\(\int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx\) [1653]

Optimal result

Integrand size = 18, antiderivative size = 112 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=-\arctan \left (\frac {\sqrt [4]{-1+x^4}}{x}\right )-\frac {3 \arctan \left (\frac {-\frac {1}{\sqrt {2}}+\frac {\sqrt {-1+x^4}}{\sqrt {2}}}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {\sqrt [4]{-1+x^4}}{x}\right )+\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}} \]

[Out]

-arctan((x^4-1)^(1/4)/x)-3/4*arctan((-1/2*2^(1/2)+1/2*(x^4-1)^(1/2)*2^(1/2))/(x^4-1)^(1/4))*2^(1/2)+arctanh((x
^4-1)^(1/4)/x)+3/4*arctanh(2^(1/2)*(x^4-1)^(1/4)/(1+(x^4-1)^(1/2)))*2^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.37, number of steps used = 17, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {1847, 246, 218, 212, 209, 272, 65, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=\arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {3 \arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{2 \sqrt {2}}-\frac {3 \arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {3 \log \left (\sqrt {x^4-1}-\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{4 \sqrt {2}}+\frac {3 \log \left (\sqrt {x^4-1}+\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{4 \sqrt {2}} \]

[In]

Int[(-3 + 2*x)/(x*(-1 + x^4)^(1/4)),x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1847

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{\sqrt [4]{-1+x^4}}-\frac {3}{x \sqrt [4]{-1+x^4}}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx-3 \int \frac {1}{x \sqrt [4]{-1+x^4}} \, dx \\ & = -\left (\frac {3}{4} \text {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x} x} \, dx,x,x^4\right )\right )+2 \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = -\left (3 \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )\right )+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {3}{2} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right ) \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{4 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{4 \sqrt {2}} \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3 \log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}} \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {3 \arctan \left (1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}-\frac {3 \arctan \left (1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3 \log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.62 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=-\arctan \left (\frac {\sqrt [4]{-1+x^4}}{x}\right )+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{-1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {\sqrt [4]{-1+x^4}}{x}\right )+\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}} \]

[In]

Integrate[(-3 + 2*x)/(x*(-1 + x^4)^(1/4)),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 9.69 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98

method	result	size
meijerg	\(-\frac {3 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \left (\frac {\pi \sqrt {2}\, x^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], x^{4}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+4 \ln \left (x \right )+i \pi \right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{8 \pi \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {2 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], x^{4}\right )}{\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}\)	\(110\)
trager	\(\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 {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {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}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{2}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) x^{4}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}-2 \left (x^{4}-1\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{x^{4}}\right )}{4}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}-2 \sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{x^{4}}\right )}{4}\)	\(339\)

[In]

int((-3+2*x)/x/(x^4-1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-3/8/Pi*2^(1/2)*GAMMA(3/4)/signum(x^4-1)^(1/4)*(-signum(x^4-1))^(1/4)*(1/4*Pi*2^(1/2)/GAMMA(3/4)*x^4*hypergeom
([1,1,5/4],[2,2],x^4)+(-3*ln(2)-1/2*Pi+4*ln(x)+I*Pi)*Pi*2^(1/2)/GAMMA(3/4))+2/signum(x^4-1)^(1/4)*(-signum(x^4
-1))^(1/4)*x*hypergeom([1/4,1/4],[5/4],x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.77 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.49 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=\left (\frac {3}{16} i + \frac {3}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x^{4} - 2 i - 2\right )} - \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{4} - 1} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} - 4 i \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{4}}\right ) - \left (\frac {3}{16} i - \frac {3}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x^{4} + 2 i - 2\right )} + \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{4} - 1} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} + 4 i \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{4}}\right ) + \left (\frac {3}{16} i - \frac {3}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x^{4} - 2 i + 2\right )} - \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{4} - 1} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} + 4 i \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{4}}\right ) - \left (\frac {3}{16} i + \frac {3}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x^{4} + 2 i + 2\right )} + \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{4} - 1} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} - 4 i \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{4}}\right ) - \frac {1}{2} \, \arctan \left (2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + \frac {1}{2} \, \log \left (2 \, x^{4} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} - 1} x^{2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right ) \]

