\(\int \frac {x^4}{4+(1+x^2)^4} \, dx\) [152]

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

Optimal result

Integrand size = 15, antiderivative size = 167 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=-\frac {1}{16} \sqrt {-11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )}-2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {1}{16} \sqrt {-11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )}+2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}}+\frac {1}{16} \sqrt {11+5 \sqrt {5}} \text {arctanh}\left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )} x}{\sqrt {5}+x^2}\right ) \] Output:

-1/16*(-11+5*5^(1/2))^(1/2)*arctan(((-4+2*5^(1/2))^(1/2)-2*x)/(4+2*5^(1/2) 
)^(1/2))+1/16*(-11+5*5^(1/2))^(1/2)*arctan(((-4+2*5^(1/2))^(1/2)+2*x)/(4+2 
*5^(1/2))^(1/2))-1/16*arctanh(2^(1/2)*x/(x^2+1))*2^(1/2)+1/16*(11+5*5^(1/2 
))^(1/2)*arctanh((-4+2*5^(1/2))^(1/2)*x/(5^(1/2)+x^2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.50 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=\frac {1}{32} \left ((2-6 i) \sqrt {2-i} \arctan \left (\frac {x}{\sqrt {2-i}}\right )+(2+6 i) \sqrt {2+i} \arctan \left (\frac {x}{\sqrt {2+i}}\right )+\sqrt {2} \left (\log \left (-1+\sqrt {2} x-x^2\right )-\log \left (1+\sqrt {2} x+x^2\right )\right )\right ) \] Input:

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

Output:

((2 - 6*I)*Sqrt[2 - I]*ArcTan[x/Sqrt[2 - I]] + (2 + 6*I)*Sqrt[2 + I]*ArcTa 
n[x/Sqrt[2 + I]] + Sqrt[2]*(Log[-1 + Sqrt[2]*x - x^2] - Log[1 + Sqrt[2]*x 
+ x^2]))/32
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.37, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2460, 2009}

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.

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

\(\Big \downarrow \) 2460

\(\displaystyle \int \left (\frac {5-x^2}{8 \left (x^4+4 x^2+5\right )}+\frac {x^2-1}{8 \left (x^4+1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{16} \sqrt {5 \sqrt {5}-11} \arctan \left (\frac {\sqrt {2 \left (\sqrt {5}-2\right )}-2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {1}{16} \sqrt {5 \sqrt {5}-11} \arctan \left (\frac {2 x+\sqrt {2 \left (\sqrt {5}-2\right )}}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {\log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {1}{32} \sqrt {11+5 \sqrt {5}} \log \left (x^2-\sqrt {2 \left (\sqrt {5}-2\right )} x+\sqrt {5}\right )+\frac {1}{32} \sqrt {11+5 \sqrt {5}} \log \left (x^2+\sqrt {2 \left (\sqrt {5}-2\right )} x+\sqrt {5}\right )\)

Input:

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

Output:

-1/16*(Sqrt[-11 + 5*Sqrt[5]]*ArcTan[(Sqrt[2*(-2 + Sqrt[5])] - 2*x)/Sqrt[2* 
(2 + Sqrt[5])]]) + (Sqrt[-11 + 5*Sqrt[5]]*ArcTan[(Sqrt[2*(-2 + Sqrt[5])] + 
 2*x)/Sqrt[2*(2 + Sqrt[5])]])/16 + Log[1 - Sqrt[2]*x + x^2]/(16*Sqrt[2]) - 
 Log[1 + Sqrt[2]*x + x^2]/(16*Sqrt[2]) - (Sqrt[11 + 5*Sqrt[5]]*Log[Sqrt[5] 
 - Sqrt[2*(-2 + Sqrt[5])]*x + x^2])/32 + (Sqrt[11 + 5*Sqrt[5]]*Log[Sqrt[5] 
 + Sqrt[2*(-2 + Sqrt[5])]*x + x^2])/32
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2460
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, 
Int[ExpandIntegrand[u*(Qx /. x -> x^2)^p, x], x] /;  !SumQ[NonfreeFactors[Q 
x, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] &&  !BinomialQ[Px, x] && 
 !TrinomialQ[Px, x] && ILtQ[p, 0] && RationalFunctionQ[u, x]
 
Maple [C] (verified)

