\(\int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx\) [7]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [F]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 17, antiderivative size = 185 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\frac {1}{2} \arctan \left (\frac {1}{2} \left (-1+\left (1+\frac {1}{x}\right )^2\right )\right )-\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2-\sqrt {2 \left (1+\sqrt {5}\right )}+\frac {2}{x}}{\sqrt {2 \left (-1+\sqrt {5}\right )}}\right )}{4 \sqrt {10}}-\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2+\sqrt {2 \left (1+\sqrt {5}\right )}+\frac {2}{x}}{\sqrt {2 \left (-1+\sqrt {5}\right )}}\right )}{4 \sqrt {10}}+\frac {1}{2} \sqrt {\frac {1}{5} \left (-2+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )}{\sqrt {5}+\left (1+\frac {1}{x}\right )^2}\right ) \] Output:

1/2*arctan(-1/2+1/2*(1+1/x)^2)-1/40*(5^(1/2)+1)^(3/2)*arctan((2-(2+2*5^(1/ 
2))^(1/2)+2/x)/(-2+2*5^(1/2))^(1/2))*10^(1/2)-1/40*(5^(1/2)+1)^(3/2)*arcta 
n((2+(2+2*5^(1/2))^(1/2)+2/x)/(-2+2*5^(1/2))^(1/2))*10^(1/2)+1/10*(-10+5*5 
^(1/2))^(1/2)*arctanh((2+2*5^(1/2))^(1/2)*(1+1/x)/(5^(1/2)+(1+1/x)^2))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.25 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\frac {1}{4} \text {RootSum}\left [1+4 \text {$\#$1}+4 \text {$\#$1}^2+4 \text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{1+2 \text {$\#$1}+4 \text {$\#$1}^3}\&\right ] \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.46, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.765, Rules used = {2504, 27, 2202, 27, 1432, 1083, 217, 1483, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2504

\(\displaystyle -16 \int \frac {1}{16 \left (\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5\right ) x^2}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \frac {1}{\left (\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5\right ) x^2}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 2202

\(\displaystyle -\int \frac {\left (1+\frac {1}{x}\right )^2+1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )-\int -\frac {2 \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \frac {1+\frac {1}{x}}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )-\int \frac {\left (1+\frac {1}{x}\right )^2+1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 1432

\(\displaystyle \int \frac {1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )^2-\int \frac {\left (1+\frac {1}{x}\right )^2+1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle -2 \int \frac {1}{-\left (1+\frac {1}{x}\right )^4-16}d\left (2 \left (1+\frac {1}{x}\right )^2-2\right )-\int \frac {\left (1+\frac {1}{x}\right )^2+1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )-\int \frac {\left (1+\frac {1}{x}\right )^2+1}{\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}d\left (1+\frac {1}{x}\right )\)

\(\Big \downarrow \) 1483

\(\displaystyle -\frac {\int \frac {\sqrt {2 \left (1+\sqrt {5}\right )}-\left (1-\sqrt {5}\right ) \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\int \frac {\left (1-\sqrt {5}\right ) \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle -\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{\sqrt {2}}-\frac {1}{2} \left (1-\sqrt {5}\right ) \int -\frac {\sqrt {2 \left (1+\sqrt {5}\right )}-2 \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{\sqrt {2}}+\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{\sqrt {2}}+\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {5}\right )}-2 \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{\sqrt {2}}+\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {5}\right )}-2 \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )-\sqrt {2} \left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{2 \left (1-\sqrt {5}\right )-\left (2 \left (1+\frac {1}{x}\right )-\sqrt {2 \left (1+\sqrt {5}\right )}\right )^2}d\left (2 \left (1+\frac {1}{x}\right )-\sqrt {2 \left (1+\sqrt {5}\right )}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )-\sqrt {2} \left (1+\sqrt {5}\right )^{3/2} \int \frac {1}{2 \left (1-\sqrt {5}\right )-\left (2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}\right )^2}d\left (2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {5}\right )}-2 \left (1+\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )+\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2 \left (\frac {1}{x}+1\right )-\sqrt {2 \left (1+\sqrt {5}\right )}}{\sqrt {2 \left (\sqrt {5}-1\right )}}\right )}{\sqrt {\sqrt {5}-1}}}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {2 \left (1+\frac {1}{x}\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\left (1+\frac {1}{x}\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (1+\frac {1}{x}\right )+\sqrt {5}}d\left (1+\frac {1}{x}\right )+\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2 \left (\frac {1}{x}+1\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\sqrt {2 \left (\sqrt {5}-1\right )}}\right )}{\sqrt {\sqrt {5}-1}}}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \arctan \left (\frac {1}{4} \left (2 \left (\frac {1}{x}+1\right )^2-2\right )\right )-\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2 \left (\frac {1}{x}+1\right )-\sqrt {2 \left (1+\sqrt {5}\right )}}{\sqrt {2 \left (\sqrt {5}-1\right )}}\right )}{\sqrt {\sqrt {5}-1}}-\frac {1}{2} \left (1-\sqrt {5}\right ) \log \left (\left (\frac {1}{x}+1\right )^2-\sqrt {2 \left (1+\sqrt {5}\right )} \left (\frac {1}{x}+1\right )+\sqrt {5}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}-\frac {\frac {\left (1+\sqrt {5}\right )^{3/2} \arctan \left (\frac {2 \left (\frac {1}{x}+1\right )+\sqrt {2 \left (1+\sqrt {5}\right )}}{\sqrt {2 \left (\sqrt {5}-1\right )}}\right )}{\sqrt {\sqrt {5}-1}}+\frac {1}{2} \left (1-\sqrt {5}\right ) \log \left (\left (\frac {1}{x}+1\right )^2+\sqrt {2 \left (1+\sqrt {5}\right )} \left (\frac {1}{x}+1\right )+\sqrt {5}\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}\)

