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

Optimal. Leaf size=141 \[ \frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x-1}{\sqrt {1-x^2}}\right )+\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x+1}{\sqrt {1-x^2}}\right )-\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {2 x^2-1}{\sqrt {3}}\right )+x \tan ^{-1}\left (\sqrt {1-x^2}+x\right )-\frac {1}{4} \tanh ^{-1}\left (x \sqrt {1-x^2}\right )-\frac {1}{8} \log \left (x^4-x^2+1\right )-\frac {1}{2} \sin ^{-1}(x) \]

[Out]

-1/2*arcsin(x)+x*arctan(x+(-x^2+1)^(1/2))-1/4*arctanh(x*(-x^2+1)^(1/2))-1/8*ln(x^4-x^2+1)-1/4*arctan(1/3*(2*x^
2-1)*3^(1/2))*3^(1/2)+1/4*arctan((-1+x*3^(1/2))/(-x^2+1)^(1/2))*3^(1/2)+1/4*arctan((1+x*3^(1/2))/(-x^2+1)^(1/2
))*3^(1/2)

________________________________________________________________________________________

Rubi [C]  time = 0.75, antiderivative size = 269, normalized size of antiderivative = 1.91, number of steps used = 40, number of rules used = 15, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {5203, 12, 6742, 216, 1114, 634, 618, 204, 628, 1174, 402, 377, 205, 1293, 1107} \[ -\frac {1}{8} \log \left (x^4-x^2+1\right )+\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )+\frac {1}{12} \left (-\sqrt {3}+3 i\right ) \tan ^{-1}\left (\frac {x}{\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt {1-x^2}}\right )+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt {1-x^2}}\right )}{\sqrt {3}}-\frac {1}{12} \left (\sqrt {3}+3 i\right ) \tan ^{-1}\left (\frac {\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt {1-x^2}}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt {1-x^2}}\right )}{\sqrt {3}}+x \tan ^{-1}\left (\sqrt {1-x^2}+x\right )-\frac {1}{2} \sin ^{-1}(x) \]

Warning: Unable to verify antiderivative.

[In]

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

Rule 12

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

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 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 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 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]

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 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 1114

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

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 5203

Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 + u^2), x], x] /; Inv
erseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