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

Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.41

method result size
risch \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}-44 \textit {\_Z}^{2}+125\right )}{\sum }\textit {\_R} \ln \left (6 \textit {\_R}^{3}-31 \textit {\_R} +10 x \right )\right )}{16}+\frac {\sqrt {2}\, \ln \left (x^{2}-x \sqrt {2}+1\right )}{32}-\frac {\sqrt {2}\, \ln \left (x^{2}+x \sqrt {2}+1\right )}{32}\) \(68\)
default \(-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}+x \sqrt {2}+1}{x^{2}-x \sqrt {2}+1}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{64}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-x \sqrt {2}+1}{x^{2}+x \sqrt {2}+1}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{64}-\frac {\left (3 \sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+7 \sqrt {-4+2 \sqrt {5}}\right ) \ln \left (x^{2}-x \sqrt {-4+2 \sqrt {5}}+\sqrt {5}\right )}{64}-\frac {\left (-2 \sqrt {5}+\frac {\left (3 \sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+7 \sqrt {-4+2 \sqrt {5}}\right ) \sqrt {-4+2 \sqrt {5}}}{2}\right ) \arctan \left (\frac {2 x -\sqrt {-4+2 \sqrt {5}}}{\sqrt {4+2 \sqrt {5}}}\right )}{16 \sqrt {4+2 \sqrt {5}}}+\frac {\left (3 \sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+7 \sqrt {-4+2 \sqrt {5}}\right ) \ln \left (x^{2}+x \sqrt {-4+2 \sqrt {5}}+\sqrt {5}\right )}{64}+\frac {\left (2 \sqrt {5}-\frac {\left (3 \sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+7 \sqrt {-4+2 \sqrt {5}}\right ) \sqrt {-4+2 \sqrt {5}}}{2}\right ) \arctan \left (\frac {\sqrt {-4+2 \sqrt {5}}+2 x}{\sqrt {4+2 \sqrt {5}}}\right )}{16 \sqrt {4+2 \sqrt {5}}}\) \(357\)

Input:

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

Output:

1/16*sum(_R*ln(6*_R^3-31*_R+10*x),_R=RootOf(4*_Z^4-44*_Z^2+125))+1/32*2^(1 
/2)*ln(x^2-x*2^(1/2)+1)-1/32*2^(1/2)*ln(x^2+x*2^(1/2)+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.40 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=\frac {1}{16} \, \sqrt {5 \, \sqrt {5} - 11} \arctan \left (\frac {1}{2} \, \sqrt {5 \, \sqrt {5} + 11} \sqrt {5 \, \sqrt {5} - 11} {\left (\sqrt {5} - 2\right )} + \frac {1}{2} \, {\left (\sqrt {5} x + x\right )} \sqrt {5 \, \sqrt {5} - 11}\right ) - \frac {1}{16} \, \sqrt {5 \, \sqrt {5} - 11} \arctan \left (\frac {1}{2} \, \sqrt {5 \, \sqrt {5} + 11} \sqrt {5 \, \sqrt {5} - 11} {\left (\sqrt {5} - 2\right )} - \frac {1}{2} \, {\left (\sqrt {5} x + x\right )} \sqrt {5 \, \sqrt {5} - 11}\right ) - \frac {1}{32} \, \sqrt {5 \, \sqrt {5} + 11} \log \left (2 \, x^{2} + {\left (3 \, \sqrt {5} x - 7 \, x\right )} \sqrt {5 \, \sqrt {5} + 11} + 2 \, \sqrt {5}\right ) + \frac {1}{32} \, \sqrt {5 \, \sqrt {5} + 11} \log \left (2 \, x^{2} - {\left (3 \, \sqrt {5} x - 7 \, x\right )} \sqrt {5 \, \sqrt {5} + 11} + 2 \, \sqrt {5}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} - 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \] Input:

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

Output:

1/16*sqrt(5*sqrt(5) - 11)*arctan(1/2*sqrt(5*sqrt(5) + 11)*sqrt(5*sqrt(5) - 
 11)*(sqrt(5) - 2) + 1/2*(sqrt(5)*x + x)*sqrt(5*sqrt(5) - 11)) - 1/16*sqrt 
(5*sqrt(5) - 11)*arctan(1/2*sqrt(5*sqrt(5) + 11)*sqrt(5*sqrt(5) - 11)*(sqr 
t(5) - 2) - 1/2*(sqrt(5)*x + x)*sqrt(5*sqrt(5) - 11)) - 1/32*sqrt(5*sqrt(5 
) + 11)*log(2*x^2 + (3*sqrt(5)*x - 7*x)*sqrt(5*sqrt(5) + 11) + 2*sqrt(5)) 
+ 1/32*sqrt(5*sqrt(5) + 11)*log(2*x^2 - (3*sqrt(5)*x - 7*x)*sqrt(5*sqrt(5) 
 + 11) + 2*sqrt(5)) + 1/32*sqrt(2)*log((x^4 + 4*x^2 - 2*sqrt(2)*(x^3 + x) 
+ 1)/(x^4 + 1))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1182 vs. \(2 (141) = 282\).

