\(\int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx\) [697]

Optimal result

Integrand size = 14, antiderivative size = 54 \[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\frac {\left (\left (1+x^4\right )^2\right )^{7/8} \left (\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )\right )}{\left (1+x^4\right )^{7/4}} \]

[Out]

((x^4+1)^2)^(7/8)*(1/2*arctan(x/(x^4+1)^(1/4))+1/2*arctanh(x/(x^4+1)^(1/4)))/(x^4+1)^(7/4)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1357, 246, 218, 212, 209} \[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\frac {\sqrt [4]{x^4+1} \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [8]{x^8+2 x^4+1}}+\frac {\sqrt [4]{x^4+1} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [8]{x^8+2 x^4+1}} \]

[In]

Int[(1 + 2*x^4 + x^8)^(-1/8),x]

[Out]

((1 + x^4)^(1/4)*ArcTan[x/(1 + x^4)^(1/4)])/(2*(1 + 2*x^4 + x^8)^(1/8)) + ((1 + x^4)^(1/4)*ArcTanh[x/(1 + x^4)
^(1/4)])/(2*(1 + 2*x^4 + x^8)^(1/8))

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{2+2 x^4} \int \frac {1}{\sqrt [4]{2+2 x^4}} \, dx}{\sqrt [8]{1+2 x^4+x^8}} \\ & = \frac {\sqrt [4]{2+2 x^4} \text {Subst}\left (\int \frac {1}{1-2 x^4} \, dx,x,\frac {x}{\sqrt [4]{2+2 x^4}}\right )}{\sqrt [8]{1+2 x^4+x^8}} \\ & = \frac {\sqrt [4]{2+2 x^4} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{2+2 x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}}+\frac {\sqrt [4]{2+2 x^4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{2+2 x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}} \\ & = \frac {\sqrt [4]{1+x^4} \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}}+\frac {\sqrt [4]{1+x^4} \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\frac {\sqrt [4]{1+x^4} \left (2 \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\log \left (1-\frac {x}{\sqrt [4]{1+x^4}}\right )+\log \left (1+\frac {x}{\sqrt [4]{1+x^4}}\right )\right )}{4 \sqrt [8]{\left (1+x^4\right )^2}} \]

[In]

Integrate[(1 + 2*x^4 + x^8)^(-1/8),x]

[Out]

((1 + x^4)^(1/4)*(2*ArcTan[x/(1 + x^4)^(1/4)] - Log[1 - x/(1 + x^4)^(1/4)] + Log[1 + x/(1 + x^4)^(1/4)]))/(4*(
(1 + x^4)^2)^(1/8))

Maple [F]

\[\int \frac {1}{\left (x^{8}+2 x^{4}+1\right )^{\frac {1}{8}}}d x\]

[In]

int(1/(x^8+2*x^4+1)^(1/8),x)

[Out]

int(1/(x^8+2*x^4+1)^(1/8),x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {{\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {x + {\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}}{x}\right ) - \frac {1}{4} \, \log \left (-\frac {x - {\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}}{x}\right ) \]

[In]

integrate(1/(x^8+2*x^4+1)^(1/8),x, algorithm="fricas")

[Out]

-1/2*arctan((x^8 + 2*x^4 + 1)^(1/8)/x) + 1/4*log((x + (x^8 + 2*x^4 + 1)^(1/8))/x) - 1/4*log(-(x - (x^8 + 2*x^4
+ 1)^(1/8))/x)

Sympy [F]

\[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\int \frac {1}{\sqrt [8]{x^{8} + 2 x^{4} + 1}}\, dx \]

[In]

integrate(1/(x**8+2*x**4+1)**(1/8),x)

[Out]

Integral((x**8 + 2*x**4 + 1)**(-1/8), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\int { \frac {1}{{\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}} \,d x } \]

[In]

integrate(1/(x^8+2*x^4+1)^(1/8),x, algorithm="maxima")

[Out]

x/(x^4 + 1)^(1/4) + integrate(x^4/(x^4 + 1)^(5/4), x)

Giac [F]

\[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\int { \frac {1}{{\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}} \,d x } \]

[In]

integrate(1/(x^8+2*x^4+1)^(1/8),x, algorithm="giac")

[Out]

integrate((x^8 + 2*x^4 + 1)^(-1/8), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx=\int \frac {1}{{\left (x^8+2\,x^4+1\right )}^{1/8}} \,d x \]

[In]

int(1/(2*x^4 + x^8 + 1)^(1/8),x)

[Out]

int(1/(2*x^4 + x^8 + 1)^(1/8), x)