\(\int \frac {1}{x^5 (1-x^4+x^8)} \, dx\) [356]

Optimal result

Integrand size = 16, antiderivative size = 48 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}+\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right ) \]

[Out]

-1/4/x^4+ln(x)-1/8*ln(x^8-x^4+1)+1/12*arctan(1/3*(-2*x^4+1)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1371, 723, 814, 648, 632, 210, 642} \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4 x^4}-\frac {1}{8} \log \left (x^8-x^4+1\right )+\log (x) \]

[In]

Int[1/(x^5*(1 - x^4 + x^8)),x]

[Out]

-1/4*1/x^4 + ArcTan[(1 - 2*x^4)/Sqrt[3]]/(4*Sqrt[3]) + Log[x] - Log[1 - x^4 + x^8]/8

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 723

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m + 1)/((m
+ 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d + e*x)^(m + 1)*(Simp[c*d - b*e - c
*e*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && LtQ[m, -1]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (1-x+x^2\right )} \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\frac {1}{4} \text {Subst}\left (\int \frac {1-x}{x \left (1-x+x^2\right )} \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{x}-\frac {x}{1-x+x^2}\right ) \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\log (x)-\frac {1}{4} \text {Subst}\left (\int \frac {x}{1-x+x^2} \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\log (x)-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^4\right )-\frac {1}{8} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^4\right ) \\ & = -\frac {1}{4 x^4}+\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right ) \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}+\log (x)-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^4}{-1+2 \text {$\#$1}^4}\&\right ] \]

[In]

Integrate[1/(x^5*(1 - x^4 + x^8)),x]

[Out]

-1/4*1/x^4 + Log[x] - RootSum[1 - #1^4 + #1^8 & , (Log[x - #1]*#1^4)/(-1 + 2*#1^4) & ]/4

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {1}{4 x^{4}}+\ln \left (x \right )-\frac {\ln \left (x^{8}-x^{4}+1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{4}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}\)	\(38\)
default	\(-\frac {1}{4 x^{4}}+\ln \left (x \right )-\frac {\ln \left (x^{8}-x^{4}+1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right )}{12}\)	\(40\)

[In]

int(1/x^5/(x^8-x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/4/x^4+ln(x)-1/8*ln(x^8-x^4+1)-1/12*3^(1/2)*arctan(2/3*(x^4-1/2)*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=-\frac {2 \, \sqrt {3} x^{4} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + 3 \, x^{4} \log \left (x^{8} - x^{4} + 1\right ) - 24 \, x^{4} \log \left (x\right ) + 6}{24 \, x^{4}} \]

[In]

integrate(1/x^5/(x^8-x^4+1),x, algorithm="fricas")

[Out]

-1/24*(2*sqrt(3)*x^4*arctan(1/3*sqrt(3)*(2*x^4 - 1)) + 3*x^4*log(x^8 - x^4 + 1) - 24*x^4*log(x) + 6)/x^4

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{8} - x^{4} + 1 \right )}}{8} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} - \frac {1}{4 x^{4}} \]

[In]

integrate(1/x**5/(x**8-x**4+1),x)

[Out]

log(x) - log(x**8 - x**4 + 1)/8 - sqrt(3)*atan(2*sqrt(3)*x**4/3 - sqrt(3)/3)/12 - 1/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{4 \, x^{4}} - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

integrate(1/x^5/(x^8-x^4+1),x, algorithm="maxima")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 - 1)) - 1/4/x^4 - 1/8*log(x^8 - x^4 + 1) + 1/4*log(x^4)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {x^{4} + 1}{4 \, x^{4}} - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

integrate(1/x^5/(x^8-x^4+1),x, algorithm="giac")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 - 1)) - 1/4*(x^4 + 1)/x^4 - 1/8*log(x^8 - x^4 + 1) + 1/4*log(x^4)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^5 \left (1-x^4+x^8\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^8-x^4+1\right )}{8}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^4}{3}\right )}{12}-\frac {1}{4\,x^4} \]

[In]

int(1/(x^5*(x^8 - x^4 + 1)),x)

[Out]

log(x) - log(x^8 - x^4 + 1)/8 + (3^(1/2)*atan(3^(1/2)/3 - (2*3^(1/2)*x^4)/3))/12 - 1/(4*x^4)