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

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

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {x^2}{a^2}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a^4 + x^4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.01, size = 15, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {x^2}{a^2}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(a^4 + x^4),x]

[Out]

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

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fricas [A]  time = 0.95, size = 13, normalized size = 0.87 \[ \frac {\arctan \left (\frac {x^{2}}{a^{2}}\right )}{2 \, a^{2}} \]

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

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giac [A]  time = 0.97, size = 13, normalized size = 0.87 \[ \frac {\arctan \left (\frac {x^{2}}{a^{2}}\right )}{2 \, a^{2}} \]

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

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maple [A]  time = 0.29, size = 14, normalized size = 0.93

method	result	size
default	\(\frac {\arctan \left (\frac {x^{2}}{a^{2}}\right )}{2 a^{2}}\)	\(14\)
risch	\(\frac {\arctan \left (\frac {x^{2}}{a^{2}}\right )}{2 a^{2}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 13, normalized size = 0.87 \[ \frac {\arctan \left (\frac {x^{2}}{a^{2}}\right )}{2 \, a^{2}} \]

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

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mupad [B]  time = 0.19, size = 13, normalized size = 0.87 \[ \frac {\mathrm {atan}\left (\frac {x^2}{a^2}\right )}{2\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^4 + x^4),x)

[Out]

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

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sympy [C]  time = 0.15, size = 29, normalized size = 1.93 \[ \frac {- \frac {i \log {\left (- i a^{2} + x^{2} \right )}}{4} + \frac {i \log {\left (i a^{2} + x^{2} \right )}}{4}}{a^{2}} \]

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

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