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

Optimal. Leaf size=120 \[ -\frac {1}{4} \sqrt {3} \log \left (x^2-\sqrt {3} \sqrt {x^2+1}+2\right )+\frac {1}{4} \sqrt {3} \log \left (x^2+\sqrt {3} \sqrt {x^2+1}+2\right )+x \tan ^{-1}\left (x \sqrt {x^2+1}\right )+\frac {1}{2} \tan ^{-1}\left (\sqrt {3}-2 \sqrt {x^2+1}\right )-\frac {1}{2} \tan ^{-1}\left (2 \sqrt {x^2+1}+\sqrt {3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5203, 1685, 826, 1169, 634, 618, 204, 628} \[ -\frac {1}{4} \sqrt {3} \log \left (x^2-\sqrt {3} \sqrt {x^2+1}+2\right )+\frac {1}{4} \sqrt {3} \log \left (x^2+\sqrt {3} \sqrt {x^2+1}+2\right )+x \tan ^{-1}\left (x \sqrt {x^2+1}\right )+\frac {1}{2} \tan ^{-1}\left (\sqrt {3}-2 \sqrt {x^2+1}\right )-\frac {1}{2} \tan ^{-1}\left (2 \sqrt {x^2+1}+\sqrt {3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x*Sqrt[1 + x^2]],x]

[Out]

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

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 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 826

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

Rule 1169

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

Rule 1685

Int[(Px_)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[(Px /. x -> Sqrt[x])*(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] && PolyQ[Px, x^2]

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]

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

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Mathematica [C]  time = 0.31, size = 136, normalized size = 1.13 \[ \frac {1}{4} \left (4 x \tan ^{-1}\left (x \sqrt {x^2+1}\right )+\left (1+i \sqrt {3}\right ) \sqrt {2-2 i \sqrt {3}} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x^2+1}}{\sqrt {1-i \sqrt {3}}}\right )+\left (1-i \sqrt {3}\right ) \sqrt {2+2 i \sqrt {3}} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x^2+1}}{\sqrt {1+i \sqrt {3}}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x*Sqrt[1 + x^2]],x]

[Out]

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

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fricas [B]  time = 0.48, size = 287, normalized size = 2.39 \[ x \arctan \left (\sqrt {x^{2} + 1} x\right ) - \frac {1}{4} \, \sqrt {3} \log \left (32 \, x^{4} + 80 \, x^{2} + 32 \, \sqrt {3} {\left (x^{3} + x\right )} - 16 \, {\left (2 \, x^{3} + \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 32\right ) + \frac {1}{4} \, \sqrt {3} \log \left (32 \, x^{4} + 80 \, x^{2} - 32 \, \sqrt {3} {\left (x^{3} + x\right )} - 16 \, {\left (2 \, x^{3} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 32\right ) + \arctan \left (2 \, \sqrt {2 \, x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} + \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2} {\left (x + \sqrt {x^{2} + 1}\right )} + \sqrt {3} - 2 \, \sqrt {x^{2} + 1}\right ) + \arctan \left (2 \, \sqrt {2 \, x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2} {\left (x + \sqrt {x^{2} + 1}\right )} - \sqrt {3} - 2 \, \sqrt {x^{2} + 1}\right ) \]

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(sqrt(x^2 + 1)*x) - 1/4*sqrt(3)*log(32*x^4 + 80*x^2 + 32*sqrt(3)*(x^3 + x) - 16*(2*x^3 + sqrt(3)*(2*x^
2 + 1) + 4*x)*sqrt(x^2 + 1) + 32) + 1/4*sqrt(3)*log(32*x^4 + 80*x^2 - 32*sqrt(3)*(x^3 + x) - 16*(2*x^3 - sqrt(
3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 32) + arctan(2*sqrt(2*x^4 + 5*x^2 + 2*sqrt(3)*(x^3 + x) - (2*x^3 + sqrt(
3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 2)*(x + sqrt(x^2 + 1)) + sqrt(3) - 2*sqrt(x^2 + 1)) + arctan(2*sqrt(2*x^
4 + 5*x^2 - 2*sqrt(3)*(x^3 + x) - (2*x^3 - sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 2)*(x + sqrt(x^2 + 1)) -
sqrt(3) - 2*sqrt(x^2 + 1))

