*a*b^3*c*e*f)/c^4 + (a*(b^3*f + 2*a*c^2
*e + b*c^2*d - b^2*c*e - 3*a*b*c*f)^2)/(c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))))/(b^6*f^2 + 4*a^2*c^4*
e^2 + b^2*c^4*d^2 + b^4*c^2*e^2 - 4*a*b^2*c^3*e^2 - 2*b^5*c*e*f + 9*a^2*b^2*c^2*f^2 - 6*a*b^4*c*f^2 - 2*b^3*c^
3*d*e + 2*b^4*c^2*d*f - 6*a*b^2*c^3*d*f + 10*a*b^3*c^2*e*f - 12*a^2*b*c^3*e*f + 4*a*b*c^4*d*e))*(b^3*f + 2*a*c
^2*e + b*c^2*d - b^2*c*e - 3*a*b*c*f))/(2*c^3*(4*a*c - b^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(f*x**4+e*x**2+d)/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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