\(\int \frac {1-x^4}{1-2 x^4+x^8} \, dx\) [26]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2} \]

[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 = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {28, 21, 218, 212, 209} \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {\arctan (x)}{2}+\frac {\text {arctanh}(x)}{2} \]

[In]

Int[(1 - x^4)/(1 - 2*x^4 + x^8),x]

[Out]

ArcTan[x]/2 + ArcTanh[x]/2

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.92 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {\arctan (x)}{2}-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (1+x) \]

[In]

Integrate[(1 - x^4)/(1 - 2*x^4 + x^8),x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (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	\(-\frac {\ln \left (x -1\right )}{4}+\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{4}\)	\(18\)
parallelrisch	\(-\frac {\ln \left (x -1\right )}{4}+\frac {i \ln \left (x +i\right )}{4}-\frac {i \ln \left (x -i\right )}{4}+\frac {\ln \left (x +1\right )}{4}\)	\(30\)

[In]

int((-x^4+1)/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

integrate((-x^4+1)/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (8) = 16\).

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=- \frac {\log {\left (x - 1 \right )}}{4} + \frac {\log {\left (x + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

integrate((-x^4+1)/(x^8-2*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 = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate((-x^4+1)/(x^8-2*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 = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1-x^4}{1-2 x^4+x^8} \, dx=\frac {\mathrm {atan}\left (x\right )}{2}+\frac {\mathrm {atanh}\left (x\right )}{2} \]

[In]

int(-(x^4 - 1)/(x^8 - 2*x^4 + 1),x)

[Out]

atan(x)/2 + atanh(x)/2