3.55 \(\int \frac{1}{1+4 x+4 x^2+4 x^4} \, dx\)

Optimal. Leaf size=234 \[ -\frac{1}{4} \sqrt{\frac{1}{5} \left (\sqrt{5}-2\right )} \log \left (\left (\frac{1}{x}+1\right )^2-\sqrt{2 \left (1+\sqrt{5}\right )} \left (\frac{1}{x}+1\right )+\sqrt{5}\right )+\frac{1}{4} \sqrt{\frac{1}{5} \left (\sqrt{5}-2\right )} \log \left (\left (\frac{1}{x}+1\right )^2+\sqrt{2 \left (1+\sqrt{5}\right )} \left (\frac{1}{x}+1\right )+\sqrt{5}\right )+\frac{1}{2} \tan ^{-1}\left (\frac{1}{2} \left (\left (\frac{1}{x}+1\right )^2-1\right )\right )-\frac{1}{2} \sqrt{\frac{1}{5} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\frac{2}{x}-\sqrt{2 \left (1+\sqrt{5}\right )}+2}{\sqrt{2 \left (\sqrt{5}-1\right )}}\right )-\frac{1}{2} \sqrt{\frac{1}{5} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\frac{2}{x}+\sqrt{2 \left (1+\sqrt{5}\right )}+2}{\sqrt{2 \left (\sqrt{5}-1\right )}}\right ) \]

[Out]

ArcTan[(-1 + (1 + x^(-1))^2)/2]/2 - (Sqrt[(2 + Sqrt[5])/5]*ArcTan[(2 - Sqrt[2*(1 + Sqrt[5])] + 2/x)/Sqrt[2*(-1
+ Sqrt[5])]])/2 - (Sqrt[(2 + Sqrt[5])/5]*ArcTan[(2 + Sqrt[2*(1 + Sqrt[5])] + 2/x)/Sqrt[2*(-1 + Sqrt[5])]])/2
- (Sqrt[(-2 + Sqrt[5])/5]*Log[Sqrt[5] - Sqrt[2*(1 + Sqrt[5])]*(1 + x^(-1)) + (1 + x^(-1))^2])/4 + (Sqrt[(-2 +
Sqrt[5])/5]*Log[Sqrt[5] + Sqrt[2*(1 + Sqrt[5])]*(1 + x^(-1)) + (1 + x^(-1))^2])/4

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Rubi [A]  time = 0.315422, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {2069, 1673, 1169, 634, 618, 204, 628, 12, 1107} \[ -\frac{1}{4} \sqrt{\frac{1}{5} \left (\sqrt{5}-2\right )} \log \left (\left (\frac{1}{x}+1\right )^2-\sqrt{2 \left (1+\sqrt{5}\right )} \left (\frac{1}{x}+1\right )+\sqrt{5}\right )+\frac{1}{4} \sqrt{\frac{1}{5} \left (\sqrt{5}-2\right )} \log \left (\left (\frac{1}{x}+1\right )^2+\sqrt{2 \left (1+\sqrt{5}\right )} \left (\frac{1}{x}+1\right )+\sqrt{5}\right )+\frac{1}{2} \tan ^{-1}\left (\frac{1}{2} \left (\left (\frac{1}{x}+1\right )^2-1\right )\right )-\frac{1}{2} \sqrt{\frac{1}{5} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\frac{2}{x}-\sqrt{2 \left (1+\sqrt{5}\right )}+2}{\sqrt{2 \left (\sqrt{5}-1\right )}}\right )-\frac{1}{2} \sqrt{\frac{1}{5} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\frac{2}{x}+\sqrt{2 \left (1+\sqrt{5}\right )}+2}{\sqrt{2 \left (\sqrt{5}-1\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4*x + 4*x^2 + 4*x^4)^(-1),x]

[Out]

ArcTan[(-1 + (1 + x^(-1))^2)/2]/2 - (Sqrt[(2 + Sqrt[5])/5]*ArcTan[(2 - Sqrt[2*(1 + Sqrt[5])] + 2/x)/Sqrt[2*(-1
+ Sqrt[5])]])/2 - (Sqrt[(2 + Sqrt[5])/5]*ArcTan[(2 + Sqrt[2*(1 + Sqrt[5])] + 2/x)/Sqrt[2*(-1 + Sqrt[5])]])/2
- (Sqrt[(-2 + Sqrt[5])/5]*Log[Sqrt[5] - Sqrt[2*(1 + Sqrt[5])]*(1 + x^(-1)) + (1 + x^(-1))^2])/4 + (Sqrt[(-2 +
Sqrt[5])/5]*Log[Sqrt[5] + Sqrt[2*(1 + Sqrt[5])]*(1 + x^(-1)) + (1 + x^(-1))^2])/4

