3.32.41 $\int \frac{(-32+2x^2-16x^3+24x^4)\log (\frac{-16-x^2+4x^3-4x^4}{x})+(-16-x^2+4x^3-4x^4){\log }^2(\frac{-16-x^2+4x^3-4x^4}{x})}{64x^2+4x^4-16x^5+16x^6}dx$

Optimal. Leaf size=29 $$\frac{{\log }^2(-\frac{16}{x}-x+4x(x-x^2))}{4x}$$

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Rubi [A]  time = 5.67, antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 51, number of rules used = 19, integrand size = 107, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.178, Rules used = {6741, 12, 6742, 2528, 2525, 1680, 1662, 1107, 618, 204, 1127, 1161, 1164, 628, 1106, 1094, 634, 1673, 1169} $$\begin{array}{c}\frac{{\log }^2(-\frac{4x^4-4x^3+x^2+16}{x})}{4x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[((-32 + 2*x^2 - 16*x^3 + 24*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4)/x] + (-16 - x^2 + 4*x^3 - 4*x^4)*Log[(-16
- x^2 + 4*x^3 - 4*x^4)/x]^2)/(64*x^2 + 4*x^4 - 16*x^5 + 16*x^6),x]

[Out]

Log[-((16 + x^2 - 4*x^3 + 4*x^4)/x)]^2/(4*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 1094

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

Rule 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

Rule 1107

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

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1164

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

Rule 1169

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

Rule 1662

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x],
k}, Int[(d*x)^m*Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(a + b*x^2 + c*x^4)^p, x] + Dist[1/d, Int[(d*
x)^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q - 1)/2 + 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]

