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

Optimal. Leaf size=107 \[ \frac {\left (x^4-x^2\right )^{3/4} \left (67 x^4+2 x^2-85\right )}{80 x \left (x^2-1\right )^2 \left (x^2+1\right )}+\frac {15 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^2}}\right )}{32 \sqrt [4]{2}}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-x^2}}\right )}{32 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 194, normalized size of antiderivative = 1.81, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2056, 1254, 466, 414, 527, 12, 377, 212, 206, 203} \begin {gather*} -\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{x^4-x^2}}+\frac {x}{4 \left (1-x^2\right ) \left (x^2+1\right ) \sqrt [4]{x^4-x^2}}+\frac {67 x}{80 \sqrt [4]{x^4-x^2}}+\frac {15 \sqrt [4]{x^2-1} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{32 \sqrt [4]{2} \sqrt [4]{x^4-x^2}}+\frac {15 \sqrt [4]{x^2-1} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{32 \sqrt [4]{2} \sqrt [4]{x^4-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(67*x)/(80*(-x^2 + x^4)^(1/4)) - x/(40*(1 - x^2)*(-x^2 + x^4)^(1/4)) + x/(4*(1 - x^2)*(1 + x^2)*(-x^2 + x^4)^(
1/4)) + (15*Sqrt[x]*(-1 + x^2)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(32*2^(1/4)*(-x^2 + x^4)^(1/4
)) + (15*Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[(2^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(32*2^(1/4)*(-x^2 + x^4)^(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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

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 1254

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(f*x)^m*(d +
e*x^2)^(q + p)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, c, d, e, f, q, m, q}, x] && EqQ[c*d^2 + a*e^2, 0] && Integ
erQ[p]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {1}{\left (-1+x^4\right )^2 \sqrt [4]{-x^2+x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{-1+x^2} \left (-1+x^4\right )^2} \, dx}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (-1+x^2\right )^{9/4} \left (1+x^2\right )^2} \, dx}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^4\right )^{9/4} \left (1+x^4\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {7-8 x^4}{\left (-1+x^4\right )^{9/4} \left (1+x^4\right )} \, dx,x,\sqrt {x}\right )}{4 \sqrt [4]{-x^2+x^4}}\\ &=-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {-71+4 x^4}{\left (-1+x^4\right )^{5/4} \left (1+x^4\right )} \, dx,x,\sqrt {x}\right )}{40 \sqrt [4]{-x^2+x^4}}\\ &=\frac {67 x}{80 \sqrt [4]{-x^2+x^4}}-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {75}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx,x,\sqrt {x}\right )}{80 \sqrt [4]{-x^2+x^4}}\\ &=\frac {67 x}{80 \sqrt [4]{-x^2+x^4}}-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (15 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx,x,\sqrt {x}\right )}{16 \sqrt [4]{-x^2+x^4}}\\ &=\frac {67 x}{80 \sqrt [4]{-x^2+x^4}}-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (15 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-2 x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{16 \sqrt [4]{-x^2+x^4}}\\ &=\frac {67 x}{80 \sqrt [4]{-x^2+x^4}}-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {\left (15 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \sqrt [4]{-x^2+x^4}}+\frac {\left (15 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \sqrt [4]{-x^2+x^4}}\\ &=\frac {67 x}{80 \sqrt [4]{-x^2+x^4}}-\frac {x}{40 \left (1-x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {x}{4 \left (1-x^2\right ) \left (1+x^2\right ) \sqrt [4]{-x^2+x^4}}+\frac {15 \sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}+\frac {15 \sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \sqrt [4]{2} \sqrt [4]{-x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.56, size = 92, normalized size = 0.86 \begin {gather*} \frac {x^3 \left (75 \sqrt [4]{1-x^2} \left (x^4-1\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {2 x^2}{x^2+1}\right )+\sqrt [4]{x^2+1} \left (67 x^4+2 x^2-85\right )\right )}{80 \left (x^2 \left (x^2-1\right )\right )^{5/4} \left (x^2+1\right )^{5/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*((1 + x^2)^(1/4)*(-85 + 2*x^2 + 67*x^4) + 75*(1 - x^2)^(1/4)*(-1 + x^4)*Hypergeometric2F1[1/4, 1/4, 5/4,
(2*x^2)/(1 + x^2)]))/(80*(x^2*(-1 + x^2))^(5/4)*(1 + x^2)^(5/4))

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IntegrateAlgebraic [A]  time = 0.38, size = 107, normalized size = 1.00 \begin {gather*} \frac {\left (-x^2+x^4\right )^{3/4} \left (-85+2 x^2+67 x^4\right )}{80 x \left (-1+x^2\right )^2 \left (1+x^2\right )}+\frac {15 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{32 \sqrt [4]{2}}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{32 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-x^2 + x^4)^(3/4)*(-85 + 2*x^2 + 67*x^4))/(80*x*(-1 + x^2)^2*(1 + x^2)) + (15*ArcTan[(2^(1/4)*x)/(-x^2 + x^4
)^(1/4)])/(32*2^(1/4)) + (15*ArcTanh[(2^(1/4)*x)/(-x^2 + x^4)^(1/4)])/(32*2^(1/4))

