\(\int \frac {x^{14}}{1+x^2+x^4} \, dx\) [859]

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=-\frac {x^3}{3}+\frac {x^5}{5}-\frac {x^9}{9}+\frac {x^{11}}{11}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{2} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:

-1/3*x^3+1/5*x^5-1/9*x^9+1/11*x^11-1/6*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)-1/2*arctanh(x/(x^2+1))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=-\frac {x^3}{3}+\frac {x^5}{5}-\frac {x^9}{9}+\frac {x^{11}}{11}+\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )}{\sqrt {6+6 i \sqrt {3}}}+\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{\sqrt {6-6 i \sqrt {3}}} \] Input:

Integrate[x^14/(1 + x^2 + x^4),x]
 

Output:

-1/3*x^3 + x^5/5 - x^9/9 + x^11/11 + ((-I + Sqrt[3])*ArcTan[((-I + Sqrt[3] 
)*x)/2])/Sqrt[6 + (6*I)*Sqrt[3]] + ((I + Sqrt[3])*ArcTan[((I + Sqrt[3])*x) 
/2])/Sqrt[6 - (6*I)*Sqrt[3]]
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {1442, 27, 1602, 27, 1442, 27, 1602, 27, 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.

\(\displaystyle \int \frac {x^{14}}{x^4+x^2+1} \, dx\)

\(\Big \downarrow \) 1442

\(\displaystyle \frac {x^{11}}{11}-\frac {1}{11} \int \frac {11 x^{10} \left (x^2+1\right )}{x^4+x^2+1}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^{11}}{11}-\int \frac {x^{10} \left (x^2+1\right )}{x^4+x^2+1}dx\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {1}{9} \int \frac {9 x^8}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^8}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}\)

\(\Big \downarrow \) 1442

\(\displaystyle -\frac {1}{5} \int \frac {5 x^4 \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \frac {x^4 \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {1}{3} \int \frac {3 x^2}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^2}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 1447

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 1475

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 1478

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx+\frac {1}{2} \int -\frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {x^{11}}{11}-\frac {x^9}{9}+\frac {x^5}{5}-\frac {x^3}{3}+\frac {1}{2} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\frac {1}{2} \log \left (x^2+x+1\right )\right )\)

Input:

Int[x^14/(1 + x^2 + x^4),x]
 

Output:

-1/3*x^3 + x^5/5 - x^9/9 + x^11/11 + (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 1442

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), 
x] - Simp[d^4/(c*(m + 4*p + 1))   Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 
 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x 
] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && 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 1602

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {x^{11}}{11}-\frac {x^{9}}{9}+\frac {x^{5}}{5}-\frac {x^{3}}{3}-\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\)	\(74\)
risch	\(\frac {x^{11}}{11}-\frac {x^{9}}{9}+\frac {x^{5}}{5}-\frac {x^{3}}{3}+\frac {\ln \left (4 x^{2}-4 x +4\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}\)	\(80\)

Input:

int(x^14/(x^4+x^2+1),x,method=_RETURNVERBOSE)
 

Output:

1/11*x^11-1/9*x^9+1/5*x^5-1/3*x^3-1/4*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))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {1}{11} \, x^{11} - \frac {1}{9} \, x^{9} + \frac {1}{5} \, x^{5} - \frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:

integrate(x^14/(x^4+x^2+1),x, algorithm="fricas")
 

Output:

1/11*x^11 - 1/9*x^9 + 1/5*x^5 - 1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*( 
2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/4*log(x^2 + x + 
1) + 1/4*log(x^2 - x + 1)
 

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {x^{11}}{11} - \frac {x^{9}}{9} + \frac {x^{5}}{5} - \frac {x^{3}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{4} - \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \] Input:

integrate(x**14/(x**4+x**2+1),x)
 

Output:

x**11/11 - x**9/9 + x**5/5 - x**3/3 + log(x**2 - x + 1)/4 - log(x**2 + x + 
 1)/4 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3) 
*x/3 + sqrt(3)/3)/6
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {1}{11} \, x^{11} - \frac {1}{9} \, x^{9} + \frac {1}{5} \, x^{5} - \frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:

integrate(x^14/(x^4+x^2+1),x, algorithm="maxima")
 

Output:

1/11*x^11 - 1/9*x^9 + 1/5*x^5 - 1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*( 
2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/4*log(x^2 + x + 
1) + 1/4*log(x^2 - x + 1)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {1}{11} \, x^{11} - \frac {1}{9} \, x^{9} + \frac {1}{5} \, x^{5} - \frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:

integrate(x^14/(x^4+x^2+1),x, algorithm="giac")
 

Output:

1/11*x^11 - 1/9*x^9 + 1/5*x^5 - 1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*( 
2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/4*log(x^2 + x + 
1) + 1/4*log(x^2 - x + 1)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 18.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {x^5}{5}+\mathrm {atan}\left (\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{2}{}\mathrm {i}\right )-\frac {x^3}{3}-\mathrm {atan}\left (-\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{2}{}\mathrm {i}\right )-\frac {x^9}{9}+\frac {x^{11}}{11} \] Input:

int(x^14/(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^3/3 + x^5/5 - x^9/9 + x^11/11
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int \frac {x^{14}}{1+x^2+x^4} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{6}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{6}+\frac {\mathrm {log}\left (x^{2}-x +1\right )}{4}-\frac {\mathrm {log}\left (x^{2}+x +1\right )}{4}+\frac {x^{11}}{11}-\frac {x^{9}}{9}+\frac {x^{5}}{5}-\frac {x^{3}}{3} \] Input:

int(x^14/(x^4+x^2+1),x)
 

Output:

(330*sqrt(3)*atan((2*x - 1)/sqrt(3)) + 330*sqrt(3)*atan((2*x + 1)/sqrt(3)) 
 + 495*log(x**2 - x + 1) - 495*log(x**2 + x + 1) + 180*x**11 - 220*x**9 + 
396*x**5 - 660*x**3)/1980