3.1022 \(\int \frac{x \sqrt{1-x^2}}{x-x^3+\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} \sqrt{1-x^2}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} x}{\sqrt{1-x^2}}\right )}{\sqrt{3}}+\sin ^{-1}(x) \]

[Out]

ArcSin[x] - ArcTan[(1 - 2*x^2)/Sqrt[3]]/Sqrt[3] - ArcTan[x/(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*Sqrt[1 - x^2]
)]/Sqrt[3] - ArcTan[(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*x)/Sqrt[1 - x^2]]/Sqrt[3]

________________________________________________________________________________________

Rubi [A]  time = 0.387729, antiderivative size = 149, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.303, Rules used = {6742, 1293, 216, 1174, 377, 205, 1251, 773, 618, 204} \[ -\frac{x^2}{2}-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} \sqrt{1-x^2}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} x}{\sqrt{1-x^2}}\right )}{\sqrt{3}}+\frac{1}{4} (1-x)^2+\frac{1}{4} (x+1)^2+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x)^2/4 - x^2/2 + (1 + x)^2/4 + ArcSin[x] - ArcTan[(1 - 2*x^2)/Sqrt[3]]/Sqrt[3] - ArcTan[x/(Sqrt[-((I - Sq
rt[3])/(I + Sqrt[3]))]*Sqrt[1 - x^2])]/Sqrt[3] - ArcTan[(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*x)/Sqrt[1 - x^2]
]/Sqrt[3]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 1293

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
e*f^2)/c, Int[(f*x)^(m - 2)*(d + e*x^2)^(q - 1), x], x] - Dist[f^2/c, Int[((f*x)^(m - 2)*(d + e*x^2)^(q - 1)*S
imp[a*e - (c*d - b*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c,
0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, 1] && LeQ[m, 3]

Rule 216

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

Rule 1174

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]
}, Dist[(2*c)/r, Int[(d + e*x^2)^q/(b - r + 2*c*x^2), x], x] - Dist[(2*c)/r, Int[(d + e*x^2)^q/(b + r + 2*c*x^
2), x], x]] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Integ
erQ[q]

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 205

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

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{x \sqrt{1-x^2}}{x-x^3+\sqrt{1-x^2}} \, dx &=\int \left (\frac{1}{2} (-1+x)+\frac{1+x}{2}-\frac{x^2 \sqrt{1-x^2}}{1-x^2+x^4}+\frac{x^3 \left (1-x^2\right )}{1-x^2+x^4}\right ) \, dx\\ &=\frac{1}{4} (1-x)^2+\frac{1}{4} (1+x)^2-\int \frac{x^2 \sqrt{1-x^2}}{1-x^2+x^4} \, dx+\int \frac{x^3 \left (1-x^2\right )}{1-x^2+x^4} \, dx\\ &=\frac{1}{4} (1-x)^2+\frac{1}{4} (1+x)^2+\frac{1}{2} \operatorname{Subst}\left (\int \frac{(1-x) x}{1-x+x^2} \, dx,x,x^2\right )+\int \frac{1}{\sqrt{1-x^2}} \, dx-\int \frac{1}{\sqrt{1-x^2} \left (1-x^2+x^4\right )} \, dx\\ &=\frac{1}{4} (1-x)^2-\frac{x^2}{2}+\frac{1}{4} (1+x)^2+\sin ^{-1}(x)+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )+\frac{(2 i) \int \frac{1}{\sqrt{1-x^2} \left (-1-i \sqrt{3}+2 x^2\right )} \, dx}{\sqrt{3}}-\frac{(2 i) \int \frac{1}{\sqrt{1-x^2} \left (-1+i \sqrt{3}+2 x^2\right )} \, dx}{\sqrt{3}}\\ &=\frac{1}{4} (1-x)^2-\frac{x^2}{2}+\frac{1}{4} (1+x)^2+\sin ^{-1}(x)-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{-1+i \sqrt{3}-\left (-1-i \sqrt{3}\right ) x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )}{\sqrt{3}}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{-1-i \sqrt{3}-\left (-1+i \sqrt{3}\right ) x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )}{\sqrt{3}}-\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )\\ &=\frac{1}{4} (1-x)^2-\frac{x^2}{2}+\frac{1}{4} (1+x)^2+\sin ^{-1}(x)-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} \sqrt{1-x^2}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} x}{\sqrt{1-x^2}}\right )}{\sqrt{3}}+\frac{\tan ^{-1}\left (\frac{-1+2 x^2}{\sqrt{3}}\right )}{\sqrt{3}}\\ \end{align*}

