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

Optimal. Leaf size=67 \[ -\frac {x}{2 \sqrt [4]{x^4-1}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1455, 527, 12, 377, 212, 206, 203} \begin {gather*} -\frac {x}{2 \sqrt [4]{x^4-1}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 1455

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :
> Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, n, q, r}, x] && Eq
Q[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx &=\int \frac {-1+2 x^4}{\left (-1+x^4\right )^{5/4} \left (1+x^4\right )} \, dx\\ &=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {1}{2} \int \frac {3}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx\\ &=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3}{2} \int \frac {1}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx\\ &=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1-2 x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 91, normalized size = 1.36 \begin {gather*} \frac {3 \left (-\log \left (1-\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}\right )+\log \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}+1\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}\right )\right )}{8 \sqrt [4]{2}}-\frac {x}{2 \sqrt [4]{x^4-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*x/(-1 + x^4)^(1/4) + (3*(2*ArcTan[(2^(1/4)*x)/(1 - x^4)^(1/4)] - Log[1 - (2^(1/4)*x)/(1 - x^4)^(1/4)] + L
og[1 + (2^(1/4)*x)/(1 - x^4)^(1/4)]))/(8*2^(1/4))

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IntegrateAlgebraic [A]  time = 0.34, size = 67, normalized size = 1.00 \begin {gather*} -\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.52, size = 243, normalized size = 3.63 \begin {gather*} -\frac {12 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \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 ) - 3 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \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 ) + 3 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \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 ) + 16 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(12*2^(3/4)*(x^4 - 1)*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)) - 3*2^(3/4)*(x^4 - 1)*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)) + 3*2^(3/4)*(
x^4 - 1)*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)) + 16*(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 {2 \, x^{4} - 1}{{\left (x^{8} - 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^4-1)/(x^4-1)^(1/4)/(x^8-1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.13, size = 215, normalized size = 3.21

method	result	size
risch	\(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}-\frac {3 \RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {-\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}\)	\(215\)
trager	\(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {3 \RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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