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

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

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

________________________________________________________________________________________

Rubi [C]  time = 0.745992, 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 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 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 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 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 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 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 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 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 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 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 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 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]

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 = 3.39686, 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

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Maple [C]  time = 0.061, size = 439, normalized size = 3.1 \begin{align*} x\arctan \left ( x+\sqrt{-{x}^{2}+1} \right ) -{\frac{\ln \left ({x}^{4}-{x}^{2}+1 \right ) }{8}}-{\frac{\sqrt{3}}{4}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{i}{8}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{-1-i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) +{\frac{1}{8}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{-1-i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) }-{\frac{i}{8}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{-1+i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) +{\frac{1}{8}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{-1+i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) }-{\frac{i}{8}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{1+i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) -{\frac{1}{8}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{1+i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) }+{\frac{i}{8}}\sqrt{3}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{1-i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) -{\frac{1}{8}\ln \left ({\frac{1}{{x}^{2}} \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}+{\frac{1-i\sqrt{3}}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-1 \right ) }+\arctan \left ({\frac{1}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) } \right ) \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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} \end{align*}

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)

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Fricas [A]  time = 2.62975, size = 529, normalized size = 3.75 \begin{align*} 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 ) \end{align*}

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)

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

[Out]

Timed out

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Giac [B]  time = 1.13898, size = 491, normalized size = 3.48 \begin{align*} x \arctan \left (x + \sqrt{-x^{2} + 1}\right ) - \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 ) - \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(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)