\(\int \frac {x^3}{1-3 x^4+x^8} \, dx\) [390]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F(-1)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 23 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {\text {arctanh}\left (\frac {3-2 x^4}{\sqrt {5}}\right )}{2 \sqrt {5}} \]

[Out]

1/10*arctanh(1/5*(-2*x^4+3)*5^(1/2))*5^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1366, 632, 212} \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {\text {arctanh}\left (\frac {3-2 x^4}{\sqrt {5}}\right )}{2 \sqrt {5}} \]

[In]

Int[x^3/(1 - 3*x^4 + x^8),x]

[Out]

ArcTanh[(3 - 2*x^4)/Sqrt[5]]/(2*Sqrt[5])

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
 c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{1-3 x+x^2} \, dx,x,x^4\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{5-x^2} \, dx,x,-3+2 x^4\right )\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {3-2 x^4}{\sqrt {5}}\right )}{2 \sqrt {5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {\log \left (3+\sqrt {5}-2 x^4\right )-\log \left (-3+\sqrt {5}+2 x^4\right )}{4 \sqrt {5}} \]

[In]

Integrate[x^3/(1 - 3*x^4 + x^8),x]

[Out]

(Log[3 + Sqrt[5] - 2*x^4] - Log[-3 + Sqrt[5] + 2*x^4])/(4*Sqrt[5])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method result size
default \(-\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x^{4}-3\right ) \sqrt {5}}{5}\right )}{10}\) \(19\)
risch \(\frac {\ln \left (2 x^{4}-\sqrt {5}-3\right ) \sqrt {5}}{20}-\frac {\ln \left (2 x^{4}+\sqrt {5}-3\right ) \sqrt {5}}{20}\) \(36\)

[In]

int(x^3/(x^8-3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/10*5^(1/2)*arctanh(1/5*(2*x^4-3)*5^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (18) = 36\).

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {2 \, x^{8} - 6 \, x^{4} - \sqrt {5} {\left (2 \, x^{4} - 3\right )} + 7}{x^{8} - 3 \, x^{4} + 1}\right ) \]

[In]

integrate(x^3/(x^8-3*x^4+1),x, algorithm="fricas")

[Out]

1/20*sqrt(5)*log((2*x^8 - 6*x^4 - sqrt(5)*(2*x^4 - 3) + 7)/(x^8 - 3*x^4 + 1))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {\sqrt {5} \log {\left (x^{4} - \frac {3}{2} - \frac {\sqrt {5}}{2} \right )}}{20} - \frac {\sqrt {5} \log {\left (x^{4} - \frac {3}{2} + \frac {\sqrt {5}}{2} \right )}}{20} \]

[In]

integrate(x**3/(x**8-3*x**4+1),x)

[Out]

sqrt(5)*log(x**4 - 3/2 - sqrt(5)/2)/20 - sqrt(5)*log(x**4 - 3/2 + sqrt(5)/2)/20

Maxima [F(-1)]

Timed out. \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\text {Timed out} \]

[In]

integrate(x^3/(x^8-3*x^4+1),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{4} - \sqrt {5} - 3 \right |}}{{\left | 2 \, x^{4} + \sqrt {5} - 3 \right |}}\right ) \]

[In]

integrate(x^3/(x^8-3*x^4+1),x, algorithm="giac")

[Out]

1/20*sqrt(5)*log(abs(2*x^4 - sqrt(5) - 3)/abs(2*x^4 + sqrt(5) - 3))

Mupad [B] (verification not implemented)

Time = 8.64 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {x^3}{1-3 x^4+x^8} \, dx=\frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {3\,\sqrt {5}-8\,\sqrt {5}\,x^4}{18\,x^4-7}\right )}{10} \]

[In]

int(x^3/(x^8 - 3*x^4 + 1),x)

[Out]

(5^(1/2)*atanh((3*5^(1/2) - 8*5^(1/2)*x^4)/(18*x^4 - 7)))/10