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

Optimal. Leaf size=85 \[ \frac {3}{4} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4+1\& ,\frac {\text {$\#$1}^3 \log \left (\sqrt [4]{x^4-1}-\text {$\#$1} x\right )-\text {$\#$1}^3 \log (x)}{2 \text {$\#$1}^4-1}\& \right ]+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right ) \]

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Rubi [C]  time = 0.50, antiderivative size = 341, normalized size of antiderivative = 4.01, number of steps used = 26, number of rules used = 11, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {6728, 240, 212, 206, 203, 21, 1428, 408, 377, 208, 205} \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {i \sqrt {3} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt [4]{x^4-1}}\right )}{2 \left (-\frac {-\sqrt {3}+i}{\sqrt {3}+i}\right )^{3/4}}-\frac {1}{2} i \sqrt {3} \left (-\frac {-\sqrt {3}+i}{\sqrt {3}+i}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt [4]{x^4-1}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {i \sqrt {3} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt [4]{x^4-1}}\right )}{2 \left (-\frac {-\sqrt {3}+i}{\sqrt {3}+i}\right )^{3/4}}-\frac {1}{2} i \sqrt {3} \left (-\frac {-\sqrt {3}+i}{\sqrt {3}+i}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt [4]{x^4-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] + ((I/2)*Sqrt[3]*ArcTan[x/((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*(-1 + x^4)^(1/4))
])/(-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4) - (I/2)*Sqrt[3]*(-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4)*ArcTan[((-((I
- Sqrt[3])/(I + Sqrt[3])))^(1/4)*x)/(-1 + x^4)^(1/4)] + ArcTanh[x/(-1 + x^4)^(1/4)] + ((I/2)*Sqrt[3]*ArcTanh[
x/((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*(-1 + x^4)^(1/4))])/(-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4) - (I/2)*S
qrt[3]*(-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4)*ArcTanh[((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*x)/(-1 + x^4)^(1
/4)]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 205

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

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 208

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

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 1428

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[(2*c)/r, Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[(2*c)/r, Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-1+x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (1-x^4+x^8\right )} \, dx &=\int \left (\frac {2}{\sqrt [4]{-1+x^4}}-\frac {3 \left (1-x^4\right )}{\sqrt [4]{-1+x^4} \left (1-x^4+x^8\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx-3 \int \frac {1-x^4}{\sqrt [4]{-1+x^4} \left (1-x^4+x^8\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+3 \int \frac {\left (-1+x^4\right )^{3/4}}{1-x^4+x^8} \, dx\\ &=-\left (\left (2 i \sqrt {3}\right ) \int \frac {\left (-1+x^4\right )^{3/4}}{-1-i \sqrt {3}+2 x^4} \, dx\right )+\left (2 i \sqrt {3}\right ) \int \frac {\left (-1+x^4\right )^{3/4}}{-1+i \sqrt {3}+2 x^4} \, dx+\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\left (-3+i \sqrt {3}\right ) \int \frac {1}{\sqrt [4]{-1+x^4} \left (-1+i \sqrt {3}+2 x^4\right )} \, dx+\left (3+i \sqrt {3}\right ) \int \frac {1}{\sqrt [4]{-1+x^4} \left (-1-i \sqrt {3}+2 x^4\right )} \, dx\\ &=\tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\left (-3+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+i \sqrt {3}-\left (1+i \sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\left (3+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-i \sqrt {3}-\left (1-i \sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-i+\sqrt {3}}-\sqrt {i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {\frac {1}{3} \left (-i+\sqrt {3}\right )}}-\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-i+\sqrt {3}}+\sqrt {i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {\frac {1}{3} \left (-i+\sqrt {3}\right )}}-\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}-\sqrt {-i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {\frac {1}{3} \left (i+\sqrt {3}\right )}}-\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}+\sqrt {-i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {\frac {1}{3} \left (i+\sqrt {3}\right )}}\\ &=\tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\left (3 i-\sqrt {3}\right ) \sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{2 \left (i-\sqrt {3}\right )}-\frac {1}{2} i \sqrt {3} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{-1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\left (3 i-\sqrt {3}\right ) \sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{2 \left (i-\sqrt {3}\right )}-\frac {1}{2} i \sqrt {3} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{-1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.57, size = 317, normalized size = 3.73 \begin {gather*} \frac {1}{2} \left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {i \sqrt {3} \tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}+i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}-i}{\sqrt {3}+i}\right )^{3/4}}-i \sqrt {3} \left (\frac {\sqrt {3}-i}{\sqrt {3}+i}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}+i}} x}{\sqrt [4]{x^4-1}}\right )+2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {i \sqrt {3} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}+i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}-i}{\sqrt {3}+i}\right )^{3/4}}-i \sqrt {3} \left (\frac {\sqrt {3}-i}{\sqrt {3}+i}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}+i}} x}{\sqrt [4]{x^4-1}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[x/(-1 + x^4)^(1/4)] + (I*Sqrt[3]*ArcTan[x/(((-I + Sqrt[3])/(I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))])/
((-I + Sqrt[3])/(I + Sqrt[3]))^(3/4) - I*Sqrt[3]*((-I + Sqrt[3])/(I + Sqrt[3]))^(3/4)*ArcTan[(((-I + Sqrt[3])/
(I + Sqrt[3]))^(1/4)*x)/(-1 + x^4)^(1/4)] + 2*ArcTanh[x/(-1 + x^4)^(1/4)] + (I*Sqrt[3]*ArcTanh[x/(((-I + Sqrt[
3])/(I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))])/((-I + Sqrt[3])/(I + Sqrt[3]))^(3/4) - I*Sqrt[3]*((-I + Sqrt[3])/
(I + Sqrt[3]))^(3/4)*ArcTanh[(((-I + Sqrt[3])/(I + Sqrt[3]))^(1/4)*x)/(-1 + x^4)^(1/4)])/2

