\(\int \frac {1}{x^3 (1-x^4+x^8)} \, dx\) [355]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{4 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{4 \sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1373, 1137, 1178, 642} \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{4 \sqrt {3}}+\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{4 \sqrt {3}} \]

[In]

Int[1/(x^3*(1 - x^4 + x^8)),x]

[Out]

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

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 1137

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] - Dist[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] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[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 1373

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {1}{12} \left (-\frac {6}{x^2}-\sqrt {3} \log \left (-1+\sqrt {3} x^2-x^4\right )+\sqrt {3} \log \left (1+\sqrt {3} x^2+x^4\right )\right ) \]

[In]

Integrate[1/(x^3*(1 - x^4 + x^8)),x]

[Out]

(-6/x^2 - Sqrt[3]*Log[-1 + Sqrt[3]*x^2 - x^4] + Sqrt[3]*Log[1 + Sqrt[3]*x^2 + x^4])/12

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {1}{2 x^{2}}-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right ) \sqrt {3}}{12}\)	\(44\)
risch	\(-\frac {1}{2 x^{2}}-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right ) \sqrt {3}}{12}\)	\(44\)

[In]

int(1/x^3/(x^8-x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {\sqrt {3} x^{2} \log \left (\frac {x^{8} + 5 \, x^{4} + 2 \, \sqrt {3} {\left (x^{6} + x^{2}\right )} + 1}{x^{8} - x^{4} + 1}\right ) - 6}{12 \, x^{2}} \]

[In]

integrate(1/x^3/(x^8-x^4+1),x, algorithm="fricas")

[Out]

1/12*(sqrt(3)*x^2*log((x^8 + 5*x^4 + 2*sqrt(3)*(x^6 + x^2) + 1)/(x^8 - x^4 + 1)) - 6)/x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{12} - \frac {1}{2 x^{2}} \]

[In]

integrate(1/x**3/(x**8-x**4+1),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} - x^{4} + 1\right )} x^{3}} \,d x } \]

[In]

integrate(1/x^3/(x^8-x^4+1),x, algorithm="maxima")

[Out]

-1/2/x^2 - integrate((x^4 - 1)*x/(x^8 - x^4 + 1), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (43) = 86\).

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) - \frac {1}{24} \, {\left (\sqrt {3} x^{4} - \sqrt {3}\right )} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, {\left (\sqrt {3} x^{4} - \sqrt {3}\right )} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - \frac {1}{2 \, x^{2}} \]

[In]

integrate(1/x^3/(x^8-x^4+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x^2}{9\,\left (\frac {2\,x^4}{9}+\frac {2}{9}\right )}\right )}{6}-\frac {1}{2\,x^2} \]

[In]

int(1/(x^3*(x^8 - x^4 + 1)),x)

[Out]

(3^(1/2)*atanh((2*3^(1/2)*x^2)/(9*((2*x^4)/9 + 2/9))))/6 - 1/(2*x^2)