Input:

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

Output:

ArcTan[(-2 + 2*(1 + x^(-1))^2)/4]/2 - (((1 + Sqrt[5])^(3/2)*ArcTan[(-Sqrt[ 
2*(1 + Sqrt[5])] + 2*(1 + x^(-1)))/Sqrt[2*(-1 + Sqrt[5])]])/Sqrt[-1 + Sqrt 
[5]] - ((1 - Sqrt[5])*Log[Sqrt[5] - Sqrt[2*(1 + Sqrt[5])]*(1 + x^(-1)) + ( 
1 + x^(-1))^2])/2)/(2*Sqrt[10*(1 + Sqrt[5])]) - (((1 + Sqrt[5])^(3/2)*ArcT 
an[(Sqrt[2*(1 + Sqrt[5])] + 2*(1 + x^(-1)))/Sqrt[2*(-1 + Sqrt[5])]])/Sqrt[ 
-1 + Sqrt[5]] + ((1 - Sqrt[5])*Log[Sqrt[5] + Sqrt[2*(1 + Sqrt[5])]*(1 + x^ 
(-1)) + (1 + x^(-1))^2])/2)/(2*Sqrt[10*(1 + Sqrt[5])])
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1432
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

rule 1483
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]}, Simp[1/(2*c*q*r)   In 
t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[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] && N 
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
 

rule 2202
Int[(Pn_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{n 
 = Expon[Pn, x], k}, Int[Sum[Coeff[Pn, x, 2*k]*x^(2*k), {k, 0, n/2}]*(a + b 
*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pn, x, 2*k + 1]*x^(2*k), {k, 0, (n - 
1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pn, x] 
 &&  !PolyQ[Pn, x^2]
 

rule 2504
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]}, Simp[-16* 
a^2   Subst[Int[(1/(b - 4*a*x)^2)*(a*((-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 2 
56*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, 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 
]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.22

method result size
default \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}+2 \textit {\_R} +1}\right )}{4}\) \(41\)
risch \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}+2 \textit {\_R} +1}\right )}{4}\) \(41\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.15 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\frac {1}{2} \, {\left ({\left (\sqrt {5} + 2\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} {\left (2 \, x + 1\right )} + \frac {1}{2} \, {\left (\sqrt {5} {\left (8 \, x + 3\right )} + 20 \, x + 5\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} + x + \frac {1}{2}\right ) + \frac {1}{2} \, {\left ({\left (\sqrt {5} + 2\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} - 1\right )} \arctan \left (-\frac {1}{2} \, \sqrt {5} {\left (2 \, x + 1\right )} + \frac {1}{2} \, {\left (\sqrt {5} {\left (8 \, x + 3\right )} + 20 \, x + 5\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} - x - \frac {1}{2}\right ) + \frac {1}{4} \, \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} \log \left (4 \, x^{2} + {\left (\sqrt {5} {\left (2 \, x - 3\right )} + 10 \, x - 5\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} + \sqrt {5} + 1\right ) - \frac {1}{4} \, \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} \log \left (4 \, x^{2} - {\left (\sqrt {5} {\left (2 \, x - 3\right )} + 10 \, x - 5\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {2}{5}} + \sqrt {5} + 1\right ) \] Input:

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

Output:

