\(\int \frac {x}{1+x^8} \, dx\) [79]

Optimal result

Integrand size = 9, antiderivative size = 93 \[ \int \frac {x}{1+x^8} \, dx=-\frac {\arctan \left (1-\sqrt {2} x^2\right )}{4 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x^2+x^4\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x^2+x^4\right )}{8 \sqrt {2}} \] Output:

1/8*2^(1/2)*arctan(-1+x^2*2^(1/2))+1/8*2^(1/2)*arctan(1+x^2*2^(1/2))-1/16* 
ln(1-x^2*2^(1/2)+x^4)*2^(1/2)+1/16*ln(1+x^2*2^(1/2)+x^4)*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.60 \[ \int \frac {x}{1+x^8} \, dx=-\frac {2 \arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right )+2 \arctan \left (\cot \left (\frac {\pi }{8}\right )-x \csc \left (\frac {\pi }{8}\right )\right )+2 \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right )-2 \arctan \left (x \sec \left (\frac {\pi }{8}\right )-\tan \left (\frac {\pi }{8}\right )\right )+\log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right )+\log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right )-\log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right )-\log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2}} \] Input:

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

Output:

-1/8*(2*ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]] + 2*ArcTan[Cot[Pi/8] - x*Csc[Pi/ 
8]] + 2*ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])] - 2*ArcTan[x*Sec[Pi/8] - Tan[Pi/ 
8]] + Log[1 + x^2 - 2*x*Cos[Pi/8]] + Log[1 + x^2 + 2*x*Cos[Pi/8]] - Log[1 
+ x^2 - 2*x*Sin[Pi/8]] - Log[1 + x^2 + 2*x*Sin[Pi/8]])/Sqrt[2]
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {807, 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.

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

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 = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.22

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x^{2}+\textit {\_R} \right )\right )}{8}\)	\(20\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, x^{2}+x^{4}}{1-\sqrt {2}\, x^{2}+x^{4}}\right )+2 \arctan \left (1+\sqrt {2}\, x^{2}\right )+2 \arctan \left (-1+\sqrt {2}\, x^{2}\right )\right )}{16}\)	\(60\)
meijerg	\(-\frac {x^{2} \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{16 \left (x^{8}\right )^{\frac {1}{4}}}+\frac {x^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{8}\right )^{\frac {1}{4}}}+\frac {x^{2} \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{16 \left (x^{8}\right )^{\frac {1}{4}}}+\frac {x^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{8}\right )^{\frac {1}{4}}}\)	\(139\)

Input:

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

Output:

1/8*sum(_R*ln(x^2+_R),_R=RootOf(_Z^4+1))
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \frac {x}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\sqrt {2} x^{2} + 1\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\sqrt {2} x^{2} - 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {x}{1+x^8} \, dx=- \frac {\sqrt {2} \log {\left (x^{4} - \sqrt {2} x^{2} + 1 \right )}}{16} + \frac {\sqrt {2} \log {\left (x^{4} + \sqrt {2} x^{2} + 1 \right )}}{16} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} - 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} + 1 \right )}}{8} \] Input:

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

Output:

-sqrt(2)*log(x**4 - sqrt(2)*x**2 + 1)/16 + sqrt(2)*log(x**4 + sqrt(2)*x**2 
 + 1)/16 + sqrt(2)*atan(sqrt(2)*x**2 - 1)/8 + sqrt(2)*atan(sqrt(2)*x**2 + 
1)/8
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {x}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {x}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 10.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.40 \[ \int \frac {x}{1+x^8} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \] Input:

int(x/(x^8 + 1),x)
 

Output:

2^(1/2)*atan(2^(1/2)*x^2*(1/2 - 1i/2))*(1/8 + 1i/8) + 2^(1/2)*atan(2^(1/2) 
*x^2*(1/2 + 1i/2))*(1/8 - 1i/8)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.27 \[ \int \frac {x}{1+x^8} \, dx=-\frac {\sqrt {\sqrt {2}+2}\, \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}-2 x}{\sqrt {\sqrt {2}+2}}\right )}{8}-\frac {\sqrt {\sqrt {2}+2}\, \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}+2 x}{\sqrt {\sqrt {2}+2}}\right )}{8}-\frac {\sqrt {\sqrt {2}+2}\, \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}-2 x}{\sqrt {-\sqrt {2}+2}}\right )}{8}-\frac {\sqrt {\sqrt {2}+2}\, \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}+2 x}{\sqrt {-\sqrt {2}+2}}\right )}{8}+\frac {\sqrt {2}\, \mathrm {log}\left (-\sqrt {-\sqrt {2}+2}\, x +x^{2}+1\right )}{16}-\frac {\sqrt {2}\, \mathrm {log}\left (-\sqrt {\sqrt {2}+2}\, x +x^{2}+1\right )}{16}+\frac {\sqrt {2}\, \mathrm {log}\left (\sqrt {-\sqrt {2}+2}\, x +x^{2}+1\right )}{16}-\frac {\sqrt {2}\, \mathrm {log}\left (\sqrt {\sqrt {2}+2}\, x +x^{2}+1\right )}{16} \] Input:

int(x/(x^8+1),x)
 

Output:

( - 2*sqrt(sqrt(2) + 2)*sqrt( - sqrt(2) + 2)*atan((sqrt( - sqrt(2) + 2) - 
2*x)/sqrt(sqrt(2) + 2)) - 2*sqrt(sqrt(2) + 2)*sqrt( - sqrt(2) + 2)*atan((s 
qrt( - sqrt(2) + 2) + 2*x)/sqrt(sqrt(2) + 2)) - 2*sqrt(sqrt(2) + 2)*sqrt( 
- sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) - 2*x)/sqrt( - sqrt(2) + 2)) - 2*sq 
rt(sqrt(2) + 2)*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) + 2*x)/sqrt( 
- sqrt(2) + 2)) + sqrt(2)*log( - sqrt( - sqrt(2) + 2)*x + x**2 + 1) - sqrt 
(2)*log( - sqrt(sqrt(2) + 2)*x + x**2 + 1) + sqrt(2)*log(sqrt( - sqrt(2) + 
 2)*x + x**2 + 1) - sqrt(2)*log(sqrt(sqrt(2) + 2)*x + x**2 + 1))/16