\(\int \frac {x^4}{-1+x^4} \, dx\) [206]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (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 = 14 \[ \int \frac {x^4}{-1+x^4} \, dx=x-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2} \]

[Out]

x-1/2*arctan(x)-1/2*arctanh(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {327, 218, 212, 209} \[ \int \frac {x^4}{-1+x^4} \, dx=-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2}+x \]

[In]

Int[x^4/(-1 + x^4),x]

[Out]

x - ArcTan[x]/2 - ArcTanh[x]/2

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]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = x+\int \frac {1}{-1+x^4} \, dx \\ & = x-\frac {1}{2} \int \frac {1}{1-x^2} \, dx-\frac {1}{2} \int \frac {1}{1+x^2} \, dx \\ & = x-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {x^4}{-1+x^4} \, dx=x-\frac {\arctan (x)}{2}+\frac {1}{4} \log (1-x)-\frac {1}{4} \log (1+x) \]

[In]

Integrate[x^4/(-1 + x^4),x]

[Out]

x - ArcTan[x]/2 + Log[1 - x]/4 - Log[1 + x]/4

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36

method result size
default \(x +\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (1+x \right )}{4}-\frac {\arctan \left (x \right )}{2}\) \(19\)
risch \(x +\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (1+x \right )}{4}-\frac {\arctan \left (x \right )}{2}\) \(19\)
parallelrisch \(x +\frac {i \ln \left (x -i\right )}{4}-\frac {i \ln \left (x +i\right )}{4}-\frac {\ln \left (1+x \right )}{4}+\frac {\ln \left (-1+x \right )}{4}\) \(31\)
meijerg \(-\frac {\left (-1\right )^{\frac {3}{4}} \left (4 \left (-1\right )^{\frac {1}{4}} x +\frac {x \left (-1\right )^{\frac {1}{4}} \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{\left (x^{4}\right )^{\frac {1}{4}}}\right )}{4}\) \(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {x^4}{-1+x^4} \, dx=x - \frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {x^4}{-1+x^4} \, dx=x + \frac {\log {\left (x - 1 \right )}}{4} - \frac {\log {\left (x + 1 \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

x + log(x - 1)/4 - log(x + 1)/4 - atan(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {x^4}{-1+x^4} \, dx=x - \frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {x^4}{-1+x^4} \, dx=x - \frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {x^4}{-1+x^4} \, dx=x-\frac {\mathrm {atan}\left (x\right )}{2}-\frac {\mathrm {atanh}\left (x\right )}{2} \]

[In]

int(x^4/(x^4 - 1),x)

[Out]

x - atan(x)/2 - atanh(x)/2