3.1021 \(\int \frac {1}{\frac {1}{x}+\sqrt {1-x^2}} \, dx\)

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

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Rubi [A]  time = 0.15, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {6742, 1107, 618, 204, 1293, 216, 1174, 377, 205} \[ -\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt {1-x^2}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt {1-x^2}}\right )}{\sqrt {3}}+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^(-1) + Sqrt[1 - x^2])^(-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]

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])

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 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 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 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 1107

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

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 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 6742

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

Rubi steps

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

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Mathematica [B]  time = 4.39, size = 1932, normalized size = 15.84 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*ArcSin[x] - (2*(-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]))])/Sqrt[(1
- I*Sqrt[3])/6] + (2*(I + Sqrt[3])*ArcTan[(x*(7 - I*Sqrt[3] - 8*Sqrt[3]*x + (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] + 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]))])/Sqrt[(1
+ I*Sqrt[3])/6] - (2*(I + Sqrt[3])*ArcTan[(x*(7 - I*Sqrt[3] + 8*Sqrt[3]*x + (7 + I*Sqrt[3])*x^2))/(6 - (2*I)*S
qrt[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]))])/Sqrt[(1 + I*Sqrt
[3])/6] + (2*(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]))])/Sqrt[(1 - I*Sqrt[
3])/6] - (4*I)*Sqrt[3]*Log[-1/2 - (I/2)*Sqrt[3] + x^2] + (4*I)*Sqrt[3]*Log[(I/2)*(I + Sqrt[3]) + x^2] - (I*(-I
+ Sqrt[3])*Log[16*(1 + Sqrt[3]*x + x^2)^2])/Sqrt[(1 - I*Sqrt[3])/6] + (I*(I + Sqrt[3])*Log[16*(1 + Sqrt[3]*x
+ x^2)^2])/Sqrt[(1 + I*Sqrt[3])/6] + ((1 - I*Sqrt[3])*Log[(4 - 4*Sqrt[3]*x + 4*x^2)^2])/Sqrt[(1 + I*Sqrt[3])/6
] + ((1 + I*Sqrt[3])*Log[(4 - 4*Sqrt[3]*x + 4*x^2)^2])/Sqrt[(1 - I*Sqrt[3])/6] + ((1 + I*Sqrt[3])*Log[3*I + Sq
rt[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])])/Sqrt[(1 - I*Sqrt[3])/6] - (I*(-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]]*Sqr
t[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*Sq
rt[6 - (6*I)*Sqrt[3]]*Sqrt[1 - x^2])])/Sqrt[(1 - I*Sqrt[3])/6] + ((1 - I*Sqrt[3])*Log[-3*I + Sqrt[3] - (I + Sq
rt[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])])/Sqrt[(1 + I*Sqrt[3])/6] + (I*(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[(1 + I*Sqrt[3])/6])/24

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fricas [A]  time = 0.65, size = 73, normalized size = 0.60 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{2} - 1\right )} \sqrt {-x^{2} + 1}}{3 \, {\left (x^{3} - x\right )}}\right ) - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1/x+(-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)

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giac [B]  time = 0.42, size = 193, normalized size = 1.58 \[ \frac {1}{2} \, \pi \mathrm {sgn}\relax (x) - \frac {1}{6} \, \sqrt {3} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (-\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \frac {1}{6} \, \sqrt {3} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1/x+(-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))