Rule 2069

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]}, Dist[-16*a^2, Subst[Int[(1*((a*(-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 256*a^3*e -
32*a^2*(3*b^2 - 8*a*c)*x^2 + 256*a^4*x^4))/(b - 4*a*x)^4)^p)/(b - 4*a*x)^2, x], x, b/(4*a) + 1/x], x] /; NeQ[a
, 0] && NeQ[b, 0] && EqQ[b^3 - 4*a*b*c + 8*a^2*d, 0]] /; FreeQ[p, x] && PolyQ[P4, x, 4] && IntegerQ[2*p] &&  !
IGtQ[p, 0]

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 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 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 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 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 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 12

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

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]

Rubi steps

\begin{align*} \int \frac{1}{1+4 x+4 x^2+4 x^4} \, dx &=-\left (16 \operatorname{Subst}\left (\int \frac{(4-4 x)^2}{1280-512 x^2+256 x^4} \, dx,x,1+\frac{1}{x}\right )\right )\\ &=-\left (16 \operatorname{Subst}\left (\int -\frac{32 x}{1280-512 x^2+256 x^4} \, dx,x,1+\frac{1}{x}\right )\right )-16 \operatorname{Subst}\left (\int \frac{16+16 x^2}{1280-512 x^2+256 x^4} \, dx,x,1+\frac{1}{x}\right )\\ &=512 \operatorname{Subst}\left (\int \frac{x}{1280-512 x^2+256 x^4} \, dx,x,1+\frac{1}{x}\right )-\frac{\operatorname{Subst}\left (\int \frac{16 \sqrt{2 \left (1+\sqrt{5}\right )}-\left (16-16 \sqrt{5}\right ) x}{\sqrt{5}-\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )}{32 \sqrt{10 \left (1+\sqrt{5}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{16 \sqrt{2 \left (1+\sqrt{5}\right )}+\left (16-16 \sqrt{5}\right ) x}{\sqrt{5}+\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )}{32 \sqrt{10 \left (1+\sqrt{5}\right )}}\\ &=256 \operatorname{Subst}\left (\int \frac{1}{1280-512 x+256 x^2} \, dx,x,\left (1+\frac{1}{x}\right )^2\right )+\frac{\left (1-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{-\sqrt{2 \left (1+\sqrt{5}\right )}+2 x}{\sqrt{5}-\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )}{4 \sqrt{10 \left (1+\sqrt{5}\right )}}-\frac{\left (1-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2 \left (1+\sqrt{5}\right )}+2 x}{\sqrt{5}+\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )}{4 \sqrt{10 \left (1+\sqrt{5}\right )}}-\frac{1}{20} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{5}-\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )-\frac{1}{20} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{5}+\sqrt{2 \left (1+\sqrt{5}\right )} x+x^2} \, dx,x,1+\frac{1}{x}\right )\\ &=-\frac{1}{4} \sqrt{-\frac{2}{5}+\frac{1}{\sqrt{5}}} \log \left (\sqrt{5}-\sqrt{2 \left (1+\sqrt{5}\right )} \left (1+\frac{1}{x}\right )+\left (1+\frac{1}{x}\right )^2\right )+\frac{1}{4} \sqrt{-\frac{2}{5}+\frac{1}{\sqrt{5}}} \log \left (\sqrt{5}+\sqrt{2 \left (1+\sqrt{5}\right )} \left (1+\frac{1}{x}\right )+\left (1+\frac{1}{x}\right )^2\right )-512 \operatorname{Subst}\left (\int \frac{1}{-1048576-x^2} \, dx,x,-512+512 \left (1+\frac{1}{x}\right )^2\right )+\frac{1}{10} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{5}\right )-x^2} \, dx,x,-\sqrt{2 \left (1+\sqrt{5}\right )}+2 \left (1+\frac{1}{x}\right )\right )+\frac{1}{10} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{5}\right )-x^2} \, dx,x,\sqrt{2 \left (1+\sqrt{5}\right )}+2 \left (1+\frac{1}{x}\right )\right )\\ &=\frac{1}{2} \tan ^{-1}\left (\frac{1}{2} \left (-1+\left (1+\frac{1}{x}\right )^2\right )\right )-\frac{\left (1+\sqrt{5}\right )^{3/2} \tan ^{-1}\left (\frac{2-\sqrt{2 \left (1+\sqrt{5}\right )}+\frac{2}{x}}{\sqrt{2 \left (-1+\sqrt{5}\right )}}\right )}{4 \sqrt{10}}-\frac{\left (1+\sqrt{5}\right )^{3/2} \tan ^{-1}\left (\frac{2+\sqrt{2 \left (1+\sqrt{5}\right )}+\frac{2}{x}}{\sqrt{2 \left (-1+\sqrt{5}\right )}}\right )}{4 \sqrt{10}}-\frac{1}{4} \sqrt{-\frac{2}{5}+\frac{1}{\sqrt{5}}} \log \left (\sqrt{5}-\sqrt{2 \left (1+\sqrt{5}\right )} \left (1+\frac{1}{x}\right )+\left (1+\frac{1}{x}\right )^2\right )+\frac{1}{4} \sqrt{-\frac{2}{5}+\frac{1}{\sqrt{5}}} \log \left (\sqrt{5}+\sqrt{2 \left (1+\sqrt{5}\right )} \left (1+\frac{1}{x}\right )+\left (1+\frac{1}{x}\right )^2\right )\\ \end{align*}

