\(\int \frac {x}{1-x^4} \, dx\) [94]

Optimal result

Integrand size = 11, antiderivative size = 8 \[ \int \frac {x}{1-x^4} \, dx=\frac {\text {arctanh}\left (x^2\right )}{2} \]

[Out]

1/2*arctanh(x^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 212} \[ \int \frac {x}{1-x^4} \, dx=\frac {\text {arctanh}\left (x^2\right )}{2} \]

[In]

Int[x/(1 - x^4),x]

[Out]

ArcTanh[x^2]/2

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]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(23\) vs. \(2(8)=16\).

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 2.88 \[ \int \frac {x}{1-x^4} \, dx=-\frac {1}{4} \log \left (1-x^2\right )+\frac {1}{4} \log \left (1+x^2\right ) \]

[In]

Integrate[x/(1 - x^4),x]

[Out]

-1/4*Log[1 - x^2] + Log[1 + x^2]/4

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
meijerg	\(\frac {\operatorname {arctanh}\left (x^{2}\right )}{2}\)	\(7\)
risch	\(\frac {\ln \left (x^{2}+1\right )}{4}-\frac {\ln \left (x^{2}-1\right )}{4}\)	\(18\)
default	\(-\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}\)	\(22\)
norman	\(-\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}\)	\(22\)
parallelrisch	\(-\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}\)	\(22\)

[In]

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

[Out]

1/2*arctanh(x^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (6) = 12\).

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12 \[ \int \frac {x}{1-x^4} \, dx=\frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

1/4*log(x^2 + 1) - 1/4*log(x^2 - 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.88 \[ \int \frac {x}{1-x^4} \, dx=- \frac {\log {\left (x^{2} - 1 \right )}}{4} + \frac {\log {\left (x^{2} + 1 \right )}}{4} \]

[In]

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

[Out]

-log(x**2 - 1)/4 + log(x**2 + 1)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (6) = 12\).

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12 \[ \int \frac {x}{1-x^4} \, dx=\frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

1/4*log(x^2 + 1) - 1/4*log(x^2 - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (6) = 12\).

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25 \[ \int \frac {x}{1-x^4} \, dx=\frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.74 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x}{1-x^4} \, dx=\frac {\mathrm {atanh}\left (x^2\right )}{2} \]

[In]

int(-x/(x^4 - 1),x)

[Out]

atanh(x^2)/2