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

Optimal. Leaf size=105 \[ \frac {1}{3} \sqrt [4]{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}+\frac {1}{3} \sqrt [4]{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6728, 377, 212, 206, 203} \begin {gather*} \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}}+\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)/((1 + x^4)^(1/4)*(-2 + x^4 + x^8)),x]

[Out]

(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)) + (2^(1/4)*A
rcTanh[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 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}{\sqrt [4]{1+x^4} \left (-2+x^4+x^8\right )} \, dx &=\int \left (-\frac {2}{3 \sqrt [4]{1+x^4} \left (-2+2 x^4\right )}+\frac {8}{3 \sqrt [4]{1+x^4} \left (4+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\\ &=-\left (\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-2+4 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\right )+\frac {8}{3} \operatorname {Subst}\left (\int \frac {1}{4-2 x^4} \, 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}{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 )\\ &=\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}}+\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 = 0.06, size = 94, normalized size = 0.90 \begin {gather*} \frac {4 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )+4 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 105, normalized size = 1.00 \begin {gather*} \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}}+\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)/((1 + x^4)^(1/4)*(-2 + x^4 + x^8)),x]

[Out]

(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)) + (2^(1/4)*A
rcTanh[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 = 9.74, size = 416, normalized size = 3.96 \begin {gather*} -\frac {1}{12} \cdot 8^{\frac {3}{4}} \arctan \left (-\frac {8^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 8^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x - 2^{\frac {1}{4}} {\left (8^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {1}{4}} {\left (3 \, x^{4} + 2\right )}\right )}}{2 \, {\left (x^{4} + 2\right )}}\right ) - \frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {1}{4}} {\left (3 \, x^{4} + 1\right )}\right )}}{2 \, {\left (x^{4} - 1\right )}}\right ) + \frac {1}{48} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - \frac {1}{48} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) + \frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - \frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*8^(3/4)*arctan(-1/2*(8^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*8^(1/4)*(x^4 + 1)^(3/4)*x - 2^(1/4)*(8^(3/4)*sqrt(x
^4 + 1)*x^2 + 8^(1/4)*(3*x^4 + 2)))/(x^4 + 2)) - 1/12*2^(3/4)*arctan(1/2*(4*2^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*2^
(1/4)*(x^4 + 1)^(3/4)*x + 2^(3/4)*(2*2^(3/4)*sqrt(x^4 + 1)*x^2 + 2^(1/4)*(3*x^4 + 1)))/(x^4 - 1)) + 1/48*8^(3/
4)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 8*8^(1/4)*sqrt(x^4 + 1)*x^2 + 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)
*x)/(x^4 + 2)) - 1/48*8^(3/4)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*x^
4 + 2) + 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) + 1/48*2^(3/4)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4)*sqrt(x
^4 + 1)*x^2 + 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 1/48*2^(3/4)*log((4*sqrt(2)*(x^4 + 1)^(1
/4)*x^3 - 4*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1))

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

[Out]

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

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maple [C]  time = 2.93, size = 457, normalized size = 4.35

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+3 \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{4}+4 \RootOf \left (\textit {\_Z}^{4}-8\right ) \left (x^{4}+1\right )^{\frac {3}{4}} x +8 x^{2} \sqrt {x^{4}+1}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{24}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{24}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{4}-4 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-8 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \sqrt {x^{4}+1}\, x^{2}+16 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}}{x^{4}+2}\right )}{24}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right )^{3} \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{4}+4 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+8 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{2}+16 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \RootOf \left (\textit {\_Z}^{4}-8\right )^{3}}{x^{4}+2}\right )}{24}\)	\(457\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*RootOf(_Z^4-8)*ln((2*RootOf(_Z^4-8)^3*(x^4+1)^(1/4)*x^3+3*RootOf(_Z^4-8)^2*x^4+4*RootOf(_Z^4-8)*(x^4+1)^(
3/4)*x+8*x^2*(x^4+1)^(1/2)+RootOf(_Z^4-8)^2)/(-1+x)/(1+x)/(x^2+1))+1/24*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(((x^4
+1)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootOf(_Z^4-8)^2*x^2+2*(x^4+1)^(1/4)*RootOf(_Z^4-8)^2*x^3-3*RootOf(_Z^
2+RootOf(_Z^4-8)^2)*x^4-4*(x^4+1)^(3/4)*x-RootOf(_Z^2+RootOf(_Z^4-8)^2))/(-1+x)/(1+x)/(x^2+1))+1/24*RootOf(_Z^
2+RootOf(_Z^4-8)^2)*RootOf(_Z^4-8)^2*ln((3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootOf(_Z^4-8)^2*x^4-4*(x^4+1)^(1/4)*
RootOf(_Z^4-8)^2*x^3-8*RootOf(_Z^2+RootOf(_Z^4-8)^2)*(x^4+1)^(1/2)*x^2+16*(x^4+1)^(3/4)*x+2*RootOf(_Z^2+RootOf
(_Z^4-8)^2)*RootOf(_Z^4-8)^2)/(x^4+2))+1/24*RootOf(_Z^4-8)^3*ln((3*RootOf(_Z^4-8)^3*x^4+4*(x^4+1)^(1/4)*RootOf
(_Z^4-8)^2*x^3+8*(x^4+1)^(1/2)*RootOf(_Z^4-8)*x^2+16*(x^4+1)^(3/4)*x+2*RootOf(_Z^4-8)^3)/(x^4+2))

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

[Out]

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

[Out]

int((x^4 - 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 {x^{4} - 2}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt [4]{x^{4} + 1} \left (x^{4} + 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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