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

Optimal. Leaf size=152 \[ \frac{1}{8} \log \left (x^4-x^2+1\right )+\frac{1}{4} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt{3} x-1}{\sqrt{1-x^2}}\right )+\frac{1}{4} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt{3} x+1}{\sqrt{1-x^2}}\right )-\frac{1}{4} \sqrt{3} \tan ^{-1}\left (\frac{2 x^2-1}{\sqrt{3}}\right )-\sqrt{1-x^2} \tan ^{-1}\left (\sqrt{1-x^2}+x\right )+\frac{1}{4} \tanh ^{-1}\left (x \sqrt{1-x^2}\right )-\frac{1}{2} \sin ^{-1}(x) \]

[Out]

-ArcSin[x]/2 + (Sqrt[3]*ArcTan[(-1 + Sqrt[3]*x)/Sqrt[1 - x^2]])/4 + (Sqrt[3]*ArcTan[(1 + Sqrt[3]*x)/Sqrt[1 - x
^2]])/4 - (Sqrt[3]*ArcTan[(-1 + 2*x^2)/Sqrt[3]])/4 - Sqrt[1 - x^2]*ArcTan[x + Sqrt[1 - x^2]] + ArcTanh[x*Sqrt[
1 - x^2]]/4 + Log[1 - x^2 + x^4]/8

________________________________________________________________________________________

Rubi [C]  time = 0.456532, antiderivative size = 286, normalized size of antiderivative = 1.88, number of steps used = 32, number of rules used = 16, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.593, Rules used = {261, 5207, 12, 6742, 1107, 618, 204, 1293, 216, 1174, 377, 205, 402, 1247, 634, 628} \[ \frac{1}{8} \log \left (x^4-x^2+1\right )+\frac{1}{4} \sqrt{3} \tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )-\frac{1}{12} \left (-\sqrt{3}+3 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} \sqrt{1-x^2}}\right )+\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} \sqrt{1-x^2}}\right )}{2 \sqrt{3}}+\frac{1}{12} \left (\sqrt{3}+3 i\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} x}{\sqrt{1-x^2}}\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{-\frac{-\sqrt{3}+i}{\sqrt{3}+i}} x}{\sqrt{1-x^2}}\right )}{2 \sqrt{3}}-\sqrt{1-x^2} \tan ^{-1}\left (\sqrt{1-x^2}+x\right )-\frac{1}{2} \sin ^{-1}(x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-ArcSin[x]/2 + (Sqrt[3]*ArcTan[(1 - 2*x^2)/Sqrt[3]])/4 + ArcTan[x/(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*Sqrt[1
- x^2])]/(2*Sqrt[3]) - ((3*I - Sqrt[3])*ArcTan[x/(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*Sqrt[1 - x^2])])/12 +
ArcTan[(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]*x)/Sqrt[1 - x^2]]/(2*Sqrt[3]) + ((3*I + Sqrt[3])*ArcTan[(Sqrt[-((
I - Sqrt[3])/(I + Sqrt[3]))]*x)/Sqrt[1 - x^2]])/12 - Sqrt[1 - x^2]*ArcTan[x + Sqrt[1 - x^2]] + Log[1 - x^2 + x
^4]/8

Rule 261

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

Rule 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[u], w, x] - Dist
[b, Int[SimplifyIntegrand[(w*D[u, x])/(1 + u^2), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x]
&& InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcTan[u]), x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

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

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 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])

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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rule 634