Mathematica [B]  time = 1.58702, size = 1910, normalized size = 15.66 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[x] - ((-I + Sqrt[3])*ArcTan[(x*(-7 - I*Sqrt[3] + 8*Sqrt[3]*x + I*(7*I + Sqrt[3])*x^2))/(-6 - (2*I)*Sqrt
[3] + 3*(-I + Sqrt[3])*x^3 - 2*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + (2*I)*x^2*(9*I + Sqrt[3] + I*Sqrt[2 - (
2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x*(3*I + 11*Sqrt[3] + 2*Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2]))])/(2*Sqrt[6 - (
6*I)*Sqrt[3]]) + ((I + Sqrt[3])*ArcTan[(x*(7 - I*Sqrt[3] - 8*Sqrt[3]*x + (7 + I*Sqrt[3])*x^2))/(-6 + (2*I)*Sqr
t[3] + 3*(I + Sqrt[3])*x^3 - 2*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + x^2*(-18 - (2*I)*Sqrt[3] - 2*Sqrt[2 + (
2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x*(-3*I + 11*Sqrt[3] + 2*Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2]))])/(2*Sqrt[6 +
(6*I)*Sqrt[3]]) - ((I + Sqrt[3])*ArcTan[(x*(7 - I*Sqrt[3] + 8*Sqrt[3]*x + (7 + I*Sqrt[3])*x^2))/(6 - (2*I)*Sqr
t[3] + 3*(I + Sqrt[3])*x^3 + 2*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + 2*x^2*(9 + I*Sqrt[3] + Sqrt[2 + (2*I)*S
qrt[3]]*Sqrt[1 - x^2]) + x*(-3*I + 11*Sqrt[3] + 2*Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2]))])/(2*Sqrt[6 + (6*I)*
Sqrt[3]]) + ((1 + I*Sqrt[3])*ArcTanh[(x*(7*I - Sqrt[3] + (8*I)*Sqrt[3]*x + (7*I + Sqrt[3])*x^2))/(6 + (2*I)*Sq
rt[3] + 3*(-I + Sqrt[3])*x^3 + 2*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + 2*x^2*(9 - I*Sqrt[3] + Sqrt[2 - (2*I)
*Sqrt[3]]*Sqrt[1 - x^2]) + x*(3*I + 11*Sqrt[3] + 2*Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2]))])/(2*Sqrt[6 - (6*I)
*Sqrt[3]]) - ((I/2)*Log[-1/2 - (I/2)*Sqrt[3] + x^2])/Sqrt[3] + ((I/2)*Log[(I/2)*(I + Sqrt[3]) + x^2])/Sqrt[3]
- ((I/4)*(-I + Sqrt[3])*Log[16*(1 + Sqrt[3]*x + x^2)^2])/Sqrt[6 - (6*I)*Sqrt[3]] + ((I/4)*(I + Sqrt[3])*Log[16
*(1 + Sqrt[3]*x + x^2)^2])/Sqrt[6 + (6*I)*Sqrt[3]] + ((1 + I*Sqrt[3])*Log[(4 - 4*Sqrt[3]*x + 4*x^2)^2])/(4*Sqr
t[6 - (6*I)*Sqrt[3]]) + ((1 - I*Sqrt[3])*Log[(4 - 4*Sqrt[3]*x + 4*x^2)^2])/(4*Sqrt[6 + (6*I)*Sqrt[3]]) + ((1 +
I*Sqrt[3])*Log[3*I + Sqrt[3] - (-I + Sqrt[3])*x^4 + (2*I)*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + (5*I)*x^2*(
2 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x*(3 + (5*I)*Sqrt[3] + (3*I)*Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2
]) + I*x^3*(3*I + 3*Sqrt[3] + Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2])])/(4*Sqrt[6 - (6*I)*Sqrt[3]]) - ((I/4)*(-
I + Sqrt[3])*Log[3*I + Sqrt[3] - (-I + Sqrt[3])*x^4 + (2*I)*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2] + (5*I)*x^2*
(2 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x^3*(3 - (3*I)*Sqrt[3] - I*Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2]
) - I*x*(-3*I + 5*Sqrt[3] + 3*Sqrt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2])])/Sqrt[6 - (6*I)*Sqrt[3]] + ((1 - I*Sqrt[
3])*Log[-3*I + Sqrt[3] - (I + Sqrt[3])*x^4 - (2*I)*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2] - (5*I)*x^2*(2 + Sqrt
[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x*(3 - (5*I)*Sqrt[3] - (3*I)*Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2]) - I*x
^3*(-3*I + 3*Sqrt[3] + Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2])])/(4*Sqrt[6 + (6*I)*Sqrt[3]]) + ((I/4)*(I + Sqrt
[3])*Log[-3*I + Sqrt[3] - (I + Sqrt[3])*x^4 - (2*I)*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2] - (5*I)*x^2*(2 + Sqr
t[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x^3*(3 + (3*I)*Sqrt[3] + I*Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2]) + I*x*
(3*I + 5*Sqrt[3] + 3*Sqrt[6 + (6*I)*Sqrt[3]]*Sqrt[1 - x^2])])/Sqrt[6 + (6*I)*Sqrt[3]]

