\(\int \frac {x^3}{1+3 x^4+x^8} \, dx\) [372]

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

[In]

Int[x^3/(1 + 3*x^4 + x^8),x]

[Out]

-1/2*ArcTanh[(3 + 2*x^4)/Sqrt[5]]/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 = 0.01 (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 = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{4}+3\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{10}\)	\(19\)
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*arctanh(1/5*(2*x^4+3)*5^(1/2))*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 = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {x^3}{1+3 x^4+x^8} \, dx=\frac {\sqrt {5} \log {\left (x^{4} - \frac {\sqrt {5}}{2} + \frac {3}{2} \right )}}{20} - \frac {\sqrt {5} \log {\left (x^{4} + \frac {\sqrt {5}}{2} + \frac {3}{2} \right )}}{20} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {x^3}{1+3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {x^3}{1+3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {x^3}{1+3 x^4+x^8} \, dx=\frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {8\,\sqrt {5}\,x^4+3\,\sqrt {5}}{18\,x^4+7}\right )}{10} \]

[In]

int(x^3/(3*x^4 + x^8 + 1),x)

[Out]

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