3.2.3 \(\int \frac {x}{1-x^4} \, dx\) [103]

Optimal. Leaf size=20 \[ \frac {1}{4} \log \left (\frac {1+x^2}{1-x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 8, normalized size of antiderivative = 0.40, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 212} \begin {gather*} \frac {\text {arctanh}\left (x^2\right )}{2} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 23, normalized size = 1.15 \begin {gather*} -\frac {1}{4} \log \left (1-x^2\right )+\frac {1}{4} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(1 - x^4),x]

[Out]

-1/4*Log[1 - x^2] + Log[1 + x^2]/4

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Maple [A]
time = 0.02, size = 22, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(x^2 + 1) - 1/4*log(x^2 - 1)

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Fricas [A]
time = 0.57, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(x^2 + 1) - 1/4*log(x^2 - 1)

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Sympy [A]
time = 0.03, size = 15, normalized size = 0.75 \begin {gather*} - \frac {\log {\left (x^{2} - 1 \right )}}{4} + \frac {\log {\left (x^{2} + 1 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x**2 - 1)/4 + log(x**2 + 1)/4

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Giac [A]
time = 0.46, size = 18, normalized size = 0.90 \begin {gather*} \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 6, normalized size = 0.30 \begin {gather*} \frac {\mathrm {atanh}\left (x^2\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x/(x^4 - 1),x)

[Out]

atanh(x^2)/2

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