\(\int \frac {1}{x^6 (1+x^4+x^8)} \, dx\) [80]

Optimal result

Integrand size = 14, antiderivative size = 77 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{5 x^5}+\frac {1}{x}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:

-1/5/x^5+1/x-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x 
)*3^(1/2))*3^(1/2)-1/6*arctanh(3^(1/2)*x/(x^2+1))*3^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {1}{60} \left (-\frac {12}{x^5}+\frac {60}{x}+10 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+10 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+5 \sqrt {3} \log \left (-1+\sqrt {3} x-x^2\right )-5 \sqrt {3} \log \left (1+\sqrt {3} x+x^2\right )\right ) \] Input:

Integrate[1/(x^6*(1 + x^4 + x^8)),x]
 

Output:

(-12/x^5 + 60/x + 10*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 10*Sqrt[3]*ArcTa 
n[(1 + 2*x)/Sqrt[3]] + 5*Sqrt[3]*Log[-1 + Sqrt[3]*x - x^2] - 5*Sqrt[3]*Log 
[1 + Sqrt[3]*x + x^2])/60
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {1704, 27, 1828, 1708, 1475, 1083, 217, 1478, 25, 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.

\(\displaystyle \int \frac {1}{x^6 \left (x^8+x^4+1\right )} \, dx\)

\(\Big \downarrow \) 1704

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1828

\(\displaystyle \int \frac {x^6}{x^8+x^4+1}dx-\frac {1}{5 x^5}+\frac {1}{x}\)

\(\Big \downarrow \) 1708

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

\(\Big \downarrow \) 1475

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1478

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/(x^6*(1 + x^4 + x^8)),x]
 

Output:

-1/5*1/x^5 + x^(-1) + (ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 2* 
x)/Sqrt[3]]/Sqrt[3])/2 + (Log[1 - Sqrt[3]*x + x^2]/(2*Sqrt[3]) - Log[1 + S 
qrt[3]*x + x^2]/(2*Sqrt[3]))/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 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 1475

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

rule 1478

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

rule 1704

Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_ 
Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1) 
)), x] - Simp[1/(a*d^n*(m + 1))   Int[(d*x)^(m + n)*(b*(m + n*(p + 1) + 1) 
+ c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && 
LtQ[m, -1] && IntegerQ[p]
 

rule 1708

Int[(x_)^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> W 
ith[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, -Simp[1/(2*c*r)   Int[x^ 
(m - 3*(n/2))*((q - r*x^(n/2))/(q - r*x^(n/2) + x^n)), x], x] + Simp[1/(2*c 
*r)   Int[x^(m - 3*(n/2))*((q + r*x^(n/2))/(q + r*x^(n/2) + x^n)), x], x]]] 
 /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 
 0] && IGtQ[m, 0] && GeQ[m, 3*(n/2)] && LtQ[m, 2*n] && NegQ[b^2 - 4*a*c]
 

rule 1828

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( 
c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^n + c*x^ 
(2*n))^(p + 1)/(a*f*(m + 1))), x] + Simp[1/(a*f^n*(m + 1))   Int[(f*x)^(m + 
 n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - 
c*d*(m + 2*n*(p + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x 
] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && Int 
egerQ[p]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}+\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}-\frac {1}{5 x^{5}}+\frac {1}{x}\)	\(75\)
risch	\(\frac {x^{4}-\frac {1}{5}}{x^{5}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, x}{3}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, x^{3}}{3}+\frac {2 \sqrt {3}\, x}{3}\right )}{6}+\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}\)	\(77\)

Input:

int(1/x^6/(x^8+x^4+1),x,method=_RETURNVERBOSE)
 

Output:

1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/6*arctan(1/3*(2*x+1)*3^(1/2))*3^ 
(1/2)-1/12*3^(1/2)*ln(x^2+3^(1/2)*x+1)+1/12*3^(1/2)*ln(x^2-3^(1/2)*x+1)-1/ 
5/x^5+1/x
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) + 10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + 5 \, \sqrt {3} x^{5} \log \left (\frac {x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) + 60 \, x^{4} - 12}{60 \, x^{5}} \] Input:

integrate(1/x^6/(x^8+x^4+1),x, algorithm="fricas")
 

Output:

1/60*(10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*(x^3 + 2*x)) + 10*sqrt(3)*x^5*arct 
an(1/3*sqrt(3)*x) + 5*sqrt(3)*x^5*log((x^4 + 5*x^2 - 2*sqrt(3)*(x^3 + x) + 
 1)/(x^4 - x^2 + 1)) + 60*x^4 - 12)/x^5
 

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {\sqrt {3} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{12} + \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} + \frac {5 x^{4} - 1}{5 x^{5}} \] Input:

integrate(1/x**6/(x**8+x**4+1),x)
 

Output:

sqrt(3)*(2*atan(sqrt(3)*x/3) + 2*atan(sqrt(3)*x**3/3 + 2*sqrt(3)*x/3))/12 
+ sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1) 
/12 + (5*x**4 - 1)/(5*x**5)
 

Maxima [F]

\[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + x^{4} + 1\right )} x^{6}} \,d x } \] Input:

integrate(1/x^6/(x^8+x^4+1),x, algorithm="maxima")
 

Output:

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) 
*(2*x - 1)) + 1/5*(5*x^4 - 1)/x^5 + 1/2*integrate((x^2 - 1)/(x^4 - x^2 + 1 
), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {5 \, x^{4} - 1}{5 \, x^{5}} + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) \] Input:

integrate(1/x^6/(x^8+x^4+1),x, algorithm="giac")
 

Output:

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) 
*(2*x - 1)) - 1/24*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/24*sqrt(3)*log(x^2 
 - sqrt(3)*x + 1) + 1/5*(5*x^4 - 1)/x^5 + 1/4*arctan(2*x + sqrt(3)) + 1/4* 
arctan(2*x - sqrt(3))
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {x^4-\frac {1}{5}}{x^5}-\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )}{6}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}-\frac {2}{3}\right )}\right )}{6} \] Input:

int(1/(x^6*(x^4 + x^8 + 1)),x)
 

Output:

(x^4 - 1/5)/x^5 - (3^(1/2)*atanh((2*3^(1/2)*x)/(3*((2*x^2)/3 + 2/3))))/6 - 
 (3^(1/2)*atan((2*3^(1/2)*x)/(3*((2*x^2)/3 - 2/3))))/6
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx=\frac {10 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{5}+10 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x^{5}+5 \sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x^{5}-5 \sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x^{5}+60 x^{4}-12}{60 x^{5}} \] Input:

int(1/x^6/(x^8+x^4+1),x)
 

Output:

(10*sqrt(3)*atan((2*x - 1)/sqrt(3))*x**5 + 10*sqrt(3)*atan((2*x + 1)/sqrt( 
3))*x**5 + 5*sqrt(3)*log( - sqrt(3)*x + x**2 + 1)*x**5 - 5*sqrt(3)*log(sqr 
t(3)*x + x**2 + 1)*x**5 + 60*x**4 - 12)/(60*x**5)