\(\int \frac {1}{-1+x^4} \, dx\) [136]

Optimal result

Integrand size = 7, antiderivative size = 13 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2} \]

[Out]

-1/2*arctan(x)-1/2*arctanh(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {218, 212, 209} \[ \int \frac {1}{-1+x^4} \, dx=-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2} \]

[In]

Int[(-1 + x^4)^(-1),x]

[Out]

-1/2*ArcTan[x] - ArcTanh[x]/2

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 218

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.92 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {\arctan (x)}{2}+\frac {1}{4} \log (1-x)-\frac {1}{4} \log (1+x) \]

[In]

Integrate[(-1 + x^4)^(-1),x]

[Out]

-1/2*ArcTan[x] + Log[1 - x]/4 - Log[1 + x]/4

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {\arctan \left (x \right )}{2}-\frac {\operatorname {arctanh}\left (x \right )}{2}\)	\(10\)
risch	\(-\frac {\arctan \left (x \right )}{2}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (-1+x \right )}{4}\)	\(18\)
parallelrisch	\(\frac {i \ln \left (x -i\right )}{4}-\frac {i \ln \left (x +i\right )}{4}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (-1+x \right )}{4}\)	\(30\)
meijerg	\(\frac {x \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(38\)

[In]

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

[Out]

-1/2*arctan(x)-1/2*arctanh(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1}{-1+x^4} \, dx=\frac {\log {\left (x - 1 \right )}}{4} - \frac {\log {\left (x + 1 \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

integrate(1/(x**4-1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19 vs. \(2 (9) = 18\).

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1}{-1+x^4} \, dx=-\frac {\mathrm {atan}\left (x\right )}{2}-\frac {\mathrm {atanh}\left (x\right )}{2} \]

[In]

int(1/(x^4 - 1),x)

[Out]

- atan(x)/2 - atanh(x)/2