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IntegrateAlgebraic [A]  time = 0.31, size = 85, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {3}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}^3+\log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-1+2 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] + ArcTanh[x/(-1 + x^4)^(1/4)] + (3*RootSum[1 - #1^4 + #1^8 & , (-(Log[x]*#1^3) + Lo
g[(-1 + x^4)^(1/4) - x*#1]*#1^3)/(-1 + 2*#1^4) & ])/4

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fricas [B]  time = 0.61, size = 565, normalized size = 6.65 \begin {gather*} -\frac {1}{2} \, \sqrt {3} \sqrt {2} \arctan \left (\frac {3 \, x^{5} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} + 2 \, \sqrt {x^{4} - 1} x^{3} - \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (x^{4} - 3\right )} + {\left (2 \, \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{4} + 4 \, x^{5} + 3 \, \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} + 6 \, \sqrt {x^{4} - 1} x^{3}\right )} \sqrt {-\frac {\sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 2 \, x^{4} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 3 \, \sqrt {x^{4} - 1} x^{2} + 1}{x^{4}}} - 3 \, x}{5 \, x^{5} - 9 \, x}\right ) - \frac {1}{2} \, \sqrt {3} \sqrt {2} \arctan \left (-\frac {3 \, x^{5} - \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} + 2 \, \sqrt {x^{4} - 1} x^{3} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (x^{4} - 3\right )} - {\left (2 \, \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{4} - 4 \, x^{5} + 3 \, \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} - 6 \, \sqrt {x^{4} - 1} x^{3}\right )} \sqrt {\frac {\sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, x^{4} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x + 3 \, \sqrt {x^{4} - 1} x^{2} - 1}{x^{4}}} - 3 \, x}{5 \, x^{5} - 9 \, x}\right ) - \frac {1}{8} \, \sqrt {3} \sqrt {2} \log \left (\frac {9 \, {\left (\sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, x^{4} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x + 3 \, \sqrt {x^{4} - 1} x^{2} - 1\right )}}{x^{4}}\right ) + \frac {1}{8} \, \sqrt {3} \sqrt {2} \log \left (-\frac {9 \, {\left (\sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 2 \, x^{4} + \sqrt {3} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 3 \, \sqrt {x^{4} - 1} x^{2} + 1\right )}}{x^{4}}\right ) - \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*sqrt(2)*arctan((3*x^5 + sqrt(3)*sqrt(2)*(x^4 - 1)^(3/4)*x^2 + 2*sqrt(x^4 - 1)*x^3 - sqrt(3)*sqrt(
2)*(x^4 - 1)^(1/4)*(x^4 - 3) + (2*sqrt(3)*sqrt(2)*(x^4 - 1)^(1/4)*x^4 + 4*x^5 + 3*sqrt(3)*sqrt(2)*(x^4 - 1)^(3
/4)*x^2 + 6*sqrt(x^4 - 1)*x^3)*sqrt(-(sqrt(3)*sqrt(2)*(x^4 - 1)^(1/4)*x^3 - 2*x^4 + sqrt(3)*sqrt(2)*(x^4 - 1)^
(3/4)*x - 3*sqrt(x^4 - 1)*x^2 + 1)/x^4) - 3*x)/(5*x^5 - 9*x)) - 1/2*sqrt(3)*sqrt(2)*arctan(-(3*x^5 - sqrt(3)*s
qrt(2)*(x^4 - 1)^(3/4)*x^2 + 2*sqrt(x^4 - 1)*x^3 + sqrt(3)*sqrt(2)*(x^4 - 1)^(1/4)*(x^4 - 3) - (2*sqrt(3)*sqrt
(2)*(x^4 - 1)^(1/4)*x^4 - 4*x^5 + 3*sqrt(3)*sqrt(2)*(x^4 - 1)^(3/4)*x^2 - 6*sqrt(x^4 - 1)*x^3)*sqrt((sqrt(3)*s
qrt(2)*(x^4 - 1)^(1/4)*x^3 + 2*x^4 + sqrt(3)*sqrt(2)*(x^4 - 1)^(3/4)*x + 3*sqrt(x^4 - 1)*x^2 - 1)/x^4) - 3*x)/
(5*x^5 - 9*x)) - 1/8*sqrt(3)*sqrt(2)*log(9*(sqrt(3)*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 2*x^4 + sqrt(3)*sqrt(2)*(x^4
- 1)^(3/4)*x + 3*sqrt(x^4 - 1)*x^2 - 1)/x^4) + 1/8*sqrt(3)*sqrt(2)*log(-9*(sqrt(3)*sqrt(2)*(x^4 - 1)^(1/4)*x^
3 - 2*x^4 + sqrt(3)*sqrt(2)*(x^4 - 1)^(3/4)*x - 3*sqrt(x^4 - 1)*x^2 + 1)/x^4) - arctan((x^4 - 1)^(1/4)/x) + 1/
2*log((x + (x^4 - 1)^(1/4))/x) - 1/2*log(-(x - (x^4 - 1)^(1/4))/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 5.63, size = 474, normalized size = 5.58

