3.5.11 \(\int \frac {1}{x+x^6+x^{11}} \, dx\) [411]

3.5.11.1 Optimal result
3.5.11.2 Mathematica [C] (verified)
3.5.11.3 Rubi [A] (verified)
3.5.11.4 Maple [A] (verified)
3.5.11.5 Fricas [A] (verification not implemented)
3.5.11.6 Sympy [A] (verification not implemented)
3.5.11.7 Maxima [F]
3.5.11.8 Giac [A] (verification not implemented)
3.5.11.9 Mupad [B] (verification not implemented)

3.5.11.1 Optimal result

Integrand size = 10, antiderivative size = 39 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=-\frac {\arctan \left (\frac {1+2 x^5}{\sqrt {3}}\right )}{5 \sqrt {3}}+\log (x)-\frac {1}{10} \log \left (1+x^5+x^{10}\right ) \]

output
ln(x)-1/10*ln(x^10+x^5+1)-1/15*arctan(1/3*(2*x^5+1)*3^(1/2))*3^(1/2)
 
3.5.11.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 197, normalized size of antiderivative = 5.05 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{5 \sqrt {3}}+\log (x)-\frac {1}{10} \log \left (1+x+x^2\right )-\frac {1}{5} \text {RootSum}\left [1-\text {$\#$1}+\text {$\#$1}^3-\text {$\#$1}^4+\text {$\#$1}^5-\text {$\#$1}^7+\text {$\#$1}^8\&,\frac {-\log (x-\text {$\#$1}) \text {$\#$1}+2 \log (x-\text {$\#$1}) \text {$\#$1}^2-\log (x-\text {$\#$1}) \text {$\#$1}^3+3 \log (x-\text {$\#$1}) \text {$\#$1}^4-\log (x-\text {$\#$1}) \text {$\#$1}^5-3 \log (x-\text {$\#$1}) \text {$\#$1}^6+4 \log (x-\text {$\#$1}) \text {$\#$1}^7}{-1+3 \text {$\#$1}^2-4 \text {$\#$1}^3+5 \text {$\#$1}^4-7 \text {$\#$1}^6+8 \text {$\#$1}^7}\&\right ] \]

input
Integrate[(x + x^6 + x^11)^(-1),x]
 
output
ArcTan[(1 + 2*x)/Sqrt[3]]/(5*Sqrt[3]) + Log[x] - Log[1 + x + x^2]/10 - Roo 
tSum[1 - #1 + #1^3 - #1^4 + #1^5 - #1^7 + #1^8 & , (-(Log[x - #1]*#1) + 2* 
Log[x - #1]*#1^2 - Log[x - #1]*#1^3 + 3*Log[x - #1]*#1^4 - Log[x - #1]*#1^ 
5 - 3*Log[x - #1]*#1^6 + 4*Log[x - #1]*#1^7)/(-1 + 3*#1^2 - 4*#1^3 + 5*#1^ 
4 - 7*#1^6 + 8*#1^7) & ]/5
 
3.5.11.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {1949, 1693, 1144, 25, 1142, 1083, 217, 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.

\(\displaystyle \int \frac {1}{x^{11}+x^6+x} \, dx\)

\(\Big \downarrow \) 1949

\(\displaystyle \int \frac {1}{x \left (x^{10}+x^5+1\right )}dx\)

\(\Big \downarrow \) 1693

\(\displaystyle \frac {1}{5} \int \frac {1}{x^5 \left (x^{10}+x^5+1\right )}dx^5\)

\(\Big \downarrow \) 1144

\(\displaystyle \frac {1}{5} \left (\int -\frac {x^5+1}{x^{10}+x^5+1}dx^5+\log \left (x^5\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\log \left (x^5\right )-\int \frac {x^5+1}{x^{10}+x^5+1}dx^5\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{5} \left (-\frac {1}{2} \int \frac {1}{x^{10}+x^5+1}dx^5-\frac {1}{2} \int \frac {2 x^5+1}{x^{10}+x^5+1}dx^5+\log \left (x^5\right )\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{5} \left (\int \frac {1}{-x^{10}-3}d\left (2 x^5+1\right )-\frac {1}{2} \int \frac {2 x^5+1}{x^{10}+x^5+1}dx^5+\log \left (x^5\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (-\frac {1}{2} \int \frac {2 x^5+1}{x^{10}+x^5+1}dx^5-\frac {\arctan \left (\frac {2 x^5+1}{\sqrt {3}}\right )}{\sqrt {3}}+\log \left (x^5\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{5} \left (-\frac {\arctan \left (\frac {2 x^5+1}{\sqrt {3}}\right )}{\sqrt {3}}+\log \left (x^5\right )-\frac {1}{2} \log \left (x^{10}+x^5+1\right )\right )\)

input
Int[(x + x^6 + x^11)^(-1),x]
 
output
(-(ArcTan[(1 + 2*x^5)/Sqrt[3]]/Sqrt[3]) + Log[x^5] - Log[1 + x^5 + x^10]/2 
)/5
 

