3.127 \(\int \frac{1}{a^4-x^4} \, dx\)

Optimal. Leaf size=27 \[ \frac{\tan ^{-1}\left (\frac{x}{a}\right )}{2 a^3}+\frac{\tanh ^{-1}\left (\frac{x}{a}\right )}{2 a^3} \]

[Out]

ArcTan[x/a]/(2*a^3) + ArcTanh[x/a]/(2*a^3)

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Rubi [A]  time = 0.0080693, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {212, 206, 203} \[ \frac{\tan ^{-1}\left (\frac{x}{a}\right )}{2 a^3}+\frac{\tanh ^{-1}\left (\frac{x}{a}\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a^4 - x^4)^(-1),x]

[Out]

ArcTan[x/a]/(2*a^3) + ArcTanh[x/a]/(2*a^3)

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
 !GtQ[a/b, 0]

Rule 206

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

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{1}{a^4-x^4} \, dx &=\frac{\int \frac{1}{a^2-x^2} \, dx}{2 a^2}+\frac{\int \frac{1}{a^2+x^2} \, dx}{2 a^2}\\ &=\frac{\tan ^{-1}\left (\frac{x}{a}\right )}{2 a^3}+\frac{\tanh ^{-1}\left (\frac{x}{a}\right )}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.004278, size = 38, normalized size = 1.41 \[ -\frac{\log (a-x)}{4 a^3}+\frac{\log (a+x)}{4 a^3}+\frac{\tan ^{-1}\left (\frac{x}{a}\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^4 - x^4)^(-1),x]

[Out]

ArcTan[x/a]/(2*a^3) - Log[a - x]/(4*a^3) + Log[a + x]/(4*a^3)

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Maple [A]  time = 0.008, size = 33, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( -a+x \right ) }{4\,{a}^{3}}}+{\frac{1}{2\,{a}^{3}}\arctan \left ({\frac{x}{a}} \right ) }+{\frac{\ln \left ( a+x \right ) }{4\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^4-x^4),x)

[Out]

-1/4/a^3*ln(-a+x)+1/2*arctan(x/a)/a^3+1/4*ln(a+x)/a^3

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Maxima [A]  time = 1.40099, size = 43, normalized size = 1.59 \begin{align*} \frac{\arctan \left (\frac{x}{a}\right )}{2 \, a^{3}} + \frac{\log \left (a + x\right )}{4 \, a^{3}} - \frac{\log \left (-a + x\right )}{4 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(x/a)/a^3 + 1/4*log(a + x)/a^3 - 1/4*log(-a + x)/a^3

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Fricas [A]  time = 2.12863, size = 70, normalized size = 2.59 \begin{align*} \frac{2 \, \arctan \left (\frac{x}{a}\right ) + \log \left (a + x\right ) - \log \left (-a + x\right )}{4 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*arctan(x/a) + log(a + x) - log(-a + x))/a^3

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Sympy [C]  time = 0.13063, size = 37, normalized size = 1.37 \begin{align*} - \frac{\frac{\log{\left (- a + x \right )}}{4} - \frac{\log{\left (a + x \right )}}{4} + \frac{i \log{\left (- i a + x \right )}}{4} - \frac{i \log{\left (i a + x \right )}}{4}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(-a + x)/4 - log(a + x)/4 + I*log(-I*a + x)/4 - I*log(I*a + x)/4)/a**3

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Giac [A]  time = 1.04852, size = 46, normalized size = 1.7 \begin{align*} \frac{\arctan \left (\frac{x}{a}\right )}{2 \, a^{3}} + \frac{\log \left ({\left | a + x \right |}\right )}{4 \, a^{3}} - \frac{\log \left ({\left | -a + x \right |}\right )}{4 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(x/a)/a^3 + 1/4*log(abs(a + x))/a^3 - 1/4*log(abs(-a + x))/a^3