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fricas [B]  time = 2.36, size = 331, normalized size = 3.09 \begin {gather*} -\frac {300 \cdot 2^{\frac {3}{4}} {\left (x^{7} - x^{5} - x^{3} + x\right )} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} - x^{2}} x + 2^{\frac {1}{4}} {\left (3 \, x^{3} - x\right )}\right )} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{3} + x\right )}}\right ) - 75 \cdot 2^{\frac {3}{4}} {\left (x^{7} - x^{5} - x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} - x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x^{2}} x + 4 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x^{3} + x}\right ) + 75 \cdot 2^{\frac {3}{4}} {\left (x^{7} - x^{5} - x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} - x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x^{2}} x + 4 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x^{3} + x}\right ) - 16 \, {\left (67 \, x^{4} + 2 \, x^{2} - 85\right )} {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{1280 \, {\left (x^{7} - x^{5} - x^{3} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(300*2^(3/4)*(x^7 - x^5 - x^3 + x)*arctan(1/2*(4*2^(3/4)*(x^4 - x^2)^(1/4)*x^2 + 2^(3/4)*(2*2^(3/4)*sq
rt(x^4 - x^2)*x + 2^(1/4)*(3*x^3 - x)) + 4*2^(1/4)*(x^4 - x^2)^(3/4))/(x^3 + x)) - 75*2^(3/4)*(x^7 - x^5 - x^3
+ x)*log((4*sqrt(2)*(x^4 - x^2)^(1/4)*x^2 + 2^(3/4)*(3*x^3 - x) + 4*2^(1/4)*sqrt(x^4 - x^2)*x + 4*(x^4 - x^2)
^(3/4))/(x^3 + x)) + 75*2^(3/4)*(x^7 - x^5 - x^3 + x)*log((4*sqrt(2)*(x^4 - x^2)^(1/4)*x^2 - 2^(3/4)*(3*x^3 -
x) - 4*2^(1/4)*sqrt(x^4 - x^2)*x + 4*(x^4 - x^2)^(3/4))/(x^3 + x)) - 16*(67*x^4 + 2*x^2 - 85)*(x^4 - x^2)^(3/4
))/(x^7 - x^5 - x^3 + x)

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giac [A]  time = 0.32, size = 104, normalized size = 0.97 \begin {gather*} \frac {15}{64} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {15}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {15}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} - {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {\frac {10}{x^{2}} - 9}{10 \, {\left (\frac {1}{x^{2}} - 1\right )} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} + \frac {{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}}}{16 \, {\left (\frac {1}{x^{2}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/64*2^(3/4)*arctan(1/2*2^(3/4)*(-1/x^2 + 1)^(1/4)) - 15/128*2^(3/4)*log(2^(1/4) + (-1/x^2 + 1)^(1/4)) + 15/1
28*2^(3/4)*log(2^(1/4) - (-1/x^2 + 1)^(1/4)) - 1/10*(10/x^2 - 9)/((1/x^2 - 1)*(-1/x^2 + 1)^(1/4)) + 1/16*(-1/x
^2 + 1)^(3/4)/(1/x^2 + 1)

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maple [C]  time = 8.35, size = 273, normalized size = 2.55

method	result	size
risch	\(\frac {x \left (67 x^{4}+2 x^{2}-85\right )}{80 \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}} \left (x^{2}-1\right ) \left (x^{2}+1\right )}-\frac {15 \RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {-\sqrt {x^{4}-x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}-x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{4}-8\right ) x}{x \left (x^{2}+1\right )}\right )}{128}-\frac {15 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}-x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x -2 \left (x^{4}-x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{x \left (x^{2}+1\right )}\right )}{128}\)	\(273\)
trager	\(\frac {\left (x^{4}-x^{2}\right )^{\frac {3}{4}} \left (67 x^{4}+2 x^{2}-85\right )}{80 x \left (x^{2}-1\right )^{2} \left (x^{2}+1\right )}+\frac {15 \RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}-x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{4}-8\right ) x}{x \left (x^{2}+1\right )}\right )}{128}-\frac {15 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}-x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x -2 \left (x^{4}-x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{x \left (x^{2}+1\right )}\right )}{128}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*x*(67*x^4+2*x^2-85)/(x^2*(x^2-1))^(1/4)/(x^2-1)/(x^2+1)-15/128*RootOf(_Z^4-8)*ln((-(x^4-x^2)^(1/2)*RootOf
(_Z^4-8)^3*x+2*(x^4-x^2)^(1/4)*RootOf(_Z^4-8)^2*x^2-3*RootOf(_Z^4-8)*x^3+4*(x^4-x^2)^(3/4)+RootOf(_Z^4-8)*x)/x
/(x^2+1))-15/128*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(((x^4-x^2)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootOf(_Z^4-8
)^2*x-2*(x^4-x^2)^(1/4)*RootOf(_Z^4-8)^2*x^2-3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^3+4*(x^4-x^2)^(3/4)+RootOf(_Z^2
+RootOf(_Z^4-8)^2)*x)/x/(x^2+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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