3.5.11.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1144
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] 
 :> Simp[e*(Log[RemoveContent[d + e*x, x]]/(c*d^2 - b*d*e + a*e^2)), x] + S 
imp[1/(c*d^2 - b*d*e + a*e^2)   Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), 
x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1693
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, 
x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ 
[Simplify[(m + 1)/n]]
 

rule 1949
Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol 
] :> Int[x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p, x] /; FreeQ[{a, b, 
c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && IntegerQ[p]
 
3.5.11.4 Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79

method result size
risch \(\ln \left (x \right )-\frac {\ln \left (x^{10}+x^{5}+1\right )}{10}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{5}+\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{15}\) \(31\)
default \(\ln \left (x \right )-\frac {\left (\frac {1}{2}+\frac {i \sqrt {3}}{6}\right ) \ln \left (2 x^{4}+\left (-1+i \sqrt {3}\right ) x^{3}+\left (-1-i \sqrt {3}\right ) x^{2}+2 x -1+i \sqrt {3}\right )}{5}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}\right ) \ln \left (2 x^{4}+\left (-1-i \sqrt {3}\right ) x^{3}+\left (-1+i \sqrt {3}\right ) x^{2}+2 x -1-i \sqrt {3}\right )}{5}-\frac {\ln \left (x^{2}+x +1\right )}{10}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{15}\) \(131\)

input
int(1/(x^11+x^6+x),x,method=_RETURNVERBOSE)
 
output
ln(x)-1/10*ln(x^10+x^5+1)-1/15*3^(1/2)*arctan(2/3*(x^5+1/2)*3^(1/2))
 
3.5.11.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=-\frac {1}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{5} + 1\right )}\right ) - \frac {1}{10} \, \log \left (x^{10} + x^{5} + 1\right ) + \log \left (x\right ) \]

input
integrate(1/(x^11+x^6+x),x, algorithm="fricas")
 
output
-1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^5 + 1)) - 1/10*log(x^10 + x^5 + 1) + 
 log(x)
 
3.5.11.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{10} + x^{5} + 1 \right )}}{10} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{5}}{3} + \frac {\sqrt {3}}{3} \right )}}{15} \]

input
integrate(1/(x**11+x**6+x),x)
 
output
log(x) - log(x**10 + x**5 + 1)/10 - sqrt(3)*atan(2*sqrt(3)*x**5/3 + sqrt(3 
)/3)/15
 
3.5.11.7 Maxima [F]

\[ \int \frac {1}{x+x^6+x^{11}} \, dx=\int { \frac {1}{x^{11} + x^{6} + x} \,d x } \]

input
integrate(1/(x^11+x^6+x),x, algorithm="maxima")
 
output
1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/5*integrate((4*x^7 - 3*x^6 
- x^5 + 3*x^4 - x^3 + 2*x^2 - x)/(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1), x) 
 - 1/10*log(x^2 + x + 1) + log(x)
 
3.5.11.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=-\frac {1}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{5} + 1\right )}\right ) - \frac {1}{10} \, \log \left (x^{10} + x^{5} + 1\right ) + \log \left ({\left | x \right |}\right ) \]

input
integrate(1/(x^11+x^6+x),x, algorithm="giac")
 
output
-1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^5 + 1)) - 1/10*log(x^10 + x^5 + 1) + 
 log(abs(x))
 
3.5.11.9 Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x+x^6+x^{11}} \, dx=\ln \left (x\right )-\frac {\ln \left (x^{10}+x^5+1\right )}{10}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x^5}{3}+\frac {\sqrt {3}}{3}\right )}{15} \]

input
int(1/(x + x^6 + x^11),x)
 
output
log(x) - log(x^5 + x^10 + 1)/10 - (3^(1/2)*atan(3^(1/2)/3 + (2*3^(1/2)*x^5 
)/3))/15