Mathematica [C]  time = 0.0145809, size = 47, normalized size = 0.2 \[ \frac{1}{4} \text{RootSum}\left [4 \text{$\#$1}^4+4 \text{$\#$1}^2+4 \text{$\#$1}+1\& ,\frac{\log (x-\text{$\#$1})}{4 \text{$\#$1}^3+2 \text{$\#$1}+1}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*x + 4*x^2 + 4*x^4)^(-1),x]

[Out]

RootSum[1 + 4*#1 + 4*#1^2 + 4*#1^4 & , Log[x - #1]/(1 + 2*#1 + 4*#1^3) & ]/4

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Maple [C]  time = 0.004, size = 41, normalized size = 0.2 \begin{align*}{\frac{1}{4}\sum _{{\it \_R}={\it RootOf} \left ( 4\,{{\it \_Z}}^{4}+4\,{{\it \_Z}}^{2}+4\,{\it \_Z}+1 \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{4\,{{\it \_R}}^{3}+2\,{\it \_R}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x^4+4*x^2+4*x+1),x)

[Out]

1/4*sum(1/(4*_R^3+2*_R+1)*ln(x-_R),_R=RootOf(4*_Z^4+4*_Z^2+4*_Z+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1),x, algorithm="maxima")

[Out]

integrate(1/(4*x^4 + 4*x^2 + 4*x + 1), x)

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Fricas [C]  time = 8.50128, size = 2303, normalized size = 9.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1),x, algorithm="fricas")

[Out]

-1/20*(sqrt(10)*sqrt(-15/8*(2*sqrt(1/10*I - 1/5) - I)^2 - 5/4*(2*sqrt(1/10*I - 1/5) - I)*(2*sqrt(-1/10*I - 1/5
) + I) - 15/8*(2*sqrt(-1/10*I - 1/5) + I)^2 - 9) - 5*sqrt(1/10*I - 1/5) - 5*sqrt(-1/10*I - 1/5))*log(5/2*(2*sq
rt(1/10*I - 1/5) - I)^2*(12*sqrt(-1/10*I - 1/5) + 6*I - 1) + 15*(2*sqrt(1/10*I - 1/5) - I)*(2*sqrt(-1/10*I - 1
/5) + I)^2 - 5/2*(2*sqrt(-1/10*I - 1/5) + I)^2 + ((6*sqrt(10)*(2*sqrt(-1/10*I - 1/5) + I) - sqrt(10))*(2*sqrt(
1/10*I - 1/5) - I) - sqrt(10)*(2*sqrt(-1/10*I - 1/5) + I))*sqrt(-15/8*(2*sqrt(1/10*I - 1/5) - I)^2 - 5/4*(2*sq
rt(1/10*I - 1/5) - I)*(2*sqrt(-1/10*I - 1/5) + I) - 15/8*(2*sqrt(-1/10*I - 1/5) + I)^2 - 9) + 8*x + 3) + 1/20*
(sqrt(10)*sqrt(-15/8*(2*sqrt(1/10*I - 1/5) - I)^2 - 5/4*(2*sqrt(1/10*I - 1/5) - I)*(2*sqrt(-1/10*I - 1/5) + I)
- 15/8*(2*sqrt(-1/10*I - 1/5) + I)^2 - 9) + 5*sqrt(1/10*I - 1/5) + 5*sqrt(-1/10*I - 1/5))*log(5/2*(2*sqrt(1/1
0*I - 1/5) - I)^2*(12*sqrt(-1/10*I - 1/5) + 6*I - 1) + 15*(2*sqrt(1/10*I - 1/5) - I)*(2*sqrt(-1/10*I - 1/5) +
I)^2 - 5/2*(2*sqrt(-1/10*I - 1/5) + I)^2 - ((6*sqrt(10)*(2*sqrt(-1/10*I - 1/5) + I) - sqrt(10))*(2*sqrt(1/10*I
- 1/5) - I) - sqrt(10)*(2*sqrt(-1/10*I - 1/5) + I))*sqrt(-15/8*(2*sqrt(1/10*I - 1/5) - I)^2 - 5/4*(2*sqrt(1/1
0*I - 1/5) - I)*(2*sqrt(-1/10*I - 1/5) + I) - 15/8*(2*sqrt(-1/10*I - 1/5) + I)^2 - 9) + 8*x + 3) - 1/4*(2*sqrt
(1/10*I - 1/5) - I)*log(-5*(2*sqrt(1/10*I - 1/5) - I)^2*(12*sqrt(-1/10*I - 1/5) + 6*I - 1) - 30*(2*sqrt(1/10*I
- 1/5) - I)*(2*sqrt(-1/10*I - 1/5) + I)^2 - 30*(2*sqrt(-1/10*I - 1/5) + I)^3 + 8*x - 216*sqrt(-1/10*I - 1/5)
- 108*I + 21) - 1/4*(2*sqrt(-1/10*I - 1/5) + I)*log(30*(2*sqrt(-1/10*I - 1/5) + I)^3 + 5*(2*sqrt(-1/10*I - 1/5
) + I)^2 + 8*x + 216*sqrt(-1/10*I - 1/5) + 108*I - 27)