\begin {align*} \int \tan ^{-1}\left (x+\sqrt {1-x^2}\right ) \, dx &=x \tan ^{-1}\left (x+\sqrt {1-x^2}\right )-\int \frac {x \left (1-\frac {x}{\sqrt {1-x^2}}\right )}{2 \left (1+x \sqrt {1-x^2}\right )} \, dx\\ &=x \tan ^{-1}\left (x+\sqrt {1-x^2}\right )-\frac {1}{2} \int \frac {x \left (1-\frac {x}{\sqrt {1-x^2}}\right )}{1+x \sqrt {1-x^2}} \, dx\\ &=x \tan ^{-1}\left (x+\sqrt {1-x^2}\right )-\frac {1}{2} \int \left (\frac {x^2}{-x+x^3-\sqrt {1-x^2}}+\frac {x}{1+x \sqrt {1-x^2}}\right ) \, dx\\ &=x \tan ^{-1}\left (x+\sqrt {1-x^2}\right )-\frac {1}{2} \int \frac {x^2}{-x+x^3-\sqrt {1-x^2}} \, dx-\frac {1}{2} \int \frac {x}{1+x \sqrt {1-x^2}} \, dx\\ &=x \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 {1}{\sqrt {1-x^2}}+\frac {x^3}{1-x^2+x^4}+\frac {\sqrt {1-x^2}}{1-x^2+x^4}-\frac {x^2 \sqrt {1-x^2}}{1-x^2+x^4}\right ) \, dx\\ &=x \tan ^{-1}\left (x+\sqrt {1-x^2}\right )+\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx-\frac {1}{2} \int \frac {x}{1-x^2+x^4} \, dx-\frac {1}{2} \int \frac {x^3}{1-x^2+x^4} \, dx-\frac {1}{2} \int \frac {\sqrt {1-x^2}}{1-x^2+x^4} \, dx+2 \left (\frac {1}{2} \int \frac {x^2 \sqrt {1-x^2}}{1-x^2+x^4} \, dx\right )\\ &=\frac {1}{2} \sin ^{-1}(x)+x \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 {x}{1-x+x^2} \, dx,x,x^2\right )+2 \left (-\left (\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx\right )+\frac {1}{2} \int \frac {1}{\sqrt {1-x^2} \left (1-x^2+x^4\right )} \, dx\right )+\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)+x \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 )+2 \left (-\frac {1}{2} \sin ^{-1}(x)-\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}}\right )+\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}}+x \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 )+2 \left (-\frac {1}{2} \sin ^{-1}(x)+\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}}\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}{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 {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 {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 )+2 \left (-\frac {1}{2} \sin ^{-1}(x)+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt {1-x^2}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt {1-x^2}}\right )}{2 \sqrt {3}}\right )+x \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 = 3.80, size = 1822, normalized size = 12.92 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[x + Sqrt[1 - x^2]],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 191, normalized size = 1.35 \[ x \arctan \left (x + \sqrt {-x^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} \sqrt {-x^{2} + 1} x + \sqrt {3}}{3 \, {\left (2 \, x^{2} - 1\right )}}\right ) - \frac {1}{8} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} \sqrt {-x^{2} + 1} x - \sqrt {3}}{3 \, {\left (2 \, x^{2} - 1\right )}}\right ) + \frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{2} + 1} x}{x^{2} - 1}\right ) - \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) - \frac {1}{16} \, \log \left (-x^{4} + x^{2} + 2 \, \sqrt {-x^{2} + 1} x + 1\right ) + \frac {1}{16} \, \log \left (-x^{4} + x^{2} - 2 \, \sqrt {-x^{2} + 1} x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.47, size = 364, normalized size = 2.58 \[ x \arctan \left (x + \sqrt {-x^{2} + 1}\right ) - \frac {1}{4} \, \pi \mathrm {sgn}\relax (x) + \frac {1}{8} \, \sqrt {3} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (-\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} + \frac {1}{8} \, \sqrt {3} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{2} \, \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) - \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{8} \, \log \left ({\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} + \frac {2 \, x}{\sqrt {-x^{2} + 1} - 1} - \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} + 4\right ) - \frac {1}{8} \, \log \left ({\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} - \frac {2 \, x}{\sqrt {-x^{2} + 1} - 1} + \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(x + sqrt(-x^2 + 1)) - 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*sqr
t(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)) - 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*(sq
rt(-x^2 + 1) - 1)/x + 4) - 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)