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{2}-\frac {\ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-2 x^{4}+1\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (-\frac {-6 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}} x^{7}+5 \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x^{8}+6 \left (x^{4}-1\right )^{\frac {3}{4}} x^{5}+6 \RootOf \left (\textit {\_Z}^{2}+1\right ) \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \sqrt {x^{4}-1}\, x^{2}+12 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}} x^{3}-5 \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x^{4}+6 \left (x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{x^{8}-x^{4}+1}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (-\frac {-5 \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{8}+6 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}} x^{7}+6 \left (x^{4}-1\right )^{\frac {3}{4}} x^{5}+5 \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-6 \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \sqrt {x^{4}-1}\, x^{2}-12 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}} x^{3}+6 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+1\right ) \RootOf \left (\textit {\_Z}^{2}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{x^{8}-x^{4}+1}\right )}{4}\)	\(474\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8+x^4-1)/(x^4-1)^(1/4)/(x^8-x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/2*RootOf(_Z^2+1)*ln(-2*RootOf(_Z^2+1)*(x^4-1)^(1/2)*x^2+2*RootOf(_Z^2+1)*x^4+2*(x^4-1)^(3/4)*x-2*x^3*(x^4-1)
^(1/4)-RootOf(_Z^2+1))-1/2*ln(2*(x^4-1)^(3/4)*x-2*x^2*(x^4-1)^(1/2)+2*x^3*(x^4-1)^(1/4)-2*x^4+1)-1/4*RootOf(_Z
^2+3*RootOf(_Z^2+1))*ln(-(-6*RootOf(_Z^2+1)*(x^4-1)^(1/4)*x^7+5*RootOf(_Z^2+3*RootOf(_Z^2+1))*x^8+6*(x^4-1)^(3
/4)*x^5+6*RootOf(_Z^2+1)*RootOf(_Z^2+3*RootOf(_Z^2+1))*(x^4-1)^(1/2)*x^2+12*RootOf(_Z^2+1)*(x^4-1)^(1/4)*x^3-5
*RootOf(_Z^2+3*RootOf(_Z^2+1))*x^4+6*(x^4-1)^(3/4)*x-RootOf(_Z^2+3*RootOf(_Z^2+1)))/(x^8-x^4+1))+1/4*RootOf(_Z
^2+1)*RootOf(_Z^2+3*RootOf(_Z^2+1))*ln(-(-5*RootOf(_Z^2+3*RootOf(_Z^2+1))*RootOf(_Z^2+1)*x^8+6*RootOf(_Z^2+1)*
(x^4-1)^(1/4)*x^7+6*(x^4-1)^(3/4)*x^5+5*RootOf(_Z^2+3*RootOf(_Z^2+1))*RootOf(_Z^2+1)*x^4-6*RootOf(_Z^2+3*RootO
f(_Z^2+1))*(x^4-1)^(1/2)*x^2-12*RootOf(_Z^2+1)*(x^4-1)^(1/4)*x^3+6*(x^4-1)^(3/4)*x+RootOf(_Z^2+1)*RootOf(_Z^2+
3*RootOf(_Z^2+1)))/(x^8-x^4+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {2\,x^8+x^4-1}{{\left (x^4-1\right )}^{1/4}\,\left (x^8-x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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