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Sympy [A]  time = 0.724326, size = 36, normalized size = 0.15 \begin{align*} \operatorname{RootSum}{\left (1280 t^{4} + 288 t^{2} + 32 t + 1, \left ( t \mapsto t \log{\left (- 240 t^{3} + 10 t^{2} - 54 t + x - \frac{27}{8} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x**4+4*x**2+4*x+1),x)

[Out]

RootSum(1280*_t**4 + 288*_t**2 + 32*_t + 1, Lambda(_t, _t*log(-240*_t**3 + 10*_t**2 - 54*_t + x - 27/8)))

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Giac [A]  time = 1.15943, size = 374, normalized size = 1.6 \begin{align*} \frac{{\left (\sqrt{\sqrt{5} - 2}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} + 2 \, i + 1\right )} \log \left (2 \,{\left (7 \, i + 3\right )} x + \sqrt{29 \, \sqrt{5} + 62}{\left (\frac{19 \, i}{29 \, \sqrt{5} + 62} + 1\right )} + 3 \, i - 7\right )}{4 \,{\left (i - 2\right )}} - \frac{{\left (\sqrt{\sqrt{5} - 2}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} - 2 \, i - 1\right )} \log \left (2 \,{\left (7 \, i + 3\right )} x - \sqrt{29 \, \sqrt{5} + 62}{\left (\frac{19 \, i}{29 \, \sqrt{5} + 62} + 1\right )} + 3 \, i - 7\right )}{4 \,{\left (i - 2\right )}} - \frac{{\left (\sqrt{\sqrt{5} + 2}{\left (\frac{i}{\sqrt{5} + 2} + 1\right )} + i + 2\right )} \log \left (2 \,{\left (i + 5\right )} x + \sqrt{13 \, \sqrt{5} - 22}{\left (\frac{19 \, i}{13 \, \sqrt{5} - 22} + 1\right )} - 5 \, i + 1\right )}{4 \,{\left (2 \, i - 1\right )}} + \frac{{\left (\sqrt{\sqrt{5} + 2}{\left (\frac{i}{\sqrt{5} + 2} + 1\right )} - i - 2\right )} \log \left (2 \,{\left (i + 5\right )} x - \sqrt{13 \, \sqrt{5} - 22}{\left (\frac{19 \, i}{13 \, \sqrt{5} - 22} + 1\right )} - 5 \, i + 1\right )}{4 \,{\left (2 \, i - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1),x, algorithm="giac")

[Out]

1/4*(sqrt(sqrt(5) - 2)*(i/(sqrt(5) - 2) + 1) + 2*i + 1)*log(2*(7*i + 3)*x + sqrt(29*sqrt(5) + 62)*(19*i/(29*sq
rt(5) + 62) + 1) + 3*i - 7)/(i - 2) - 1/4*(sqrt(sqrt(5) - 2)*(i/(sqrt(5) - 2) + 1) - 2*i - 1)*log(2*(7*i + 3)*
x - sqrt(29*sqrt(5) + 62)*(19*i/(29*sqrt(5) + 62) + 1) + 3*i - 7)/(i - 2) - 1/4*(sqrt(sqrt(5) + 2)*(i/(sqrt(5)
+ 2) + 1) + i + 2)*log(2*(i + 5)*x + sqrt(13*sqrt(5) - 22)*(19*i/(13*sqrt(5) - 22) + 1) - 5*i + 1)/(2*i - 1)
+ 1/4*(sqrt(sqrt(5) + 2)*(i/(sqrt(5) + 2) + 1) - i - 2)*log(2*(i + 5)*x - sqrt(13*sqrt(5) - 22)*(19*i/(13*sqrt
(5) - 22) + 1) - 5*i + 1)/(2*i - 1)