[In]

integrate((-3+2*x)/x/(x^4-1)^(1/4),x, algorithm="fricas")

[Out]

(3/16*I + 3/16)*sqrt(2)*log((sqrt(2)*((I + 1)*x^4 - 2*I - 2) - (2*I - 2)*sqrt(2)*sqrt(x^4 - 1) + 4*(x^4 - 1)^(
3/4) - 4*I*(x^4 - 1)^(1/4))/x^4) - (3/16*I - 3/16)*sqrt(2)*log((sqrt(2)*(-(I - 1)*x^4 + 2*I - 2) + (2*I + 2)*s
qrt(2)*sqrt(x^4 - 1) + 4*(x^4 - 1)^(3/4) + 4*I*(x^4 - 1)^(1/4))/x^4) + (3/16*I - 3/16)*sqrt(2)*log((sqrt(2)*((
I - 1)*x^4 - 2*I + 2) - (2*I + 2)*sqrt(2)*sqrt(x^4 - 1) + 4*(x^4 - 1)^(3/4) + 4*I*(x^4 - 1)^(1/4))/x^4) - (3/1
6*I + 3/16)*sqrt(2)*log((sqrt(2)*(-(I + 1)*x^4 + 2*I + 2) + (2*I - 2)*sqrt(2)*sqrt(x^4 - 1) + 4*(x^4 - 1)^(3/4
) - 4*I*(x^4 - 1)^(1/4))/x^4) - 1/2*arctan(2*(x^4 - 1)^(1/4)*x^3 + 2*(x^4 - 1)^(3/4)*x) + 1/2*log(2*x^4 + 2*(x
^4 - 1)^(1/4)*x^3 + 2*sqrt(x^4 - 1)*x^2 + 2*(x^4 - 1)^(3/4)*x - 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=\frac {x e^{- \frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4}} \right )}}{2 \Gamma \left (\frac {5}{4}\right )} + \frac {3 \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate((-3+2*x)/x/(x**4-1)**(1/4),x)

[Out]

x*exp(-I*pi/4)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), x**4)/(2*gamma(5/4)) + 3*gamma(1/4)*hyper((1/4, 1/4), (5/4
,), exp_polar(2*I*pi)/x**4)/(4*x*gamma(5/4))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.32 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=-\frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \frac {3}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{2} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \]

[In]

integrate((-3+2*x)/x/(x^4-1)^(1/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=\int { \frac {2 \, x - 3}{{\left (x^{4} - 1\right )}^{\frac {1}{4}} x} \,d x } \]

[In]

integrate((-3+2*x)/x/(x^4-1)^(1/4),x, algorithm="giac")

[Out]

integrate((2*x - 3)/((x^4 - 1)^(1/4)*x), x)

Mupad [B] (verification not implemented)

Time = 5.79 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.64 \[ \int \frac {-3+2 x}{x \sqrt [4]{-1+x^4}} \, dx=\frac {2\,x\,{\left (1-x^4\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ x^4\right )}{{\left (x^4-1\right )}^{1/4}}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{4}+\frac {3}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{4}-\frac {3}{4}{}\mathrm {i}\right ) \]

[In]

int((2*x - 3)/(x*(x^4 - 1)^(1/4)),x)

[Out]

(2*x*(1 - x^4)^(1/4)*hypergeom([1/4, 1/4], 5/4, x^4))/(x^4 - 1)^(1/4) - 2^(1/2)*atan(2^(1/2)*(x^4 - 1)^(1/4)*(
1/2 + 1i/2))*(3/4 + 3i/4) - 2^(1/2)*atan(2^(1/2)*(x^4 - 1)^(1/4)*(1/2 - 1i/2))*(3/4 - 3i/4)