\(\int \frac {1}{a^4-x^4} \, dx\) [127]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 27 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\arctan \left (\frac {x}{a}\right )}{2 a^3}+\frac {\text {arctanh}\left (\frac {x}{a}\right )}{2 a^3} \]

[Out]

1/2*arctan(x/a)/a^3+1/2*arctanh(x/a)/a^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {218, 212, 209} \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\arctan \left (\frac {x}{a}\right )}{2 a^3}+\frac {\text {arctanh}\left (\frac {x}{a}\right )}{2 a^3} \]

[In]

Int[(a^4 - x^4)^(-1),x]

[Out]

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

Rule 209

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

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 218

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] &&  !Gt
Q[a/b, 0]

Rubi steps \begin{align*} \text {integral}& = \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 {\arctan \left (\frac {x}{a}\right )}{2 a^3}+\frac {\text {arctanh}\left (\frac {x}{a}\right )}{2 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\arctan \left (\frac {x}{a}\right )}{2 a^3}-\frac {\log (a-x)}{4 a^3}+\frac {\log (a+x)}{4 a^3} \]

[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)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22

method result size
default \(-\frac {\ln \left (a -x \right )}{4 a^{3}}+\frac {\arctan \left (\frac {x}{a}\right )}{2 a^{3}}+\frac {\ln \left (a +x \right )}{4 a^{3}}\) \(33\)
parallelrisch \(-\frac {i \ln \left (-i a +x \right )-i \ln \left (i a +x \right )+\ln \left (-a +x \right )-\ln \left (a +x \right )}{4 a^{3}}\) \(39\)
risch \(\frac {\ln \left (a +x \right )}{4 a^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{6} \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R} \,a^{4}+x \right )\right )}{4}-\frac {\ln \left (-a +x \right )}{4 a^{3}}\) \(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {2 \, \arctan \left (\frac {x}{a}\right ) + \log \left (a + x\right ) - \log \left (-a + x\right )}{4 \, a^{3}} \]

[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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {1}{a^4-x^4} \, dx=- \frac {\frac {\log {\left (- a + x \right )}}{4} - \frac {\log {\left (a + x \right )}}{4} + \frac {i \log {\left (- i a + x \right )}}{4} - \frac {i \log {\left (i a + x \right )}}{4}}{a^{3}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\arctan \left (\frac {x}{a}\right )}{2 \, a^{3}} + \frac {\log \left (a + x\right )}{4 \, a^{3}} - \frac {\log \left (-a + x\right )}{4 \, a^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\arctan \left (\frac {x}{a}\right )}{2 \, a^{3}} + \frac {\log \left ({\left | a + x \right |}\right )}{4 \, a^{3}} - \frac {\log \left ({\left | -a + x \right |}\right )}{4 \, a^{3}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {1}{a^4-x^4} \, dx=\frac {\mathrm {atan}\left (\frac {x}{a}\right )+\mathrm {atanh}\left (\frac {x}{a}\right )}{2\,a^3} \]

[In]

int(1/(a^4 - x^4),x)

[Out]

(atan(x/a) + atanh(x/a))/(2*a^3)