\(\int \frac {1}{x (1+x^5+x^{10})} \, dx\) [409]

Optimal result

Integrand size = 14, antiderivative size = 39 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=-\frac {\arctan \left (\frac {1+2 x^5}{\sqrt {3}}\right )}{5 \sqrt {3}}+\log (x)-\frac {1}{10} \log \left (1+x^5+x^{10}\right ) \]

[Out]

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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1371, 719, 29, 648, 632, 210, 642} \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=-\frac {\arctan \left (\frac {2 x^5+1}{\sqrt {3}}\right )}{5 \sqrt {3}}-\frac {1}{10} \log \left (x^{10}+x^5+1\right )+\log (x) \]

[In]

Int[1/(x*(1 + x^5 + x^10)),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 210

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 719

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \text {Subst}\left (\int \frac {1}{x \left (1+x+x^2\right )} \, dx,x,x^5\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^5\right )+\frac {1}{5} \text {Subst}\left (\int \frac {-1-x}{1+x+x^2} \, dx,x,x^5\right ) \\ & = \log (x)-\frac {1}{10} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^5\right )-\frac {1}{10} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,x^5\right ) \\ & = \log (x)-\frac {1}{10} \log \left (1+x^5+x^{10}\right )+\frac {1}{5} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^5\right ) \\ & = -\frac {\tan ^{-1}\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 ) \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 197, normalized size of antiderivative = 5.05 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{5 \sqrt {3}}+\log (x)-\frac {1}{10} \log \left (1+x+x^2\right )-\frac {1}{5} \text {RootSum}\left [1-\text {$\#$1}+\text {$\#$1}^3-\text {$\#$1}^4+\text {$\#$1}^5-\text {$\#$1}^7+\text {$\#$1}^8\&,\frac {-\log (x-\text {$\#$1}) \text {$\#$1}+2 \log (x-\text {$\#$1}) \text {$\#$1}^2-\log (x-\text {$\#$1}) \text {$\#$1}^3+3 \log (x-\text {$\#$1}) \text {$\#$1}^4-\log (x-\text {$\#$1}) \text {$\#$1}^5-3 \log (x-\text {$\#$1}) \text {$\#$1}^6+4 \log (x-\text {$\#$1}) \text {$\#$1}^7}{-1+3 \text {$\#$1}^2-4 \text {$\#$1}^3+5 \text {$\#$1}^4-7 \text {$\#$1}^6+8 \text {$\#$1}^7}\&\right ] \]

[In]

Integrate[1/(x*(1 + x^5 + x^10)),x]

[Out]

ArcTan[(1 + 2*x)/Sqrt[3]]/(5*Sqrt[3]) + Log[x] - Log[1 + x + x^2]/10 - RootSum[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

Maple [A] (verified)

Time = 0.10 (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	\(\ln \left (x \right )-\frac {\left (\frac {1}{2}+\frac {i \sqrt {3}}{6}\right ) \ln \left (2 x^{4}+\left (-1+i \sqrt {3}\right ) x^{3}+\left (-1-i \sqrt {3}\right ) x^{2}+2 x -1+i \sqrt {3}\right )}{5}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}\right ) \ln \left (2 x^{4}+\left (-1-i \sqrt {3}\right ) x^{3}+\left (-1+i \sqrt {3}\right ) x^{2}+2 x -1-i \sqrt {3}\right )}{5}-\frac {\ln \left (x^{2}+x +1\right )}{10}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{15}\)	\(131\)

[In]

int(1/x/(x^10+x^5+1),x,method=_RETURNVERBOSE)

[Out]

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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=-\frac {1}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{5} + 1\right )}\right ) - \frac {1}{10} \, \log \left (x^{10} + x^{5} + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/x/(x^10+x^5+1),x, algorithm="fricas")

[Out]

-1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^5 + 1)) - 1/10*log(x^10 + x^5 + 1) + log(x)

Sympy [A] (verification not implemented)

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

[In]

integrate(1/x/(x**10+x**5+1),x)

[Out]

log(x) - log(x**10 + x**5 + 1)/10 - sqrt(3)*atan(2*sqrt(3)*x**5/3 + sqrt(3)/3)/15

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=-\frac {1}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{5} + 1\right )}\right ) - \frac {1}{10} \, \log \left (x^{10} + x^{5} + 1\right ) + \frac {1}{5} \, \log \left (x^{5}\right ) \]

[In]

integrate(1/x/(x^10+x^5+1),x, algorithm="maxima")

[Out]

-1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^5 + 1)) - 1/10*log(x^10 + x^5 + 1) + 1/5*log(x^5)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=-\frac {1}{15} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{5} + 1\right )}\right ) - \frac {1}{10} \, \log \left (x^{10} + x^{5} + 1\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(x^10+x^5+1),x, algorithm="giac")

[Out]

-1/15*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^5 + 1)) - 1/10*log(x^10 + x^5 + 1) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \left (1+x^5+x^{10}\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^{10}+x^5+1\right )}{10}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x^5}{3}+\frac {\sqrt {3}}{3}\right )}{15} \]

[In]

int(1/(x*(x^5 + x^10 + 1)),x)

[Out]

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