Time = 0.68 (sec) , antiderivative size = 1182, normalized size of antiderivative = 7.08 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx =\text {Too large to display} \] Input:

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

Output:

sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - sqrt(2)*log(x**2 + sqrt(2)*x + 1)/3 
2 + sqrt(11/1024 + 5*sqrt(5)/1024)*log(x**2 + x*(-6*sqrt(5)*sqrt(11 + 5*sq 
rt(5)) - 32*sqrt(11 + 5*sqrt(5))/5 + 9*sqrt(2)*sqrt(11 + 5*sqrt(5))*sqrt(5 
5*sqrt(5) + 123)/10) - 5497*sqrt(2)*sqrt(55*sqrt(5) + 123)/50 - 243*sqrt(1 
0)*sqrt(55*sqrt(5) + 123)/5 + 121217/50 + 10853*sqrt(5)/10) - sqrt(11/1024 
 + 5*sqrt(5)/1024)*log(x**2 + x*(-9*sqrt(2)*sqrt(11 + 5*sqrt(5))*sqrt(55*s 
qrt(5) + 123)/10 + 32*sqrt(11 + 5*sqrt(5))/5 + 6*sqrt(5)*sqrt(11 + 5*sqrt( 
5))) - 5497*sqrt(2)*sqrt(55*sqrt(5) + 123)/50 - 243*sqrt(10)*sqrt(55*sqrt( 
5) + 123)/5 + 121217/50 + 10853*sqrt(5)/10) + 2*sqrt(-sqrt(2)*sqrt(55*sqrt 
(5) + 123)/512 + 11/1024 + 15*sqrt(5)/1024)*atan(20*x/(2*sqrt(-2*sqrt(2)*s 
qrt(55*sqrt(5) + 123) + 11 + 15*sqrt(5)) + 3*sqrt(2)*sqrt(55*sqrt(5) + 123 
)*sqrt(-2*sqrt(2)*sqrt(55*sqrt(5) + 123) + 11 + 15*sqrt(5))) - 60*sqrt(5)* 
sqrt(11 + 5*sqrt(5))/(2*sqrt(-2*sqrt(2)*sqrt(55*sqrt(5) + 123) + 11 + 15*s 
qrt(5)) + 3*sqrt(2)*sqrt(55*sqrt(5) + 123)*sqrt(-2*sqrt(2)*sqrt(55*sqrt(5) 
 + 123) + 11 + 15*sqrt(5))) - 64*sqrt(11 + 5*sqrt(5))/(2*sqrt(-2*sqrt(2)*s 
qrt(55*sqrt(5) + 123) + 11 + 15*sqrt(5)) + 3*sqrt(2)*sqrt(55*sqrt(5) + 123 
)*sqrt(-2*sqrt(2)*sqrt(55*sqrt(5) + 123) + 11 + 15*sqrt(5))) + 9*sqrt(2)*s 
qrt(11 + 5*sqrt(5))*sqrt(55*sqrt(5) + 123)/(2*sqrt(-2*sqrt(2)*sqrt(55*sqrt 
(5) + 123) + 11 + 15*sqrt(5)) + 3*sqrt(2)*sqrt(55*sqrt(5) + 123)*sqrt(-2*s 
qrt(2)*sqrt(55*sqrt(5) + 123) + 11 + 15*sqrt(5)))) + 2*sqrt(-sqrt(2)*sq...
 

Maxima [F]

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

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

Output:

-1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/32*sqrt(2)*log(x^2 - sqrt(2)*x 
+ 1) - 1/8*integrate((x^2 - 5)/(x^4 + 4*x^2 + 5), x)
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.11 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=\frac {1}{80} \, \sqrt {125 \, \sqrt {5} + 29} \arctan \left (\frac {5^{\frac {3}{4}} {\left (x + 5^{\frac {1}{4}} \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}\right )}}{5 \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{80} \, \sqrt {125 \, \sqrt {5} + 29} \arctan \left (\frac {5^{\frac {3}{4}} {\left (x - 5^{\frac {1}{4}} \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}\right )}}{5 \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{160} \, \sqrt {125 \, \sqrt {5} - 29} \log \left (x^{2} + 2 \cdot 5^{\frac {1}{4}} x \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}} + \sqrt {5}\right ) + \frac {1}{160} \, \sqrt {125 \, \sqrt {5} - 29} \log \left (x^{2} - 2 \cdot 5^{\frac {1}{4}} x \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}} + \sqrt {5}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \] Input:

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

Output:

