\(\int \frac {x^4}{(1-x^4) (1+x^4)^2} \, dx\) [451]

Optimal result

Integrand size = 20, antiderivative size = 88 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=-\frac {x}{8 \left (1+x^4\right )}+\frac {\arctan (x)}{8}+\frac {\arctan \left (1-\sqrt {2} x\right )}{16 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} x\right )}{16 \sqrt {2}}+\frac {\text {arctanh}(x)}{8}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{16 \sqrt {2}} \] Output:

-1/8*x/(x^4+1)+1/8*arctan(x)-1/32*arctan(-1+x*2^(1/2))*2^(1/2)-1/32*arctan 
(1+x*2^(1/2))*2^(1/2)+1/8*arctanh(x)-1/32*arctanh(2^(1/2)*x/(x^2+1))*2^(1/ 
2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=\frac {1}{64} \left (-\frac {8 x}{1+x^4}+8 \arctan (x)+2 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )-4 \log (1-x)+4 \log (1+x)+\sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-\sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:

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

Output:

((-8*x)/(1 + x^4) + 8*ArcTan[x] + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 2*Sqrt 
[2]*ArcTan[1 + Sqrt[2]*x] - 4*Log[1 - x] + 4*Log[1 + x] + Sqrt[2]*Log[1 - 
Sqrt[2]*x + x^2] - Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/64
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.39, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {971, 1020, 755, 756, 216, 219, 1476, 1082, 217, 1479, 25, 27, 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.

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

\(\Big \downarrow \) 971

\(\displaystyle \frac {1}{8} \int \frac {3 x^4+1}{\left (1-x^4\right ) \left (x^4+1\right )}dx-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 1020

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{8} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{8} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{8} \left (2 \left (\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {x}{8 \left (x^4+1\right )}\)

Input:

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

Output:

-1/8*x/(1 + x^4) + ((ArcTan[1 - Sqrt[2]*x]/Sqrt[2] - ArcTan[1 + Sqrt[2]*x] 
/Sqrt[2])/2 + 2*(ArcTan[x]/2 + ArcTanh[x]/2) + (Log[1 - Sqrt[2]*x + x^2]/( 
2*Sqrt[2]) - Log[1 + Sqrt[2]*x + x^2]/(2*Sqrt[2]))/2)/8
 

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 216

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

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 219

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

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 971

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) 
*(p + 1))   Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - 
 n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e 
, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 
 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1020

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^( 
n_))), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^n), x], x 
] - Simp[(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b 
, c, d, e, f, n}, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{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 1476

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

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

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

Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.53

method	result	size
risch	\(-\frac {x}{8 \left (x^{4}+1\right )}+\frac {\ln \left (x +1\right )}{16}+\frac {\arctan \left (x \right )}{8}-\frac {\ln \left (x -1\right )}{16}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{32}\)	\(47\)
default	\(-\frac {x}{8 \left (x^{4}+1\right )}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}+x \sqrt {2}+1}{x^{2}-x \sqrt {2}+1}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{64}+\frac {\arctan \left (x \right )}{8}+\frac {\ln \left (x +1\right )}{16}-\frac {\ln \left (x -1\right )}{16}\)	\(79\)

Input:

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

Output:

-1/8*x/(x^4+1)+1/16*ln(x+1)+1/8*arctan(x)-1/16*ln(x-1)+1/32*sum(_R*ln(x-_R 
),_R=RootOf(_Z^4+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.39 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x + 1\right ) + 2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x - 1\right ) + \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) - 8 \, {\left (x^{4} + 1\right )} \arctan \left (x\right ) - 4 \, {\left (x^{4} + 1\right )} \log \left (x + 1\right ) + 4 \, {\left (x^{4} + 1\right )} \log \left (x - 1\right ) + 8 \, x}{64 \, {\left (x^{4} + 1\right )}} \] Input:

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

Output:

