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

Optimal. Leaf size=54 \[ \frac {\left (\left (x^4+1\right )^2\right )^{7/8} \left (\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )\right )}{\left (x^4+1\right )^{7/4}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1343, 240, 212, 206, 203} \begin {gather*} \frac {\sqrt [4]{x^4+1} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [8]{x^8+2 x^4+1}}+\frac {\sqrt [4]{x^4+1} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [8]{x^8+2 x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[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 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 240

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 1343

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*} \int \frac {1}{\sqrt [8]{1+2 x^4+x^8}} \, dx &=\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} \operatorname {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} \operatorname {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} \operatorname {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} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}}+\frac {\sqrt [4]{1+x^4} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [8]{1+2 x^4+x^8}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 70, normalized size = 1.30 \begin {gather*} \frac {\sqrt [4]{x^4+1} \left (-\log \left (1-\frac {x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {x}{\sqrt [4]{x^4+1}}+1\right )+2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )\right )}{4 \sqrt [8]{\left (x^4+1\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 16.20, size = 54, normalized size = 1.00 \begin {gather*} \frac {\left (\left (1+x^4\right )^2\right )^{7/8} \left (\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )\right )}{\left (1+x^4\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 2*x^4 + x^8)^(-1/8),x]

[Out]

(((1 + x^4)^2)^(7/8)*(ArcTan[x/(1 + x^4)^(1/4)]/2 + ArcTanh[x/(1 + x^4)^(1/4)]/2))/(1 + x^4)^(7/4)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 65, normalized size = 1.20 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{8} + 2 \, x^{4} + 1\right )}^{\frac {1}{8}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (x^{8}+2 x^{4}+1\right )^{\frac {1}{8}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x}{{\left (x^{4} + 1\right )}^{\frac {1}{4}}} + \int \frac {x^{4}}{{\left (x^{4} + 1\right )}^{\frac {5}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (x^8+2\,x^4+1\right )}^{1/8}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4 + x^8 + 1)^(1/8),x)

[Out]

int(1/(2*x^4 + x^8 + 1)^(1/8), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [8]{x^{8} + 2 x^{4} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________