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(-32+2x^2-16x^3+24x^4)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)+(-16-x^2+4x^3-4x^4){\log }^2\left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{4x^2(16+x^2-4x^3+4x^4)}dx\\ & =\frac{1}{4}\int \frac{(-32+2x^2-16x^3+24x^4)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)+(-16-x^2+4x^3-4x^4){\log }^2\left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2(16+x^2-4x^3+4x^4)}dx\\ & =\frac{1}{4}\int (\frac{2(-16+x^2-8x^3+12x^4)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2(16+x^2-4x^3+4x^4)}-\frac{{\log }^2\left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2})dx\\ & =-(\frac{1}{4}\int \frac{{\log }^2\left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2}dx)+\frac{1}{2}\int \frac{(-16+x^2-8x^3+12x^4)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2(16+x^2-4x^3+4x^4)}dx\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-\frac{1}{2}\int \frac{(-16+x^2-8x^3+12x^4)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2(16+x^2-4x^3+4x^4)}dx+\frac{1}{2}\int (-\frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2}+\frac{2(1-6x+8x^2)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4})dx\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-\frac{1}{2}\int \frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2}dx-\frac{1}{2}\int (-\frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2}+\frac{2(1-6x+8x^2)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4})dx+\int \frac{(1-6x+8x^2)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}dx\\ & =\frac{\log (-\frac{16+x^2-4x^3+4x^4}{x})}{2x}+\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-\frac{1}{2}\int \frac{-16+x^2-8x^3+12x^4}{x^2(16+x^2-4x^3+4x^4)}dx+\frac{1}{2}\int \frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{x^2}dx-\int \frac{(1-6x+8x^2)\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}dx+\int (\frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}-\frac{6x\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}+\frac{8x^2\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4})dx\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+\frac{1}{2}\int \frac{-16+x^2-8x^3+12x^4}{x^2(16+x^2-4x^3+4x^4)}dx-\frac{1}{2}\int (-\frac{1}{x^2}+\frac{2(1-6x+8x^2)}{16+x^2-4x^3+4x^4})dx-6\int \frac{x\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}dx+8\int \frac{x^2\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}dx+\int \frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}dx-\int (\frac{\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}-\frac{6x\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4}+\frac{8x^2\log \left(\frac{-16-x^2+4x^3-4x^4}{x}\right)}{16+x^2-4x^3+4x^4})dx\\ & =-\frac{1}{2x}+\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+\frac{1}{2}\int (-\frac{1}{x^2}+\frac{2(1-6x+8x^2)}{16+x^2-4x^3+4x^4})dx-\int \frac{1-6x+8x^2}{16+x^2-4x^3+4x^4}dx\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+\int \frac{1-6x+8x^2}{16+x^2-4x^3+4x^4}dx-\mathrm{Subst}(\int \frac{128x(-1+4x)}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-128\mathrm{Subst}(\int \frac{x(-1+4x)}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)+\mathrm{Subst}(\int \frac{128x(-1+4x)}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+128\mathrm{Subst}(\int \frac{x}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)-128\mathrm{Subst}(\int \frac{4x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)+128\mathrm{Subst}(\int \frac{x(-1+4x)}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+64\mathrm{Subst}(\int \frac{1}{1025-32x+256x^2}dx,x,{(-\frac{1}{4}+x)}^2)-128\mathrm{Subst}(\int \frac{x}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)+128\mathrm{Subst}(\int \frac{4x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)-512\mathrm{Subst}(\int \frac{x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-64\mathrm{Subst}(\int \frac{1}{1025-32x+256x^2}dx,x,{(-\frac{1}{4}+x)}^2)-128\mathrm{Subst}(\int \frac{1}{-1048576-x^2}dx,x,256x(-1+2x))+256\mathrm{Subst}(\int \frac{\frac{5\sqrt{41}}{16}-x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)-256\mathrm{Subst}(\int \frac{\frac{5\sqrt{41}}{16}+x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)+512\mathrm{Subst}(\int \frac{x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)\\ & =-\frac{1}{8}{\tan }^{-1}(\frac{1}{4}(1-2x)x)+\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-\frac{1}{2}\mathrm{Subst}(\int \frac{1}{\frac{5\sqrt{41}}{16}-\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x+x^2}dx,x,-\frac{1}{4}+x)-\frac{1}{2}\mathrm{Subst}(\int \frac{1}{\frac{5\sqrt{41}}{16}+\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x+x^2}dx,x,-\frac{1}{4}+x)+128\mathrm{Subst}(\int \frac{1}{-1048576-x^2}dx,x,256x(-1+2x))-256\mathrm{Subst}(\int \frac{\frac{5\sqrt{41}}{16}-x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)+256\mathrm{Subst}(\int \frac{\frac{5\sqrt{41}}{16}+x^2}{1025-32x^2+256x^4}dx,x,-\frac{1}{4}+x)-\sqrt{\frac{2}{1+5\sqrt{41}}}\mathrm{Subst}(\int \frac{\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}+2x}{-\frac{5\sqrt{41}}{16}-\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x-x^2}dx,x,-\frac{1}{4}+x)-\sqrt{\frac{2}{1+5\sqrt{41}}}\mathrm{Subst}(\int \frac{\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}-2x}{-\frac{5\sqrt{41}}{16}+\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x-x^2}dx,x,-\frac{1}{4}+x)\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}+\sqrt{\frac{2}{1+5\sqrt{41}}}\log (5\sqrt{41}-\sqrt{2(1+5\sqrt{41})}(1-4x)+(-1+4x{)}^2)-\sqrt{\frac{2}{1+5\sqrt{41}}}\log (5\sqrt{41}+\sqrt{2(1+5\sqrt{41})}(1-4x)+(-1+4x{)}^2)+\frac{1}{2}\mathrm{Subst}(\int \frac{1}{\frac{5\sqrt{41}}{16}-\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x+x^2}dx,x,-\frac{1}{4}+x)+\frac{1}{2}\mathrm{Subst}(\int \frac{1}{\frac{5\sqrt{41}}{16}+\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x+x^2}dx,x,-\frac{1}{4}+x)+\sqrt{\frac{2}{1+5\sqrt{41}}}\mathrm{Subst}(\int \frac{\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}+2x}{-\frac{5\sqrt{41}}{16}-\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x-x^2}dx,x,-\frac{1}{4}+x)+\sqrt{\frac{2}{1+5\sqrt{41}}}\mathrm{Subst}(\int \frac{\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}-2x}{-\frac{5\sqrt{41}}{16}+\frac{1}{2}\sqrt{\frac{1}{2}(1+5\sqrt{41})}x-x^2}dx,x,-\frac{1}{4}+x)+\mathrm{Subst}(\int \frac{1}{\frac{1}{8}(1-5\sqrt{41})-x^2}dx,x,\frac{1}{2}(-1-\sqrt{\frac{1}{2}(1+5\sqrt{41})}+4x))+\mathrm{Subst}(\int \frac{1}{\frac{1}{8}(1-5\sqrt{41})-x^2}dx,x,\frac{1}{4}(-2+\sqrt{2+10\sqrt{41}}+8x))\\ & =-2\sqrt{\frac{2}{-1+5\sqrt{41}}}{\tan }^{-1}\left(\frac{-2-\sqrt{2+10\sqrt{41}}+8x}{\sqrt{-2+10\sqrt{41}}}\right)-2\sqrt{\frac{2}{-1+5\sqrt{41}}}{\tan }^{-1}\left(\frac{-2+\sqrt{2+10\sqrt{41}}+8x}{\sqrt{-2+10\sqrt{41}}}\right)+\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}-\mathrm{Subst}(\int \frac{1}{\frac{1}{8}(1-5\sqrt{41})-x^2}dx,x,\frac{1}{2}(-1-\sqrt{\frac{1}{2}(1+5\sqrt{41})}+4x))-\mathrm{Subst}(\int \frac{1}{\frac{1}{8}(1-5\sqrt{41})-x^2}dx,x,\frac{1}{4}(-2+\sqrt{2+10\sqrt{41}}+8x))\\ & =\frac{{\log }^2(-\frac{16+x^2-4x^3+4x^4}{x})}{4x}\end{array}\end{array}$$

