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

Optimal. Leaf size=129 \[ \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.35, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6728, 240, 212, 206, 203, 377} \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTan[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTan[(2^(1/4)*x)/(-1 + x^4)^
(1/4)]/(6*2^(1/4)) + ArcTanh[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTanh[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTanh
[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(6*2^(1/4))

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 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 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 {-2-x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-2-x^4+x^8\right )} \, dx &=\int \left (\frac {2}{\sqrt [4]{-1+x^4}}+\frac {2+x^4}{\sqrt [4]{-1+x^4} \left (-2-x^4+x^8\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx+\int \frac {2+x^4}{\sqrt [4]{-1+x^4} \left (-2-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 )+\int \left (\frac {8}{3 \sqrt [4]{-1+x^4} \left (-4+2 x^4\right )}-\frac {2}{3 \sqrt [4]{-1+x^4} \left (2+2 x^4\right )}\right ) \, dx\\ &=-\left (\frac {2}{3} \int \frac {1}{\sqrt [4]{-1+x^4} \left (2+2 x^4\right )} \, dx\right )+\frac {8}{3} \int \frac {1}{\sqrt [4]{-1+x^4} \left (-4+2 x^4\right )} \, 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 )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{2-4 x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {8}{3} \operatorname {Subst}\left (\int \frac {1}{-4+2 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 {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt {2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt {2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{6 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{6 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 129, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4-1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{6 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTan[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTan[(2^(1/4)*x)/(-1 + x^4)^
(1/4)]/(6*2^(1/4)) + ArcTanh[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTanh[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTanh
[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(6*2^(1/4))

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IntegrateAlgebraic [A]  time = 0.45, size = 129, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{6 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{-1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{6 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTan[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTan[(2^(1/4)*x)/(-1 + x^4)^
(1/4)]/(6*2^(1/4)) + ArcTanh[x/(-1 + x^4)^(1/4)] - (2^(1/4)*ArcTanh[x/(2^(1/4)*(-1 + x^4)^(1/4))])/3 - ArcTanh
[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(6*2^(1/4))

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fricas [B]  time = 0.60, size = 255, normalized size = 1.98 \begin {gather*} -\frac {1}{6} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {8^{\frac {3}{4}} \sqrt {2} x \sqrt {\frac {\sqrt {2} x^{2} + 2 \, \sqrt {x^{4} - 1}}{x^{2}}} - 2 \cdot 8^{\frac {3}{4}} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{8 \, x}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} x \sqrt {\frac {\sqrt {2} x^{2} + \sqrt {x^{4} - 1}}{x^{2}}} - 2^{\frac {3}{4}} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (\frac {8^{\frac {1}{4}} x + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (-\frac {8^{\frac {1}{4}} x - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{24} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\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^4-2)/(x^4-1)^(1/4)/(x^8-x^4-2),x, algorithm="fricas")

[Out]

-1/6*8^(3/4)*arctan(1/8*(8^(3/4)*sqrt(2)*x*sqrt((sqrt(2)*x^2 + 2*sqrt(x^4 - 1))/x^2) - 2*8^(3/4)*(x^4 - 1)^(1/
4))/x) - 1/6*2^(3/4)*arctan(1/2*(2^(3/4)*x*sqrt((sqrt(2)*x^2 + sqrt(x^4 - 1))/x^2) - 2^(3/4)*(x^4 - 1)^(1/4))/
x) - 1/24*8^(3/4)*log((8^(1/4)*x + 2*(x^4 - 1)^(1/4))/x) + 1/24*8^(3/4)*log(-(8^(1/4)*x - 2*(x^4 - 1)^(1/4))/x
) - 1/24*2^(3/4)*log((2^(1/4)*x + (x^4 - 1)^(1/4))/x) + 1/24*2^(3/4)*log(-(2^(1/4)*x - (x^4 - 1)^(1/4))/x) - a
rctan((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} - 2}{{\left (x^{8} - x^{4} - 2\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^4-2)/(x^4-1)^(1/4)/(x^8-x^4-2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-x^{4}-2}{\left (x^{4}-1\right )^{\frac {1}{4}} \left (x^{8}-x^{4}-2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((2*x^8 - x^4 - 2)/((x^8 - x^4 - 2)*(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+2}{{\left (x^4-1\right )}^{1/4}\,\left (-x^8+x^4+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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