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maple [B]  time = 0.04, size = 234, normalized size = 1.92 \[ -2 \arctan \left (\frac {-1+\sqrt {-x^{2}+1}}{x}\right )+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{3}-\frac {i \sqrt {3}\, \ln \left (\frac {\left (-1-i \sqrt {3}\right ) \left (-1+\sqrt {-x^{2}+1}\right )}{x}+\frac {\left (-1+\sqrt {-x^{2}+1}\right )^{2}}{x^{2}}-1\right )}{6}-\frac {i \sqrt {3}\, \ln \left (\frac {\left (1-i \sqrt {3}\right ) \left (-1+\sqrt {-x^{2}+1}\right )}{x}+\frac {\left (-1+\sqrt {-x^{2}+1}\right )^{2}}{x^{2}}-1\right )}{6}+\frac {i \sqrt {3}\, \ln \left (\frac {\left (1+i \sqrt {3}\right ) \left (-1+\sqrt {-x^{2}+1}\right )}{x}+\frac {\left (-1+\sqrt {-x^{2}+1}\right )^{2}}{x^{2}}-1\right )}{6}+\frac {i \sqrt {3}\, \ln \left (\frac {\left (i \sqrt {3}-1\right ) \left (-1+\sqrt {-x^{2}+1}\right )}{x}+\frac {\left (-1+\sqrt {-x^{2}+1}\right )^{2}}{x^{2}}-1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-x^{2} + 1} + \frac {1}{x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.92, size = 549, normalized size = 4.50 \[ \mathrm {asin}\relax (x)-\frac {\ln \left (\frac {\frac {\left (x\,\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )-1\right )\,1{}\mathrm {i}}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^2}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{\frac {\sqrt {3}}{2}-x+\frac {1}{2}{}\mathrm {i}}\right )}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^2}\,\left (\sqrt {3}-4\,{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}\right )}+\frac {\ln \left (\frac {\frac {\left (x\,\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )-1\right )\,1{}\mathrm {i}}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^2}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}}\right )}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^2}\,\left (-\sqrt {3}+4\,{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}\right )}-\frac {\ln \left (\frac {\frac {\left (x\,\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )+1\right )\,1{}\mathrm {i}}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^2}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}}\right )}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^2}\,\left (-\sqrt {3}+4\,{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}\right )}-\frac {\ln \left (x-\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}{\sqrt {3}-4\,{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}}-\frac {\ln \left (x+\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}{\sqrt {3}-4\,{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}}+\frac {\ln \left (\frac {\frac {\left (x\,\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )+1\right )\,1{}\mathrm {i}}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^2}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}}\right )}{\sqrt {1-{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^2}\,\left (\sqrt {3}-4\,{\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}\right )}+\frac {\ln \left (x-\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}{-\sqrt {3}+4\,{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}}+\frac {\ln \left (x+\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}{-\sqrt {3}+4\,{\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )}^3+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/x + (1 - x^2)^(1/2)),x)

[Out]

asin(x) - log((((x*(3^(1/2)/2 + 1i/2) - 1)*1i)/(1 - (3^(1/2)/2 + 1i/2)^2)^(1/2) - (1 - x^2)^(1/2)*1i)/(3^(1/2)
/2 - x + 1i/2))/((1 - (3^(1/2)/2 + 1i/2)^2)^(1/2)*(3^(1/2) - 4*(3^(1/2)/2 + 1i/2)^3 + 1i)) + log((((x*(3^(1/2)
/2 - 1i/2) - 1)*1i)/(1 - (3^(1/2)/2 - 1i/2)^2)^(1/2) - (1 - x^2)^(1/2)*1i)/(x - 3^(1/2)/2 + 1i/2))/((1 - (3^(1
/2)/2 - 1i/2)^2)^(1/2)*(4*(3^(1/2)/2 - 1i/2)^3 - 3^(1/2) + 1i)) - log((((x*(3^(1/2)/2 - 1i/2) + 1)*1i)/(1 - (3
^(1/2)/2 - 1i/2)^2)^(1/2) + (1 - x^2)^(1/2)*1i)/(x + 3^(1/2)/2 - 1i/2))/((1 - (3^(1/2)/2 - 1i/2)^2)^(1/2)*(4*(
3^(1/2)/2 - 1i/2)^3 - 3^(1/2) + 1i)) - (log(x - 3^(1/2)/2 - 1i/2)*(3^(1/2)/2 + 1i/2))/(3^(1/2) - 4*(3^(1/2)/2
+ 1i/2)^3 + 1i) - (log(x + 3^(1/2)/2 + 1i/2)*(3^(1/2)/2 + 1i/2))/(3^(1/2) - 4*(3^(1/2)/2 + 1i/2)^3 + 1i) + log
((((x*(3^(1/2)/2 + 1i/2) + 1)*1i)/(1 - (3^(1/2)/2 + 1i/2)^2)^(1/2) + (1 - x^2)^(1/2)*1i)/(x + 3^(1/2)/2 + 1i/2
))/((1 - (3^(1/2)/2 + 1i/2)^2)^(1/2)*(3^(1/2) - 4*(3^(1/2)/2 + 1i/2)^3 + 1i)) + (log(x - 3^(1/2)/2 + 1i/2)*(3^
(1/2)/2 - 1i/2))/(4*(3^(1/2)/2 - 1i/2)^3 - 3^(1/2) + 1i) + (log(x + 3^(1/2)/2 - 1i/2)*(3^(1/2)/2 - 1i/2))/(4*(
3^(1/2)/2 - 1i/2)^3 - 3^(1/2) + 1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{x \sqrt {1 - x^{2}} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x*sqrt(1 - x**2) + 1), x)

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