________________________________________________________________________________________

maple [C]  time = 0.10, size = 439, normalized size = 3.11 \[ x \arctan \left (x +\sqrt {-x^{2}+1}\right )+\arctan \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{4}+\frac {i \sqrt {3}\, \ln \left (\frac {\left (-1-i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}+\frac {\ln \left (\frac {\left (-1-i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}+\frac {\ln \left (\frac {\left (-1+i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}-\frac {i \sqrt {3}\, \ln \left (\frac {\left (-1+i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}+\frac {i \sqrt {3}\, \ln \left (\frac {\left (1-i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}-\frac {\ln \left (\frac {\left (1-i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}-\frac {i \sqrt {3}\, \ln \left (\frac {\left (1+i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}-\frac {\ln \left (\frac {\left (1+i \sqrt {3}\right ) \left (\sqrt {-x^{2}+1}-1\right )}{x}+\frac {\left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-1\right )}{8}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(x+(-x^2+1)^(1/2))-1/8*ln(x^4-x^2+1)-1/4*arctan(1/3*(2*x^2-1)*3^(1/2))*3^(1/2)+1/8*I*3^(1/2)*ln(((-x^2
+1)^(1/2)-1)^2/x^2+(-1-I*3^(1/2))*((-x^2+1)^(1/2)-1)/x-1)+1/8*ln(((-x^2+1)^(1/2)-1)^2/x^2+(-1-I*3^(1/2))*((-x^
2+1)^(1/2)-1)/x-1)-1/8*I*3^(1/2)*ln(((-x^2+1)^(1/2)-1)^2/x^2+(1+I*3^(1/2))*((-x^2+1)^(1/2)-1)/x-1)+1/8*ln(((-x
^2+1)^(1/2)-1)^2/x^2+(-1+I*3^(1/2))*((-x^2+1)^(1/2)-1)/x-1)-1/8*I*3^(1/2)*ln(((-x^2+1)^(1/2)-1)^2/x^2+(-1+I*3^
(1/2))*((-x^2+1)^(1/2)-1)/x-1)-1/8*ln(((-x^2+1)^(1/2)-1)^2/x^2+(1+I*3^(1/2))*((-x^2+1)^(1/2)-1)/x-1)+1/8*I*3^(
1/2)*ln(((-x^2+1)^(1/2)-1)^2/x^2+(1-I*3^(1/2))*((-x^2+1)^(1/2)-1)/x-1)-1/8*ln(((-x^2+1)^(1/2)-1)^2/x^2+(1-I*3^
(1/2))*((-x^2+1)^(1/2)-1)/x-1)+arctan(((-x^2+1)^(1/2)-1)/x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (x + \sqrt {x + 1} \sqrt {-x + 1}\right ) - \int \frac {x^{3} + x^{2} e^{\left (\frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (-x + 1\right )\right )} - x}{x^{4} + {\left (x^{2} - 1\right )} e^{\left (\log \left (x + 1\right ) + \log \left (-x + 1\right )\right )} + 2 \, {\left (x^{3} - x\right )} e^{\left (\frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (-x + 1\right )\right )} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.37, size = 661, normalized size = 4.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x + (1 - x^2)^(1/2)),x)

[Out]

x*atan(x + (1 - x^2)^(1/2)) - asin(x)/2 + (log(x - 3^(1/2)/2 - 1i/2)*(3^(1/2)/2 + (3^(1/2)/2 + 1i/2)^3 + 1i/2)
)/(2*3^(1/2) - 8*(3^(1/2)/2 + 1i/2)^3 + 2i) - (log(x - 3^(1/2)/2 + 1i/2)*(3^(1/2)/2 + (3^(1/2)/2 - 1i/2)^3 - 1
i/2))/(8*(3^(1/2)/2 - 1i/2)^3 - 2*3^(1/2) + 2i) - (log(x + 3^(1/2)/2 - 1i/2)*(3^(1/2)/2 + (3^(1/2)/2 - 1i/2)^3
- 1i/2))/(8*(3^(1/2)/2 - 1i/2)^3 - 2*3^(1/2) + 2i) + (log(x + 3^(1/2)/2 + 1i/2)*(3^(1/2)/2 + (3^(1/2)/2 + 1i/
2)^3 + 1i/2))/(2*3^(1/2) - 8*(3^(1/2)/2 + 1i/2)^3 + 2i) + (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))*((3^(1/2)/2 + 1i/2)^2 + 1))/((1 - (3^(1/2)/2
+ 1i/2)^2)^(1/2)*(2*3^(1/2) - 8*(3^(1/2)/2 + 1i/2)^3 + 2i)) - (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))*((3^(1/2)/2 - 1i/2)^2 + 1))/((1 - (3^(1/2
)/2 - 1i/2)^2)^(1/2)*(8*(3^(1/2)/2 - 1i/2)^3 - 2*3^(1/2) + 2i)) + (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))*((3^(1/2)/2 - 1i/2)^2 + 1))/((1 - (3^
(1/2)/2 - 1i/2)^2)^(1/2)*(8*(3^(1/2)/2 - 1i/2)^3 - 2*3^(1/2) + 2i)) - (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))*((3^(1/2)/2 + 1i/2)^2 + 1))/((1 -
(3^(1/2)/2 + 1i/2)^2)^(1/2)*(2*3^(1/2) - 8*(3^(1/2)/2 + 1i/2)^3 + 2i))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________