3.134 \(\int \frac{x}{a^4+x^4} \, dx\)

Optimal. Leaf size=15 \[ \frac{\tan ^{-1}\left (\frac{x^2}{a^2}\right )}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0055853, antiderivative size = 15, normalized size of antiderivative = 1., 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 = {275, 203} \[ \frac{\tan ^{-1}\left (\frac{x^2}{a^2}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a^4 + x^4),x]

[Out]

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

Rule 275

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]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0034063, size = 15, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{x^2}{a^2}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a^4 + x^4),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 14, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{a}^{2}}\arctan \left ({\frac{{x}^{2}}{{a}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^4+x^4),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.41344, size = 18, normalized size = 1.2 \begin{align*} \frac{\arctan \left (\frac{x^{2}}{a^{2}}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.85782, size = 34, normalized size = 2.27 \begin{align*} \frac{\arctan \left (\frac{x^{2}}{a^{2}}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 0.12741, size = 29, normalized size = 1.93 \begin{align*} \frac{- \frac{i \log{\left (- i a^{2} + x^{2} \right )}}{4} + \frac{i \log{\left (i a^{2} + x^{2} \right )}}{4}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.07166, size = 18, normalized size = 1.2 \begin{align*} \frac{\arctan \left (\frac{x^{2}}{a^{2}}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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