1/2*((sqrt(5) + 2)*sqrt(1/5*sqrt(5) - 2/5) + 1)*arctan(1/2*sqrt(5)*(2*x + 
1) + 1/2*(sqrt(5)*(8*x + 3) + 20*x + 5)*sqrt(1/5*sqrt(5) - 2/5) + x + 1/2) 
 + 1/2*((sqrt(5) + 2)*sqrt(1/5*sqrt(5) - 2/5) - 1)*arctan(-1/2*sqrt(5)*(2* 
x + 1) + 1/2*(sqrt(5)*(8*x + 3) + 20*x + 5)*sqrt(1/5*sqrt(5) - 2/5) - x - 
1/2) + 1/4*sqrt(1/5*sqrt(5) - 2/5)*log(4*x^2 + (sqrt(5)*(2*x - 3) + 10*x - 
 5)*sqrt(1/5*sqrt(5) - 2/5) + sqrt(5) + 1) - 1/4*sqrt(1/5*sqrt(5) - 2/5)*l 
og(4*x^2 - (sqrt(5)*(2*x - 3) + 10*x - 5)*sqrt(1/5*sqrt(5) - 2/5) + sqrt(5 
) + 1)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3432 vs. \(2 (153) = 306\).

Time = 1.51 (sec) , antiderivative size = 3432, normalized size of antiderivative = 18.55 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

sqrt(-1/40 + sqrt(5)/80)*log(x**2 + x*(-8 - 21*sqrt(5)*sqrt(-2 + sqrt(5))/ 
10 - sqrt(-2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/2 - sqrt(5)/2 + 12 
*sqrt(-2 + sqrt(5)) + 9*sqrt(5)*sqrt(-2 + sqrt(5))*sqrt(-2*sqrt(5)*sqrt(-2 
 + sqrt(5)) + sqrt(5) + 19)/5) - 841*sqrt(5)*sqrt(-2*sqrt(5)*sqrt(-2 + sqr 
t(5)) + sqrt(5) + 19)/20 - 14351/40 - 441*sqrt(-2 + sqrt(5))/4 - 75*sqrt(5 
)*sqrt(-2 + sqrt(5))*sqrt(-2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/8 
- 3*sqrt(-2 + sqrt(5))*sqrt(-2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19) 
+ 301*sqrt(5)*sqrt(-2 + sqrt(5))/10 + 7407*sqrt(5)/40 + 3913*sqrt(-2*sqrt( 
5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/40) - sqrt(-1/40 + sqrt(5)/80)*log(x 
**2 + x*(-8 - 12*sqrt(-2 + sqrt(5)) - sqrt(5)/2 + 21*sqrt(5)*sqrt(-2 + sqr 
t(5))/10 + sqrt(2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/2 + 9*sqrt(5) 
*sqrt(-2 + sqrt(5))*sqrt(2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/5) - 
 3913*sqrt(2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)/40 - 14351/40 - 75 
*sqrt(5)*sqrt(-2 + sqrt(5))*sqrt(2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 
19)/8 - 301*sqrt(5)*sqrt(-2 + sqrt(5))/10 - 3*sqrt(-2 + sqrt(5))*sqrt(2*sq 
rt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19) + 441*sqrt(-2 + sqrt(5))/4 + 7407 
*sqrt(5)/40 + 841*sqrt(5)*sqrt(2*sqrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19 
)/20) - 2*sqrt(3/80 + 3*sqrt(5)/80 + sqrt(-2*sqrt(5)*sqrt(-2 + sqrt(5)) + 
sqrt(5) + 19)/40)*atan(-20*x/(-27*sqrt(5)*sqrt(3 + 3*sqrt(5) + 2*sqrt(-2*s 
qrt(5)*sqrt(-2 + sqrt(5)) + sqrt(5) + 19)) + 5*sqrt(-2 + sqrt(5))*sqrt(...
 

Maxima [F]

\[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\int { \frac {1}{4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 358 vs. \(2 (134) = 268\).

