3.27.76 \(\int \frac {(-1+2 k^2) x-2 k^4 x^3+k^4 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (1-d+(d-2 k^2) x^2+k^4 x^4)} \, dx\)

Optimal. Leaf size=241 \[ \frac {\log \left (d^{2/3} x^4-2 d^{2/3} x^2+d^{2/3}+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{4/3}\right )}{4 \sqrt [3]{d}}-\frac {\log \left (\sqrt [3]{d} x^2-\sqrt [3]{d}+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}\right )}{2 \sqrt [3]{d}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}{-2 \sqrt [3]{d} x^2+2 \sqrt [3]{d}+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}\right )}{2 \sqrt [3]{d}} \]

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Rubi [C]  time = 2.87, antiderivative size = 451, normalized size of antiderivative = 1.87, number of steps used = 10, number of rules used = 7, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {1594, 6715, 6719, 1586, 6728, 137, 136} \begin {gather*} \frac {3 k^2 \left (1-x^2\right ) \left (\sqrt {d}-\sqrt {d-4 k^2 \left (1-k^2\right )}\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{-2 \left (1-k^2\right ) k^2+d-\sqrt {d} \sqrt {d-4 k^2 \left (1-k^2\right )}}\right )}{2 \sqrt {d} \left (-\sqrt {d} \sqrt {d-4 k^2 \left (1-k^2\right )}+d-2 \left (1-k^2\right ) k^2\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (1-x^2\right ) \left (\sqrt {d-4 k^2 \left (1-k^2\right )}+\sqrt {d}\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{-2 \left (1-k^2\right ) k^2+d+\sqrt {d} \sqrt {d-4 k^2 \left (1-k^2\right )}}\right )}{2 \sqrt {d} \left (\sqrt {d} \sqrt {d-4 k^2 \left (1-k^2\right )}+d-2 \left (1-k^2\right ) k^2\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2)*x^2 + k^4*x^4
)),x]

[Out]

(3*k^2*(Sqrt[d] - Sqrt[d - 4*k^2*(1 - k^2)])*(1 - x^2)*(1 - k^2*x^2)*AppellF1[1/3, -1/3, 1, 4/3, -((k^2*(1 - x
^2))/(1 - k^2)), (2*k^4*(1 - x^2))/(d - 2*k^2*(1 - k^2) - Sqrt[d]*Sqrt[d - 4*k^2*(1 - k^2)])])/(2*Sqrt[d]*(d -
2*k^2*(1 - k^2) - Sqrt[d]*Sqrt[d - 4*k^2*(1 - k^2)])*((1 - k^2*x^2)/(1 - k^2))^(1/3)*((1 - x^2)*(1 - k^2*x^2)
)^(2/3)) + (3*k^2*(Sqrt[d] + Sqrt[d - 4*k^2*(1 - k^2)])*(1 - x^2)*(1 - k^2*x^2)*AppellF1[1/3, -1/3, 1, 4/3, -(
(k^2*(1 - x^2))/(1 - k^2)), (2*k^4*(1 - x^2))/(d - 2*k^2*(1 - k^2) + Sqrt[d]*Sqrt[d - 4*k^2*(1 - k^2)])])/(2*S
qrt[d]*(d - 2*k^2*(1 - k^2) + Sqrt[d]*Sqrt[d - 4*k^2*(1 - k^2)])*((1 - k^2*x^2)/(1 - k^2))^(1/3)*((1 - x^2)*(1
- k^2*x^2))^(2/3))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx &=\int \frac {x \left (-1+2 k^2-2 k^4 x^2+k^4 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x} \left (-1+2 k^2-k^2 x\right )}{(1-x)^{2/3} \left (1-d+\left (d-2 k^2\right ) x+k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {\left (-k^2+\frac {k^2 \sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )}+\frac {\left (-k^2-\frac {k^2 \sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )}\right ) \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k^2 \left (1+\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (d-2 k^2-\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}-\frac {\left (k^2 \left (1+\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (d-2 k^2+\sqrt {d} \sqrt {d-4 k^2+4 k^4}+2 k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}\\ &=\frac {3 k^2 \left (1-\frac {\sqrt {d-4 k^2+4 k^4}}{\sqrt {d}}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{d-2 k^2 \left (1-k^2\right )-\sqrt {d} \sqrt {d-4 k^2+4 k^4}}\right )}{2 \left (d-2 k^2 \left (1-k^2\right )-\sqrt {d} \sqrt {d-4 k^2+4 k^4}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (\sqrt {d}+\sqrt {d-4 k^2+4 k^4}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 k^4 \left (1-x^2\right )}{d-2 k^2 \left (1-k^2\right )+\sqrt {d} \sqrt {d-4 k^2+4 k^4}}\right )}{2 \sqrt {d} \left (d-2 k^2 \left (1-k^2\right )+\sqrt {d} \sqrt {d-4 k^2+4 k^4}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 1.89, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2)*x^2 + k
^4*x^4)),x]

[Out]

Integrate[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2)*x^2 + k
^4*x^4)), x]

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IntegrateAlgebraic [A]  time = 5.56, size = 241, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} x^2+d^{2/3} x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + 2*k^2)*x - 2*k^4*x^3 + k^4*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(1 - d + (d - 2*k^2
)*x^2 + k^4*x^4)),x]

[Out]

-1/2*(Sqrt[3]*ArcTan[(Sqrt[3]*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3))/(2*d^(1/3) - 2*d^(1/3)*x^2 + (1 + (-1 - k^
2)*x^2 + k^2*x^4)^(2/3))])/d^(1/3) - Log[-d^(1/3) + d^(1/3)*x^2 + (1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3)]/(2*d^(
1/3)) + Log[d^(2/3) - 2*d^(2/3)*x^2 + d^(2/3)*x^4 + (d^(1/3) - d^(1/3)*x^2)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/
3) + (1 + (-1 - k^2)*x^2 + k^2*x^4)^(4/3)]/(4*d^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{4} x^{5} - 2 \, k^{4} x^{3} + {\left (2 \, k^{2} - 1\right )} x}{{\left (k^{4} x^{4} - {\left (2 \, k^{2} - d\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="giac")

[Out]

integrate((k^4*x^5 - 2*k^4*x^3 + (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*((k^2*x^2 - 1)*(x^2 - 1))
^(2/3)), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (2 k^{2}-1\right ) x -2 k^{4} x^{3}+k^{4} x^{5}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {2}{3}} \left (1-d +\left (-2 k^{2}+d \right ) x^{2}+k^{4} x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x)

[Out]

int(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{4} x^{5} - 2 \, k^{4} x^{3} + {\left (2 \, k^{2} - 1\right )} x}{{\left (k^{4} x^{4} - {\left (2 \, k^{2} - d\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(-2*k^2+d)*x^2+k^4*x^4),x, algori
thm="maxima")

[Out]

integrate((k^4*x^5 - 2*k^4*x^3 + (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*((k^2*x^2 - 1)*(x^2 - 1))
^(2/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {k^4\,x^5-2\,k^4\,x^3+x\,\left (2\,k^2-1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (k^4\,x^4-d+x^2\,\left (d-2\,k^2\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((k^4*x^5 - 2*k^4*x^3 + x*(2*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(k^4*x^4 - d + x^2*(d - 2*k^2) + 1)
),x)

[Out]

int((k^4*x^5 - 2*k^4*x^3 + x*(2*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(k^4*x^4 - d + x^2*(d - 2*k^2) + 1)
), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*k**2-1)*x-2*k**4*x**3+k**4*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3)/(1-d+(-2*k**2+d)*x**2+k**4*x*
*4),x)

[Out]

Timed out

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