1/80*sqrt(125*sqrt(5) + 29)*arctan(1/5*5^(3/4)*(x + 5^(1/4)*sqrt(-1/5*sqrt 
(5) + 1/2))/sqrt(1/5*sqrt(5) + 1/2)) + 1/80*sqrt(125*sqrt(5) + 29)*arctan( 
1/5*5^(3/4)*(x - 5^(1/4)*sqrt(-1/5*sqrt(5) + 1/2))/sqrt(1/5*sqrt(5) + 1/2) 
) - 1/160*sqrt(125*sqrt(5) - 29)*log(x^2 + 2*5^(1/4)*x*sqrt(-1/5*sqrt(5) + 
 1/2) + sqrt(5)) + 1/160*sqrt(125*sqrt(5) - 29)*log(x^2 - 2*5^(1/4)*x*sqrt 
(-1/5*sqrt(5) + 1/2) + sqrt(5)) - 1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 
1/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1)
 

Mupad [B] (verification not implemented)

Time = 10.41 (sec) , antiderivative size = 1222, normalized size of antiderivative = 7.32 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((2472*x - ((1474560*x - ((111149056*x - (2147483648*x*((11/1024 - (5 
*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2)) - 335544320)*( 
(11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2)))* 
((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2)) 
- 10485760)*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/1 
024)^(1/2)))*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/ 
1024)^(1/2)) - 70400)*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1 
024 + 11/1024)^(1/2)))*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/ 
1024 + 11/1024)^(1/2))*1i + (2472*x - ((1474560*x - ((111149056*x - (21474 
83648*x*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1024 + 11/1024) 
^(1/2)) + 335544320)*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/10 
24 + 11/1024)^(1/2)))*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^(1/2))/1 
024 + 11/1024)^(1/2)) + 10485760)*((11/1024 - (5*5^(1/2))/1024)^(1/2) + (( 
5*5^(1/2))/1024 + 11/1024)^(1/2)))*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ( 
(5*5^(1/2))/1024 + 11/1024)^(1/2)) + 70400)*((11/1024 - (5*5^(1/2))/1024)^ 
(1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2)))*((11/1024 - (5*5^(1/2))/1024) 
^(1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2))*1i)/((2472*x - ((1474560*x - 
((111149056*x - (2147483648*x*((11/1024 - (5*5^(1/2))/1024)^(1/2) + ((5*5^ 
(1/2))/1024 + 11/1024)^(1/2)) + 335544320)*((11/1024 - (5*5^(1/2))/1024)^( 
1/2) + ((5*5^(1/2))/1024 + 11/1024)^(1/2)))*((11/1024 - (5*5^(1/2))/102...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56 \[ \int \frac {x^4}{4+\left (1+x^2\right )^4} \, dx=\frac {\sqrt {2}\, \left (6 \sqrt {\sqrt {5}+2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}-2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-14 \sqrt {\sqrt {5}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}-2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-6 \sqrt {\sqrt {5}+2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}+2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )+14 \sqrt {\sqrt {5}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}+2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-3 \sqrt {\sqrt {5}-2}\, \sqrt {5}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )+3 \sqrt {\sqrt {5}-2}\, \sqrt {5}\, \mathrm {log}\left (\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )-7 \sqrt {\sqrt {5}-2}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )+7 \sqrt {\sqrt {5}-2}\, \mathrm {log}\left (\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )+2 \,\mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )-2 \,\mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )\right )}{64} \] Input:

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

Output:

(sqrt(2)*(6*sqrt(sqrt(5) + 2)*sqrt(5)*atan((sqrt(sqrt(5) - 2)*sqrt(2) - 2* 
x)/(sqrt(sqrt(5) + 2)*sqrt(2))) - 14*sqrt(sqrt(5) + 2)*atan((sqrt(sqrt(5) 
- 2)*sqrt(2) - 2*x)/(sqrt(sqrt(5) + 2)*sqrt(2))) - 6*sqrt(sqrt(5) + 2)*sqr 
t(5)*atan((sqrt(sqrt(5) - 2)*sqrt(2) + 2*x)/(sqrt(sqrt(5) + 2)*sqrt(2))) + 
 14*sqrt(sqrt(5) + 2)*atan((sqrt(sqrt(5) - 2)*sqrt(2) + 2*x)/(sqrt(sqrt(5) 
 + 2)*sqrt(2))) - 3*sqrt(sqrt(5) - 2)*sqrt(5)*log( - sqrt(sqrt(5) - 2)*sqr 
t(2)*x + sqrt(5) + x**2) + 3*sqrt(sqrt(5) - 2)*sqrt(5)*log(sqrt(sqrt(5) - 
2)*sqrt(2)*x + sqrt(5) + x**2) - 7*sqrt(sqrt(5) - 2)*log( - sqrt(sqrt(5) - 
 2)*sqrt(2)*x + sqrt(5) + x**2) + 7*sqrt(sqrt(5) - 2)*log(sqrt(sqrt(5) - 2 
)*sqrt(2)*x + sqrt(5) + x**2) + 2*log( - sqrt(2)*x + x**2 + 1) - 2*log(sqr 
t(2)*x + x**2 + 1)))/64