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{\$Aborted}\end{array}$$

Verification is not applicable to the result.

[In]

Integrate[((-32 + 2*x^2 - 16*x^3 + 24*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4)/x] + (-16 - x^2 + 4*x^3 - 4*x^4)*Lo
g[(-16 - x^2 + 4*x^3 - 4*x^4)/x]^2)/(64*x^2 + 4*x^4 - 16*x^5 + 16*x^6),x]

[Out]

$Aborted

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fricas [A]  time = 0.74, size = 28, normalized size = 0.97 $$\begin{array}{c}\frac{\log {(-\frac{4x^4-4x^3+x^2+16}{x})}^2}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16*x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2
-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2),x, algorithm="fricas")

[Out]

1/4*log(-(4*x^4 - 4*x^3 + x^2 + 16)/x)^2/x

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Exception raised: TypeError}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16*x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2
-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{poly1[46450893462272555700247723121885101557651700311
39394747419

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maple [A]  time = 0.05, size = 30, normalized size = 1.03

method	result	size
norman	$\frac{\ln {\left(\frac{-4x^4+4x^3-x^2-16}{x}\right)}^2}{4x}$	$30$
risch	$\frac{\ln {\left(\frac{-4x^4+4x^3-x^2-16}{x}\right)}^2}{4x}$	$30$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^4+4*x^3-x^2-16)*ln((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16*x^3+2*x^2-32)*ln((-4*x^4+4*x^3-x^2-16)/x))
/(16*x^6-16*x^5+4*x^4+64*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*ln((-4*x^4+4*x^3-x^2-16)/x)^2/x

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maxima [A]  time = 0.78, size = 52, normalized size = 1.79 $$\begin{array}{c}\frac{\log {(-4x^4+4x^3-x^2-16)}^2-2\log (-4x^4+4x^3-x^2-16)\log (x)+\log (x{)}^2}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16*x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2
-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2),x, algorithm="maxima")

[Out]

1/4*(log(-4*x^4 + 4*x^3 - x^2 - 16)^2 - 2*log(-4*x^4 + 4*x^3 - x^2 - 16)*log(x) + log(x)^2)/x

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mupad [B]  time = 2.07, size = 28, normalized size = 0.97 $$\begin{array}{c}\frac{{\ln (-\frac{4x^4-4x^3+x^2+16}{x})}^2}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-(x^2 - 4*x^3 + 4*x^4 + 16)/x)*(2*x^2 - 16*x^3 + 24*x^4 - 32) - log(-(x^2 - 4*x^3 + 4*x^4 + 16)/x)^2*
(x^2 - 4*x^3 + 4*x^4 + 16))/(64*x^2 + 4*x^4 - 16*x^5 + 16*x^6),x)

[Out]

log(-(x^2 - 4*x^3 + 4*x^4 + 16)/x)^2/(4*x)

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sympy [A]  time = 0.22, size = 22, normalized size = 0.76 $$\begin{array}{c}\frac{\log {\left(\frac{-4x^4+4x^3-x^2-16}{x}\right)}^2}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**4+4*x**3-x**2-16)*ln((-4*x**4+4*x**3-x**2-16)/x)**2+(24*x**4-16*x**3+2*x**2-32)*ln((-4*x**4+
4*x**3-x**2-16)/x))/(16*x**6-16*x**5+4*x**4+64*x**2),x)

[Out]

log((-4*x**4 + 4*x**3 - x**2 - 16)/x)**2/(4*x)

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