\(\int \frac {-2+x+x^2}{x^2 (-1+x^2)^{3/4}} \, dx\) [1227]

Optimal result

Integrand size = 19, antiderivative size = 90 \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{-1+x^2}}{x}+\frac {\arctan \left (\frac {-\frac {1}{\sqrt {2}}+\frac {\sqrt {-1+x^2}}{\sqrt {2}}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.53, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1821, 272, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}+\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {2 \sqrt [4]{x^2-1}}{x}-\frac {\log \left (\sqrt {x^2-1}-\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\sqrt {x^2-1}+\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}} \]

[In]

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

[Out]

(-2*(-1 + x^2)^(1/4))/x - ArcTan[1 - Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2] + ArcTan[1 + Sqrt[2]*(-1 + x^2)^(1/4)]/
Sqrt[2] - Log[1 - Sqrt[2]*(-1 + x^2)^(1/4) + Sqrt[-1 + x^2]]/(2*Sqrt[2]) + Log[1 + Sqrt[2]*(-1 + x^2)^(1/4) +
Sqrt[-1 + x^2]]/(2*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 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 217

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

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 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 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.53 \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{-1+x^2}}{x}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 3.00 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {2 \left (x^{2}-1\right )^{\frac {1}{4}}}{x}+\frac {{\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{2 \Gamma \left (\frac {3}{4}\right ) \operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(76\)
meijerg	\(\frac {{\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], x^{2}\right )}{\operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}+\frac {{\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{2 \Gamma \left (\frac {3}{4}\right ) \operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}+\frac {2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}\right ], x^{2}\right )}{\operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}} x}\)	\(125\)
trager	\(-\frac {2 \left (x^{2}-1\right )^{\frac {1}{4}}}{x}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (x^{2}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{2}}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {-2 \sqrt {x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \left (x^{2}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{2}\)	\(170\)

[In]

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

[Out]

-2*(x^2-1)^(1/4)/x+1/2/GAMMA(3/4)/signum(x^2-1)^(3/4)*(-signum(x^2-1))^(3/4)*(3/4*GAMMA(3/4)*x^2*hypergeom([1,
1,7/4],[2,2],x^2)+(-3*ln(2)+1/2*Pi+2*ln(x)+I*Pi)*GAMMA(3/4))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.85 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.52 \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=\frac {\left (i - 1\right ) \, \sqrt {2} x \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x^{2} - 2 i - 2\right )} - \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{2} - 1} - 4 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}} + 4 i \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x^{2}}\right ) - \left (i + 1\right ) \, \sqrt {2} x \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x^{2} + 2 i - 2\right )} + \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{2} - 1} - 4 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}} - 4 i \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x^{2}}\right ) + \left (i + 1\right ) \, \sqrt {2} x \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x^{2} - 2 i + 2\right )} - \left (2 i + 2\right ) \, \sqrt {2} \sqrt {x^{2} - 1} - 4 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}} - 4 i \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x^{2}}\right ) - \left (i - 1\right ) \, \sqrt {2} x \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x^{2} + 2 i + 2\right )} + \left (2 i - 2\right ) \, \sqrt {2} \sqrt {x^{2} - 1} - 4 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}} + 4 i \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x^{2}}\right ) - 16 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{8 \, x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83 \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=x e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {x^{2}} \right )} - \frac {2 e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {x^{2}} \right )}}{x} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

x*exp(-3*I*pi/4)*hyper((1/2, 3/4), (3/2,), x**2) - 2*exp(I*pi/4)*hyper((-1/2, 3/4), (1/2,), x**2)/x - gamma(3/
4)*hyper((3/4, 3/4), (7/4,), exp_polar(2*I*pi)/x**2)/(2*x**(3/2)*gamma(7/4))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx=\frac {4\,{\left (\frac {1}{x^2}\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ \frac {1}{x^2}\right )}{5\,x}+\frac {x\,{\left (1-x^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {3}{2};\ x^2\right )}{{\left (x^2-1\right )}^{3/4}}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \]

[In]

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

[Out]

2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 - 1i/2))*(1/2 + 1i/2) + 2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 +
1i/2))*(1/2 - 1i/2) + (4*(1/x^2)^(3/4)*hypergeom([3/4, 5/4], 9/4, 1/x^2))/(5*x) + (x*(1 - x^2)^(3/4)*hypergeom
([1/2, 3/4], 3/2, x^2))/(x^2 - 1)^(3/4)