\(\int \frac {x}{a^4-x^4} \, dx\) [128]

Optimal result

Integrand size = 13, antiderivative size = 15 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\text {arctanh}\left (\frac {x^2}{a^2}\right )}{2 a^2} \]

[Out]

1/2*arctanh(x^2/a^2)/a^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {281, 212} \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\text {arctanh}\left (\frac {x^2}{a^2}\right )}{2 a^2} \]

[In]

Int[x/(a^4 - x^4),x]

[Out]

ArcTanh[x^2/a^2]/(2*a^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}{a^4-x^2} \, dx,x,x^2\right ) \\ & = \frac {\text {arctanh}\left (\frac {x^2}{a^2}\right )}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\text {arctanh}\left (\frac {x^2}{a^2}\right )}{2 a^2} \]

[In]

Integrate[x/(a^4 - x^4),x]

[Out]

ArcTanh[x^2/a^2]/(2*a^2)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.80

method	result	size
parallelrisch	\(-\frac {\ln \left (-a +x \right )+\ln \left (a +x \right )-\ln \left (a^{2}+x^{2}\right )}{4 a^{2}}\)	\(27\)
default	\(\frac {\ln \left (a^{2}+x^{2}\right )}{4 a^{2}}-\frac {\ln \left (a^{2}-x^{2}\right )}{4 a^{2}}\)	\(30\)
risch	\(-\frac {\ln \left (-a^{2}+x^{2}\right )}{4 a^{2}}+\frac {\ln \left (a^{2}+x^{2}\right )}{4 a^{2}}\)	\(30\)
norman	\(-\frac {\ln \left (a -x \right )}{4 a^{2}}-\frac {\ln \left (a +x \right )}{4 a^{2}}+\frac {\ln \left (a^{2}+x^{2}\right )}{4 a^{2}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\log \left (a^{2} + x^{2}\right ) - \log \left (-a^{2} + x^{2}\right )}{4 \, a^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60 \[ \int \frac {x}{a^4-x^4} \, dx=- \frac {\frac {\log {\left (- a^{2} + x^{2} \right )}}{4} - \frac {\log {\left (a^{2} + x^{2} \right )}}{4}}{a^{2}} \]

[In]

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

[Out]

-(log(-a**2 + x**2)/4 - log(a**2 + x**2)/4)/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (13) = 26\).

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.93 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\log \left (a^{2} + x^{2}\right )}{4 \, a^{2}} - \frac {\log \left (-a^{2} + x^{2}\right )}{4 \, a^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (13) = 26\).

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.00 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\log \left (a^{2} + x^{2}\right )}{4 \, a^{2}} - \frac {\log \left ({\left | -a^{2} + x^{2} \right |}\right )}{4 \, a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {x}{a^4-x^4} \, dx=\frac {\mathrm {atanh}\left (\frac {x^2}{a^2}\right )}{2\,a^2} \]

[In]

int(x/(a^4 - x^4),x)

[Out]

atanh(x^2/a^2)/(2*a^2)