-1/64*(2*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x + 1) + 2*sqrt(2)*(x^4 + 1)*arc 
tan(sqrt(2)*x - 1) + sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x + 1) - sqrt(2)* 
(x^4 + 1)*log(x^2 - sqrt(2)*x + 1) - 8*(x^4 + 1)*arctan(x) - 4*(x^4 + 1)*l 
og(x + 1) + 4*(x^4 + 1)*log(x - 1) + 8*x)/(x^4 + 1)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 145.63 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=- \frac {x}{8 x^{4} + 8} - \frac {\log {\left (x - 1 \right )}}{16} + \frac {\log {\left (x + 1 \right )}}{16} - \frac {i \log {\left (x - i \right )}}{16} + \frac {i \log {\left (x + i \right )}}{16} - \operatorname {RootSum} {\left (1048576 t^{4} + 1, \left ( t \mapsto t \log {\left (- \frac {50331648 t^{5}}{17} + \frac {496 t}{17} + x \right )} \right )\right )} \] Input:

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

Output:

-x/(8*x**4 + 8) - log(x - 1)/16 + log(x + 1)/16 - I*log(x - I)/16 + I*log( 
x + I)/16 - RootSum(1048576*_t**4 + 1, Lambda(_t, _t*log(-50331648*_t**5/1 
7 + 496*_t/17 + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.11 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=-\frac {1}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{64} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{64} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{8 \, {\left (x^{4} + 1\right )}} + \frac {1}{8} \, \arctan \left (x\right ) + \frac {1}{16} \, \log \left (x + 1\right ) - \frac {1}{16} \, \log \left (x - 1\right ) \] Input:

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

Output:

-1/32*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/32*sqrt(2)*arctan(1/ 
2*sqrt(2)*(2*x - sqrt(2))) - 1/64*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/64* 
sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/8*x/(x^4 + 1) + 1/8*arctan(x) + 1/16* 
log(x + 1) - 1/16*log(x - 1)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.14 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=-\frac {1}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{64} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{64} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{8 \, {\left (x^{4} + 1\right )}} + \frac {1}{8} \, \arctan \left (x\right ) + \frac {1}{16} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{16} \, \log \left ({\left | x - 1 \right |}\right ) \] Input:

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

Output:

-1/32*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/32*sqrt(2)*arctan(1/ 
2*sqrt(2)*(2*x - sqrt(2))) - 1/64*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/64* 
sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/8*x/(x^4 + 1) + 1/8*arctan(x) + 1/16* 
log(abs(x + 1)) - 1/16*log(abs(x - 1))
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.53 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=\frac {\mathrm {atan}\left (x\right )}{8}-\frac {x}{8\,\left (x^4+1\right )}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,x\right )\,1{}\mathrm {i}}{16}-\frac {{\left (-1\right )}^{3/4}\,\mathrm {atan}\left ({\left (-1\right )}^{3/4}\,x\right )\,1{}\mathrm {i}}{16} \] Input:

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

Output:

atan(x)/8 - (atan(x*1i)*1i)/8 + ((-1)^(1/4)*atan((-1)^(1/4)*x)*1i)/16 - (( 
-1)^(3/4)*atan((-1)^(3/4)*x)*1i)/16 - x/(8*(x^4 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.12 \[ \int \frac {x^4}{\left (1-x^4\right ) \left (1+x^4\right )^2} \, dx=\frac {2 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{4}+2 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right )-2 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{4}-2 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right )+8 \mathit {atan} \left (x \right ) x^{4}+8 \mathit {atan} \left (x \right )+\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{4}+\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )-\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{4}-\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )-4 \,\mathrm {log}\left (x -1\right ) x^{4}-4 \,\mathrm {log}\left (x -1\right )+4 \,\mathrm {log}\left (x +1\right ) x^{4}+4 \,\mathrm {log}\left (x +1\right )-8 x}{64 x^{4}+64} \] Input:

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

Output:

(2*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**4 + 2*sqrt(2)*atan((sqrt(2) - 
2*x)/sqrt(2)) - 2*sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**4 - 2*sqrt(2)*a 
tan((sqrt(2) + 2*x)/sqrt(2)) + 8*atan(x)*x**4 + 8*atan(x) + sqrt(2)*log( - 
 sqrt(2)*x + x**2 + 1)*x**4 + sqrt(2)*log( - sqrt(2)*x + x**2 + 1) - sqrt( 
2)*log(sqrt(2)*x + x**2 + 1)*x**4 - sqrt(2)*log(sqrt(2)*x + x**2 + 1) - 4* 
log(x - 1)*x**4 - 4*log(x - 1) + 4*log(x + 1)*x**4 + 4*log(x + 1) - 8*x)/( 
64*(x**4 + 1))