\(\int \frac {1}{x^3 (1-x^8)} \, dx\) [1477]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {\arctan \left (x^2\right )}{4}+\frac {\text {arctanh}\left (x^2\right )}{4} \]

[Out]

-1/2/x^2-1/4*arctan(x^2)+1/4*arctanh(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {281, 331, 304, 209, 212} \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {1}{4} \arctan \left (x^2\right )+\frac {\text {arctanh}\left (x^2\right )}{4}-\frac {1}{2 x^2} \]

[In]

Int[1/(x^3*(1 - x^8)),x]

[Out]

-1/2*1/x^2 - ArcTan[x^2]/4 + ArcTanh[x^2]/4

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 281

Int[(x_)^(m_.)*((a_) + (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))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {1}{4} \arctan \left (\frac {1}{x^2}\right )-\frac {1}{8} \log \left (1-x^2\right )+\frac {1}{8} \log \left (1+x^2\right ) \]

[In]

Integrate[1/(x^3*(1 - x^8)),x]

[Out]

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

Maple [A] (verified)

Time = 5.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
risch	\(-\frac {1}{2 x^{2}}-\frac {\arctan \left (x^{2}\right )}{4}-\frac {\ln \left (x^{2}-1\right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}\)	\(29\)
default	\(-\frac {\ln \left (-1+x \right )}{8}-\frac {\ln \left (1+x \right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}-\frac {\arctan \left (x^{2}\right )}{4}-\frac {1}{2 x^{2}}\)	\(33\)
meijerg	\(\frac {\left (-1\right )^{\frac {1}{4}} \left (\frac {4 \left (-1\right )^{\frac {3}{4}}}{x^{2}}+\frac {x^{6} \left (-1\right )^{\frac {3}{4}} \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{8}\right )^{\frac {1}{4}}\right )\right )}{\left (x^{8}\right )^{\frac {3}{4}}}\right )}{8}\)	\(56\)
parallelrisch	\(\frac {i \ln \left (x^{2}-i\right ) x^{2}-i \ln \left (x^{2}+i\right ) x^{2}-\ln \left (1+x \right ) x^{2}-\ln \left (-1+x \right ) x^{2}+\ln \left (x -i\right ) x^{2}+\ln \left (x +i\right ) x^{2}-4}{8 x^{2}}\)	\(70\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {2 \, x^{2} \arctan \left (x^{2}\right ) - x^{2} \log \left (x^{2} + 1\right ) + x^{2} \log \left (x^{2} - 1\right ) + 4}{8 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=- \frac {\log {\left (x^{2} - 1 \right )}}{8} + \frac {\log {\left (x^{2} + 1 \right )}}{8} - \frac {\operatorname {atan}{\left (x^{2} \right )}}{4} - \frac {1}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {1}{2 \, x^{2}} - \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=-\frac {1}{2 \, x^{2}} - \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/2/x^2 - 1/4*arctan(x^2) + 1/8*log(x^2 + 1) - 1/8*log(abs(x^2 - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^3 \left (1-x^8\right )} \, dx=\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {\mathrm {atan}\left (x^2\right )}{4}-\frac {1}{2\,x^2} \]

[In]

int(-1/(x^3*(x^8 - 1)),x)

[Out]

atanh(x^2)/4 - atan(x^2)/4 - 1/(2*x^2)