\(\int \frac {1}{x^{10} (1+x^2+x^4)} \, dx\) [867]

Optimal result

Integrand size = 14, antiderivative size = 81 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{2} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:

-1/9/x^9+1/7/x^7-1/3/x^3+1/x-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*a 
rctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/2*arctanh(x/(x^2+1))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}+\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )}{\sqrt {6+6 i \sqrt {3}}}+\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{\sqrt {6-6 i \sqrt {3}}} \] Input:

Integrate[1/(x^10*(1 + x^2 + x^4)),x]
 

Output:

-1/9*1/x^9 + 1/(7*x^7) - 1/(3*x^3) + x^(-1) + ((-I + Sqrt[3])*ArcTan[((-I 
+ Sqrt[3])*x)/2])/Sqrt[6 + (6*I)*Sqrt[3]] + ((I + Sqrt[3])*ArcTan[((I + Sq 
rt[3])*x)/2])/Sqrt[6 - (6*I)*Sqrt[3]]
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {1443, 27, 1604, 27, 1443, 27, 1604, 1447, 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.

\(\displaystyle \int \frac {1}{x^{10} \left (x^4+x^2+1\right )} \, dx\)

\(\Big \downarrow \) 1443

\(\displaystyle \frac {1}{9} \int -\frac {9 \left (x^2+1\right )}{x^8 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \frac {x^2+1}{x^8 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}\)

\(\Big \downarrow \) 1604

\(\displaystyle \frac {1}{7} \int \frac {7}{x^4 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {1}{x^4 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}\)

\(\Big \downarrow \) 1443

\(\displaystyle \frac {1}{3} \int -\frac {3 \left (x^2+1\right )}{x^2 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \frac {x^2+1}{x^2 \left (x^4+x^2+1\right )}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}\)

\(\Big \downarrow \) 1604

\(\displaystyle \int \frac {x^2}{x^4+x^2+1}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 1447

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 1475

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 1478

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx+\frac {1}{2} \int -\frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{x}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{9 x^9}+\frac {1}{7 x^7}-\frac {1}{3 x^3}+\frac {1}{2} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\frac {1}{2} \log \left (x^2+x+1\right )\right )+\frac {1}{x}\)

Input:

Int[1/(x^10*(1 + x^2 + x^4)),x]
 

Output:

-1/9*1/x^9 + 1/(7*x^7) - 1/(3*x^3) + x^(-1) + (ArcTan[(-1 + 2*x)/Sqrt[3]]/ 
Sqrt[3] + ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3])/2 + (Log[1 - x + x^2]/2 - Log 
[1 + x + x^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 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 1443

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim 
p[1/(a*d^2*(m + 1))   Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* 
x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 
- 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1447

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

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 1604

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) 
/(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1))   Int[(f*x)^(m + 2)*(a + b*x^2 
 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x 
], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ 
m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {1}{9 x^{9}}-\frac {1}{3 x^{3}}+\frac {1}{7 x^{7}}+\frac {1}{x}\)	\(72\)
risch	\(\frac {x^{8}-\frac {1}{3} x^{6}+\frac {1}{7} x^{2}-\frac {1}{9}}{x^{9}}-\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}\)	\(75\)

Input:

int(1/x^10/(x^4+x^2+1),x,method=_RETURNVERBOSE)
 

Output:

-1/4*ln(x^2+x+1)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/4*ln(x^2-x+1)+1 
/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-1/9/x^9-1/3/x^3+1/7/x^7+1/x
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {42 \, \sqrt {3} x^{9} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 42 \, \sqrt {3} x^{9} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 63 \, x^{9} \log \left (x^{2} + x + 1\right ) + 63 \, x^{9} \log \left (x^{2} - x + 1\right ) + 252 \, x^{8} - 84 \, x^{6} + 36 \, x^{2} - 28}{252 \, x^{9}} \] Input:

integrate(1/x^10/(x^4+x^2+1),x, algorithm="fricas")
 

Output:

1/252*(42*sqrt(3)*x^9*arctan(1/3*sqrt(3)*(2*x + 1)) + 42*sqrt(3)*x^9*arcta 
n(1/3*sqrt(3)*(2*x - 1)) - 63*x^9*log(x^2 + x + 1) + 63*x^9*log(x^2 - x + 
1) + 252*x^8 - 84*x^6 + 36*x^2 - 28)/x^9
 

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {\log {\left (x^{2} - x + 1 \right )}}{4} - \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} + \frac {63 x^{8} - 21 x^{6} + 9 x^{2} - 7}{63 x^{9}} \] Input:

integrate(1/x**10/(x**4+x**2+1),x)
 

Output:

log(x**2 - x + 1)/4 - log(x**2 + x + 1)/4 + sqrt(3)*atan(2*sqrt(3)*x/3 - s 
qrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/6 + (63*x**8 - 21*x* 
*6 + 9*x**2 - 7)/(63*x**9)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {63 \, x^{8} - 21 \, x^{6} + 9 \, x^{2} - 7}{63 \, x^{9}} - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:

integrate(1/x^10/(x^4+x^2+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)) + 1/63*(63*x^8 - 21*x^6 + 9*x^2 - 7)/x^9 - 1/4*log(x^2 + x + 1 
) + 1/4*log(x^2 - x + 1)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {63 \, x^{8} - 21 \, x^{6} + 9 \, x^{2} - 7}{63 \, x^{9}} - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:

integrate(1/x^10/(x^4+x^2+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)) + 1/63*(63*x^8 - 21*x^6 + 9*x^2 - 7)/x^9 - 1/4*log(x^2 + x + 1 
) + 1/4*log(x^2 - x + 1)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {x^8-\frac {x^6}{3}+\frac {x^2}{7}-\frac {1}{9}}{x^9}+\mathrm {atan}\left (\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{2}{}\mathrm {i}\right )-\mathrm {atan}\left (-\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(1/(x^10*(x^2 + x^4 + 1)),x)
 

Output:

atan((x*1i)/2 + (3^(1/2)*x)/2)*(3^(1/2)/6 + 1i/2) - atan((x*1i)/2 - (3^(1/ 
2)*x)/2)*(3^(1/2)/6 - 1i/2) + (x^2/7 - x^6/3 + x^8 - 1/9)/x^9
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{10} \left (1+x^2+x^4\right )} \, dx=\frac {42 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{9}+42 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x^{9}+63 \,\mathrm {log}\left (x^{2}-x +1\right ) x^{9}-63 \,\mathrm {log}\left (x^{2}+x +1\right ) x^{9}+252 x^{8}-84 x^{6}+36 x^{2}-28}{252 x^{9}} \] Input:

int(1/x^10/(x^4+x^2+1),x)
 

Output:

(42*sqrt(3)*atan((2*x - 1)/sqrt(3))*x**9 + 42*sqrt(3)*atan((2*x + 1)/sqrt( 
3))*x**9 + 63*log(x**2 - x + 1)*x**9 - 63*log(x**2 + x + 1)*x**9 + 252*x** 
8 - 84*x**6 + 36*x**2 - 28)/(252*x**9)