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

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}\left (x+\sqrt{1-x^2}\right )}{\sqrt{1-x^2}} \, dx &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\int \frac{x-\sqrt{1-x^2}}{2 \left (1+x \sqrt{1-x^2}\right )} \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \int \frac{x-\sqrt{1-x^2}}{1+x \sqrt{1-x^2}} \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \int \left (\frac{x}{1+x \sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{1+x \sqrt{1-x^2}}\right ) \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \int \frac{x}{1+x \sqrt{1-x^2}} \, dx+\frac{1}{2} \int \frac{\sqrt{1-x^2}}{1+x \sqrt{1-x^2}} \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \int \left (\frac{x}{1-x^2+x^4}-\frac{x^2 \sqrt{1-x^2}}{1-x^2+x^4}\right ) \, dx+\frac{1}{2} \int \left (\frac{\sqrt{1-x^2}}{1-x^2+x^4}-\frac{x \left (1-x^2\right )}{1-x^2+x^4}\right ) \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \int \frac{x}{1-x^2+x^4} \, dx+\frac{1}{2} \int \frac{\sqrt{1-x^2}}{1-x^2+x^4} \, dx+\frac{1}{2} \int \frac{x^2 \sqrt{1-x^2}}{1-x^2+x^4} \, dx-\frac{1}{2} \int \frac{x \left (1-x^2\right )}{1-x^2+x^4} \, dx\\ &=-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1-x}{1-x+x^2} \, dx,x,x^2\right )-\frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx+\frac{1}{2} \int \frac{1}{\sqrt{1-x^2} \left (1-x^2+x^4\right )} \, dx-\frac{i \int \frac{\sqrt{1-x^2}}{-1-i \sqrt{3}+2 x^2} \, dx}{\sqrt{3}}+\frac{i \int \frac{\sqrt{1-x^2}}{-1+i \sqrt{3}+2 x^2} \, dx}{\sqrt{3}}\\ &=-\frac{1}{2} \sin ^{-1}(x)-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac{i \int \frac{1}{\sqrt{1-x^2} \left (-1-i \sqrt{3}+2 x^2\right )} \, dx}{\sqrt{3}}+\frac{i \int \frac{1}{\sqrt{1-x^2} \left (-1+i \sqrt{3}+2 x^2\right )} \, dx}{\sqrt{3}}-\frac{1}{6} \left (3-i \sqrt{3}\right ) \int \frac{1}{\sqrt{1-x^2} \left (-1+i \sqrt{3}+2 x^2\right )} \, dx-\frac{1}{6} \left (3+i \sqrt{3}\right ) \int \frac{1}{\sqrt{1-x^2} \left (-1-i \sqrt{3}+2 x^2\right )} \, dx\\ &=-\frac{1}{2} \sin ^{-1}(x)+\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}}-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )+\frac{1}{8} \log \left (1-x^2+x^4\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )+\frac{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{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{1}{6} \left (3-i \sqrt{3}\right ) \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 )-\frac{1}{6} \left (3+i \sqrt{3}\right ) \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 )\\ &=-\frac{1}{2} \sin ^{-1}(x)+\frac{1}{4} \sqrt{3} \tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )+\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} \sqrt{1-x^2}}\right )}{2 \sqrt{3}}-\frac{1}{12} \left (3 i-\sqrt{3}\right ) \tan ^{-1}\left (\frac{x}{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} \sqrt{1-x^2}}\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} x}{\sqrt{1-x^2}}\right )}{2 \sqrt{3}}+\frac{1}{12} \left (3 i+\sqrt{3}\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{i-\sqrt{3}}{i+\sqrt{3}}} x}{\sqrt{1-x^2}}\right )-\sqrt{1-x^2} \tan ^{-1}\left (x+\sqrt{1-x^2}\right )+\frac{1}{8} \log \left (1-x^2+x^4\right )\\ \end{align*}

Mathematica [C]  time = 4.53127, size = 2180, normalized size = 14.34 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*ArcSin[x] - 48*Sqrt[1 - x^2]*ArcTan[x + Sqrt[1 - x^2]] + (2*(-3*I + Sqrt[3])*ArcTan[(3 - I*Sqrt[3] + (-3
- I*Sqrt[3])*x^4 + 2*x*(-6*I + 2*Sqrt[3] - I*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - 2*x^3*(6*I + 2*Sqrt[3] +
I*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/(I
- Sqrt[3] + (6*I)*(I + Sqrt[3])*x - 2*(-15*I + Sqrt[3])*x^2 + 6*(1 + (3*I)*Sqrt[3])*x^3 + (11*I + 3*Sqrt[3])*x
^4)])/Sqrt[(1 - I*Sqrt[3])/6] - (2*(-3*I + Sqrt[3])*ArcTan[(3 - I*Sqrt[3] + (-3 - I*Sqrt[3])*x^4 + 2*x^3*(6*I
+ 2*Sqrt[3] + I*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x*(12*I - 4*Sqrt[3] + (2*I)*Sqrt[2 - (2*I)*Sqrt[3]]*S
qrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/(I - Sqrt[3] + (6 - (6*I)*Sqrt[
3])*x - 2*(-15*I + Sqrt[3])*x^2 + (-6 - (18*I)*Sqrt[3])*x^3 + (11*I + 3*Sqrt[3])*x^4)])/Sqrt[(1 - I*Sqrt[3])/6
] - (2*(3*I + Sqrt[3])*ArcTan[(-3 - I*Sqrt[3] + (3 - I*Sqrt[3])*x^4 + 2*x^3*(-6*I + 2*Sqrt[3] - I*Sqrt[2 + (2*
I)*Sqrt[3]]*Sqrt[1 - x^2]) - 2*x*(6*I + 2*Sqrt[3] + I*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x
^2*(6 + Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/(-I - Sqrt[3] + (-6 - (6*I)*Sqrt[3])*x - 2*(15*I + Sqrt[3])*x^
2 + 6*(1 - (3*I)*Sqrt[3])*x^3 + (-11*I + 3*Sqrt[3])*x^4)])/Sqrt[(1 + I*Sqrt[3])/6] + (2*(3*I + Sqrt[3])*ArcTan
[(-3 - I*Sqrt[3] + (3 - I*Sqrt[3])*x^4 + 2*x*(6*I + 2*Sqrt[3] + I*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) + x^3
*(12*I - 4*Sqrt[3] + (2*I)*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sqrt[2 + (2*I)*Sqrt
[3]]*Sqrt[1 - x^2]))/(-I - Sqrt[3] + (6 + (6*I)*Sqrt[3])*x - 2*(15*I + Sqrt[3])*x^2 + (-6 + (18*I)*Sqrt[3])*x^
3 + (-11*I + 3*Sqrt[3])*x^4)])/Sqrt[(1 + I*Sqrt[3])/6] + 2*Sqrt[3]*(3*I + Sqrt[3])*Log[-1/2 - (I/2)*Sqrt[3] +
x^2] + 2*Sqrt[3]*(-3*I + Sqrt[3])*Log[(I/2)*(I + Sqrt[3]) + x^2] + ((3 - I*Sqrt[3])*Log[16*(1 + Sqrt[3]*x + x^
2)^2])/Sqrt[(1 + I*Sqrt[3])/6] + ((3 + I*Sqrt[3])*Log[16*(1 + Sqrt[3]*x + x^2)^2])/Sqrt[(1 - I*Sqrt[3])/6] - (
I*(-3*I + Sqrt[3])*Log[(4 - 4*Sqrt[3]*x + 4*x^2)^2])/Sqrt[(1 - I*Sqrt[3])/6] + (I*(3*I + Sqrt[3])*Log[(4 - 4*S
qrt[3]*x + 4*x^2)^2])/Sqrt[(1 + I*Sqrt[3])/6] - (I*(-3*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])])/Sqrt[(1 - I*Sqrt[3])/6] + ((3 + I*Sqrt[3])*Log[3*I + Sqrt[3] - (-I + Sqrt[3])*x^4 + (2*I)*Sqr
t[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)*Sq
rt[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] + (I*(3*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])
])/Sqrt[(1 + I*Sqrt[3])/6] + ((3 - I*Sqrt[3])*Log[-3*I + Sqrt[3] - (I + Sqrt[3])*x^4 - (2*I)*Sqrt[2 + (2*I)*Sq
rt[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])/48

