\(\int \frac {x^8}{1+2 x^4+x^8} \, dx\) [32]

Optimal result

Integrand size = 16, antiderivative size = 77 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=x+\frac {x}{4 \left (1+x^4\right )}+\frac {5 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=\frac {1}{32} \left (32 x+\frac {8 x}{1+x^4}+10 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-10 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+5 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-5 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:

Integrate[x^8/(1 + 2*x^4 + x^8),x]
 

Output:

(32*x + (8*x)/(1 + x^4) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 10*Sqrt[2]*Ar 
cTan[1 + Sqrt[2]*x] + 5*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 5*Sqrt[2]*Log[1 
 + Sqrt[2]*x + x^2])/32
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.43, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {1380, 817, 843, 755, 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.

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

\(\Big \downarrow \) 1380

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[x^8/(1 + 2*x^4 + x^8),x]
 

Output:

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

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 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 817

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

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, 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 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

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 = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.44

method	result	size
risch	\(x +\frac {x}{4 x^{4}+4}-\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16}\)	\(34\)
default	\(x +\frac {x}{4 x^{4}+4}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}+x \sqrt {2}+1}{x^{2}-x \sqrt {2}+1}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{32}\)	\(64\)

Input:

int(x^8/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)
 

Output:

x+1/4*x/(x^4+1)-5/16*sum(1/_R^3*ln(x-_R),_R=RootOf(_Z^4+1))
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=\frac {32 \, x^{5} - 10 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x + 1\right ) - 10 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x - 1\right ) - 5 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) + 5 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + 40 \, x}{32 \, {\left (x^{4} + 1\right )}} \] Input:

integrate(x^8/(x^8+2*x^4+1),x, algorithm="fricas")
 

Output:

1/32*(32*x^5 - 10*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x + 1) - 10*sqrt(2)*(x^ 
4 + 1)*arctan(sqrt(2)*x - 1) - 5*sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x + 1 
) + 5*sqrt(2)*(x^4 + 1)*log(x^2 - sqrt(2)*x + 1) + 40*x)/(x^4 + 1)
 

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=x + \frac {x}{4 x^{4} + 4} + \frac {5 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \] Input:

integrate(x**8/(x**8+2*x**4+1),x)
 

Output:

x + x/(4*x**4 + 4) + 5*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 5*sqrt(2)*lo 
g(x**2 + sqrt(2)*x + 1)/32 - 5*sqrt(2)*atan(sqrt(2)*x - 1)/16 - 5*sqrt(2)* 
atan(sqrt(2)*x + 1)/16
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + x + \frac {x}{4 \, {\left (x^{4} + 1\right )}} \] Input:

integrate(x^8/(x^8+2*x^4+1),x, algorithm="maxima")
 

Output:

-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/ 
2*sqrt(2)*(2*x - sqrt(2))) - 5/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 5/32* 
sqrt(2)*log(x^2 - sqrt(2)*x + 1) + x + 1/4*x/(x^4 + 1)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + x + \frac {x}{4 \, {\left (x^{4} + 1\right )}} \] Input:

integrate(x^8/(x^8+2*x^4+1),x, algorithm="giac")
 

Output:

-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/ 
2*sqrt(2)*(2*x - sqrt(2))) - 5/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 5/32* 
sqrt(2)*log(x^2 - sqrt(2)*x + 1) + x + 1/4*x/(x^4 + 1)
 

Mupad [B] (verification not implemented)

Time = 14.64 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.58 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=x+\frac {x}{4\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}-\frac {5}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}+\frac {5}{16}{}\mathrm {i}\right ) \] Input:

int(x^8/(2*x^4 + x^8 + 1),x)
 

Output:

x - 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(5/16 + 5i/16) - 2^(1/2)*atan(2^( 
1/2)*x*(1/2 + 1i/2))*(5/16 - 5i/16) + x/(4*(x^4 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.99 \[ \int \frac {x^8}{1+2 x^4+x^8} \, dx=\frac {10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{4}+10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right )-10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{4}-10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right )+5 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{4}+5 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )-5 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{4}-5 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )+32 x^{5}+40 x}{32 x^{4}+32} \] Input:

int(x^8/(x^8+2*x^4+1),x)
 

Output:

(10*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**4 + 10*sqrt(2)*atan((sqrt(2) 
- 2*x)/sqrt(2)) - 10*sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**4 - 10*sqrt( 
2)*atan((sqrt(2) + 2*x)/sqrt(2)) + 5*sqrt(2)*log( - sqrt(2)*x + x**2 + 1)* 
x**4 + 5*sqrt(2)*log( - sqrt(2)*x + x**2 + 1) - 5*sqrt(2)*log(sqrt(2)*x + 
x**2 + 1)*x**4 - 5*sqrt(2)*log(sqrt(2)*x + x**2 + 1) + 32*x**5 + 40*x)/(32 
*(x**4 + 1))