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

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

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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 {gather*} \frac {\log ^2\left (-\frac {4 x^4-4 x^3+x^2+16}{x}\right )}{4 x} \end {gather*}

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

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

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 {gather*} \frac {\log \left (-\frac {4 \, x^{4} - 4 \, x^{3} + x^{2} + 16}{x}\right )^{2}}{4 \, x} \end {gather*}

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 {gather*} \text {Exception raised: TypeError} \end {gather*}

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	\(\frac {\ln \left (\frac {-4 x^{4}+4 x^{3}-x^{2}-16}{x}\right )^{2}}{4 x}\)	\(30\)
risch	\(\frac {\ln \left (\frac {-4 x^{4}+4 x^{3}-x^{2}-16}{x}\right )^{2}}{4 x}\)	\(30\)

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 {gather*} \frac {\log \left (-4 \, x^{4} + 4 \, x^{3} - x^{2} - 16\right )^{2} - 2 \, \log \left (-4 \, x^{4} + 4 \, x^{3} - x^{2} - 16\right ) \log \relax (x) + \log \relax (x)^{2}}{4 \, x} \end {gather*}

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 {gather*} \frac {{\ln \left (-\frac {4\,x^4-4\,x^3+x^2+16}{x}\right )}^2}{4\,x} \end {gather*}

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 {gather*} \frac {\log {\left (\frac {- 4 x^{4} + 4 x^{3} - x^{2} - 16}{x} \right )}^{2}}{4 x} \end {gather*}

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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