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{x\arctan \left ( x+\sqrt{-{x}^{2}+1} \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\sqrt{x + 1} \sqrt{-x + 1} \arctan \left (x + \sqrt{x + 1} \sqrt{-x + 1}\right ) - \int \frac{x}{x^{2} + 2 \, x e^{\left (\frac{1}{2} \, \log \left (x + 1\right ) + \frac{1}{2} \, \log \left (-x + 1\right )\right )} + e^{\left (\log \left (x + 1\right ) + \log \left (-x + 1\right )\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(x + 1)*sqrt(-x + 1)*arctan(x + sqrt(x + 1)*sqrt(-x + 1)) - integrate(x/(x^2 + 2*x*e^(1/2*log(x + 1) + 1/
2*log(-x + 1)) + e^(log(x + 1) + log(-x + 1)) + 1), x)

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Fricas [A]  time = 2.63759, size = 548, normalized size = 3.61 \begin{align*} -\frac{1}{4} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - \sqrt{-x^{2} + 1} \arctan \left (x + \sqrt{-x^{2} + 1}\right ) - \frac{1}{8} \, \sqrt{3} \arctan \left (\frac{4 \, \sqrt{3} \sqrt{-x^{2} + 1} x + \sqrt{3}}{3 \,{\left (2 \, x^{2} - 1\right )}}\right ) - \frac{1}{8} \, \sqrt{3} \arctan \left (\frac{4 \, \sqrt{3} \sqrt{-x^{2} + 1} x - \sqrt{3}}{3 \,{\left (2 \, x^{2} - 1\right )}}\right ) + \frac{1}{2} \, \arctan \left (\frac{\sqrt{-x^{2} + 1} x}{x^{2} - 1}\right ) + \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac{1}{16} \, \log \left (-x^{4} + x^{2} + 2 \, \sqrt{-x^{2} + 1} x + 1\right ) - \frac{1}{16} \, \log \left (-x^{4} + x^{2} - 2 \, \sqrt{-x^{2} + 1} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.24644, size = 504, normalized size = 3.32 \begin{align*} -\frac{1}{4} \, \pi \mathrm{sgn}\left (x\right ) + \frac{1}{8} \, \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}{8} \, \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}{4} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - \sqrt{-x^{2} + 1} \arctan \left (x + \sqrt{-x^{2} + 1}\right ) - \frac{1}{2} \, \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 ) + \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) - \frac{1}{8} \, \log \left ({\left (\frac{x}{\sqrt{-x^{2} + 1} - 1} - \frac{\sqrt{-x^{2} + 1} - 1}{x}\right )}^{2} + \frac{2 \, x}{\sqrt{-x^{2} + 1} - 1} - \frac{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{x} + 4\right ) + \frac{1}{8} \, \log \left ({\left (\frac{x}{\sqrt{-x^{2} + 1} - 1} - \frac{\sqrt{-x^{2} + 1} - 1}{x}\right )}^{2} - \frac{2 \, x}{\sqrt{-x^{2} + 1} - 1} + \frac{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{x} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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