Time = 0.15 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.94 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\frac {1}{10} \, {\left (\frac {\sqrt {5 \, \sqrt {5} - 10}}{\sqrt {5} - 2} + 5\right )} {\left (\arctan \left (\frac {5}{3}\right ) + \arctan \left (x {\left (2 \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} + \sqrt {5} - 4 \, \sqrt {17 \, \sqrt {5} + 38} + 1\right )} - \frac {3}{2} \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} + \frac {1}{2} \, \sqrt {5} + \frac {7}{2} \, \sqrt {17 \, \sqrt {5} + 38} + \frac {1}{2}\right )\right )} - \frac {1}{10} \, {\left (\frac {\sqrt {5 \, \sqrt {5} - 10}}{\sqrt {5} - 2} - 5\right )} {\left (\arctan \left (\frac {5}{3}\right ) + \arctan \left (-x {\left (2 \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} - \sqrt {5} - 4 \, \sqrt {17 \, \sqrt {5} + 38} - 1\right )} + \frac {3}{2} \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} + \frac {1}{2} \, \sqrt {5} - \frac {7}{2} \, \sqrt {17 \, \sqrt {5} + 38} + \frac {1}{2}\right )\right )} + \frac {1}{20} \, \sqrt {5 \, \sqrt {5} - 10} \log \left (25 \, {\left (170 \, \sqrt {5} x + 380 \, x + 51 \, \sqrt {5} + \sqrt {17 \, \sqrt {5} + 38} + 114\right )}^{2} + 25 \, {\left (102 \, \sqrt {5} x + 228 \, x + 17 \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} - 85 \, \sqrt {5} + 38 \, \sqrt {17 \, \sqrt {5} + 38} - 190\right )}^{2}\right ) - \frac {1}{20} \, \sqrt {5 \, \sqrt {5} - 10} \log \left (25 \, {\left (170 \, \sqrt {5} x + 380 \, x + 51 \, \sqrt {5} - \sqrt {17 \, \sqrt {5} + 38} + 114\right )}^{2} + 25 \, {\left (102 \, \sqrt {5} x + 228 \, x - 17 \, \sqrt {5} \sqrt {17 \, \sqrt {5} + 38} - 85 \, \sqrt {5} - 38 \, \sqrt {17 \, \sqrt {5} + 38} - 190\right )}^{2}\right ) \] Input:

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

Output:

1/10*(sqrt(5*sqrt(5) - 10)/(sqrt(5) - 2) + 5)*(arctan(5/3) + arctan(x*(2*s 
qrt(5)*sqrt(17*sqrt(5) + 38) + sqrt(5) - 4*sqrt(17*sqrt(5) + 38) + 1) - 3/ 
2*sqrt(5)*sqrt(17*sqrt(5) + 38) + 1/2*sqrt(5) + 7/2*sqrt(17*sqrt(5) + 38) 
+ 1/2)) - 1/10*(sqrt(5*sqrt(5) - 10)/(sqrt(5) - 2) - 5)*(arctan(5/3) + arc 
tan(-x*(2*sqrt(5)*sqrt(17*sqrt(5) + 38) - sqrt(5) - 4*sqrt(17*sqrt(5) + 38 
) - 1) + 3/2*sqrt(5)*sqrt(17*sqrt(5) + 38) + 1/2*sqrt(5) - 7/2*sqrt(17*sqr 
t(5) + 38) + 1/2)) + 1/20*sqrt(5*sqrt(5) - 10)*log(25*(170*sqrt(5)*x + 380 
*x + 51*sqrt(5) + sqrt(17*sqrt(5) + 38) + 114)^2 + 25*(102*sqrt(5)*x + 228 
*x + 17*sqrt(5)*sqrt(17*sqrt(5) + 38) - 85*sqrt(5) + 38*sqrt(17*sqrt(5) + 
38) - 190)^2) - 1/20*sqrt(5*sqrt(5) - 10)*log(25*(170*sqrt(5)*x + 380*x + 
51*sqrt(5) - sqrt(17*sqrt(5) + 38) + 114)^2 + 25*(102*sqrt(5)*x + 228*x - 
17*sqrt(5)*sqrt(17*sqrt(5) + 38) - 85*sqrt(5) - 38*sqrt(17*sqrt(5) + 38) - 
 190)^2)
 

Mupad [B] (verification not implemented)

Time = 21.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.47 \[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\sum _{k=1}^4\ln \left (-\mathrm {root}\left (z^4+\frac {9\,z^2}{40}+\frac {z}{40}+\frac {1}{1280},z,k\right )\,\left (\frac {x}{4}+\mathrm {root}\left (z^4+\frac {9\,z^2}{40}+\frac {z}{40}+\frac {1}{1280},z,k\right )\,\left (6\,x+\mathrm {root}\left (z^4+\frac {9\,z^2}{40}+\frac {z}{40}+\frac {1}{1280},z,k\right )\,\left (36\,x+16\right )\right )\right )\right )\,\mathrm {root}\left (z^4+\frac {9\,z^2}{40}+\frac {z}{40}+\frac {1}{1280},z,k\right ) \] Input:

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

Output:

symsum(log(-root(z^4 + (9*z^2)/40 + z/40 + 1/1280, z, k)*(x/4 + root(z^4 + 
 (9*z^2)/40 + z/40 + 1/1280, z, k)*(6*x + root(z^4 + (9*z^2)/40 + z/40 + 1 
/1280, z, k)*(36*x + 16))))*root(z^4 + (9*z^2)/40 + z/40 + 1/1280, z, k), 
k, 1, 4)
 

Reduce [F]

\[ \int \frac {1}{1+4 x+4 x^2+4 x^4} \, dx=\int \frac {1}{4 x^{4}+4 x^{2}+4 x +1}d x \] Input:

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

Output:

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