________________________________________________________________________________________

Maple [B]  time = 0.052, size = 234, normalized size = 1.9 \begin{align*} -2\,\arctan \left ({\frac{-1+\sqrt{-{x}^{2}+1}}{x}} \right ) +{\frac{i}{6}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( -1+\sqrt{-{x}^{2}+1} \right ) ^{2}}+{\frac{1+i\sqrt{3}}{x} \left ( -1+\sqrt{-{x}^{2}+1} \right ) }-1 \right ) -{\frac{i}{6}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( -1+\sqrt{-{x}^{2}+1} \right ) ^{2}}+{\frac{1-i\sqrt{3}}{x} \left ( -1+\sqrt{-{x}^{2}+1} \right ) }-1 \right ) +{\frac{i}{6}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( -1+\sqrt{-{x}^{2}+1} \right ) ^{2}}+{\frac{i\sqrt{3}-1}{x} \left ( -1+\sqrt{-{x}^{2}+1} \right ) }-1 \right ) -{\frac{i}{6}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( -1+\sqrt{-{x}^{2}+1} \right ) ^{2}}+{\frac{-i\sqrt{3}-1}{x} \left ( -1+\sqrt{-{x}^{2}+1} \right ) }-1 \right ) +{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan((-1+(-x^2+1)^(1/2))/x)+1/6*I*3^(1/2)*ln((-1+(-x^2+1)^(1/2))^2/x^2+(1+I*3^(1/2))*(-1+(-x^2+1)^(1/2))/
x-1)-1/6*I*3^(1/2)*ln((-1+(-x^2+1)^(1/2))^2/x^2+(1-I*3^(1/2))*(-1+(-x^2+1)^(1/2))/x-1)+1/6*I*3^(1/2)*ln((-1+(-
x^2+1)^(1/2))^2/x^2+(I*3^(1/2)-1)*(-1+(-x^2+1)^(1/2))/x-1)-1/6*I*3^(1/2)*ln((-1+(-x^2+1)^(1/2))^2/x^2+(-I*3^(1
/2)-1)*(-1+(-x^2+1)^(1/2))/x-1)+1/3*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} + \int -\frac{x^{4} - x^{2}}{x^{3} - x - \sqrt{x + 1} \sqrt{-x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.79936, size = 204, normalized size = 1.67 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{\sqrt{3}{\left (2 \, x^{2} - 1\right )} \sqrt{-x^{2} + 1}}{3 \,{\left (x^{3} - x\right )}}\right ) - 2 \, \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x \sqrt{1 - x^{2}}}{x^{3} - x - \sqrt{1 - x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x*sqrt(1 - x**2)/(x**3 - x - sqrt(1 - x**2)), x)

________________________________________________________________________________________

Giac [B]  time = 1.23833, size = 261, normalized size = 2.14 \begin{align*} \frac{1}{2} \, \pi \mathrm{sgn}\left (x\right ) - \frac{1}{6} \, \sqrt{3}{\left (\pi \mathrm{sgn}\left (x\right ) + 2 \, \arctan \left (-\frac{\sqrt{3} x{\left (\frac{\sqrt{-x^{2} + 1} - 1}{x} + \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{3 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}\right )\right )} - \frac{1}{6} \, \sqrt{3}{\left (\pi \mathrm{sgn}\left (x\right ) + 2 \, \arctan \left (\frac{\sqrt{3} x{\left (\frac{\sqrt{-x^{2} + 1} - 1}{x} - \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{3 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}\right )\right )} + \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \arctan \left (-\frac{x{\left (\frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*pi*sgn(x) - 1/6*sqrt(3)*(pi*sgn(x) + 2*arctan(-1/3*sqrt(3)*x*((sqrt(-x^2 + 1) - 1)/x + (sqrt(-x^2 + 1) - 1
)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))) - 1/6*sqrt(3)*(pi*sgn(x) + 2*arctan(1/3*sqrt(3)*x*((sqrt(-x^2 + 1) - 1)/x
- (sqrt(-x^2 + 1) - 1)^2/x^2 + 1)/(sqrt(-x^2 + 1) - 1))) + 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + arcta
n(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))