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giac [A]  time = 1.15, size = 92, normalized size = 0.77 \[ x \arctan \left (\sqrt {x^{2} + 1} x\right ) + \frac {1}{4} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) - \frac {1}{4} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) - \frac {1}{2} \, \arctan \left (\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) - \frac {1}{2} \, \arctan \left (-\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) \]

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(sqrt(x^2 + 1)*x) + 1/4*sqrt(3)*log(x^2 + sqrt(3)*sqrt(x^2 + 1) + 2) - 1/4*sqrt(3)*log(x^2 - sqrt(3)*s
qrt(x^2 + 1) + 2) - 1/2*arctan(sqrt(3) + 2*sqrt(x^2 + 1)) - 1/2*arctan(-sqrt(3) + 2*sqrt(x^2 + 1))

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maple [B]  time = 0.05, size = 508, normalized size = 4.23 \[ x \arctan \left (\sqrt {x^{2}+1}\, x \right )+\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \sqrt {3}}{2}\right )}{3 \sqrt {\frac {\frac {\left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+1}{\left (\frac {x -1}{-x -1}+1\right )^{2}}}\, \left (\frac {x -1}{-x -1}+1\right )}+\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \sqrt {3}}{2}\right )}{3 \sqrt {\frac {\frac {\left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+1}{\left (\frac {x +1}{-x +1}+1\right )^{2}}}\, \left (\frac {x +1}{-x +1}+1\right )}-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \sqrt {3}}{2}\right )-3 \arctan \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \left (x -1\right )}{\left (\frac {\left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+1\right ) \left (-x -1\right )}\right )\right )}{12 \sqrt {\frac {\frac {\left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+1}{\left (\frac {x -1}{-x -1}+1\right )^{2}}}\, \left (\frac {x -1}{-x -1}+1\right )}-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \sqrt {3}}{2}\right )-3 \arctan \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \left (x +1\right )}{\left (\frac {\left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+1\right ) \left (-x +1\right )}\right )\right )}{12 \sqrt {\frac {\frac {\left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+1}{\left (\frac {x +1}{-x +1}+1\right )^{2}}}\, \left (\frac {x +1}{-x +1}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan((x^2+1)^(1/2)*x)+1/3*2^(1/2)/(((x-1)^2/(-1-x)^2+1)/((x-1)/(-1-x)+1)^2)^(1/2)/((x-1)/(-1-x)+1)*(2*(x-1
)^2/(-1-x)^2+2)^(1/2)*3^(1/2)*arctanh(1/2*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*3^(1/2))+1/3*2^(1/2)/(((x+1)^2/(-x+1)^2
+1)/((x+1)/(-x+1)+1)^2)^(1/2)/((x+1)/(-x+1)+1)*(2*(x+1)^2/(-x+1)^2+2)^(1/2)*3^(1/2)*arctanh(1/2*(2*(x+1)^2/(-x
+1)^2+2)^(1/2)*3^(1/2))-1/12*2^(1/2)*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*(3^(1/2)*arctanh(1/2*(2*(x-1)^2/(-1-x)^2+2)^
(1/2)*3^(1/2))-3*arctan(1/((x-1)^2/(-1-x)^2+1)*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*(x-1)/(-1-x)))/(((x-1)^2/(-1-x)^2+
1)/((x-1)/(-1-x)+1)^2)^(1/2)/((x-1)/(-1-x)+1)-1/12*2^(1/2)*(2*(x+1)^2/(-x+1)^2+2)^(1/2)*(3^(1/2)*arctanh(1/2*(
2*(x+1)^2/(-x+1)^2+2)^(1/2)*3^(1/2))-3*arctan(1/((x+1)^2/(-x+1)^2+1)*(2*(x+1)^2/(-x+1)^2+2)^(1/2)*(x+1)/(-x+1)
))/(((x+1)^2/(-x+1)^2+1)/((x+1)/(-x+1)+1)^2)^(1/2)/((x+1)/(-x+1)+1)

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

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

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mupad [B]  time = 1.09, size = 413, normalized size = 3.44 \[ x\,\mathrm {atan}\left (x\,\sqrt {x^2+1}\right )-\frac {\left (\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (\frac {x}{2}+\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}+1+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+1+\sqrt {3}\,1{}\mathrm {i}\right )}-\frac {\left (\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (1+\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}-\frac {x}{2}+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-1+\sqrt {3}\,1{}\mathrm {i}\right )}-\frac {\left (\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (\frac {x}{2}+\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}+1-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-1+\sqrt {3}\,1{}\mathrm {i}\right )}-\frac {\left (\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (1+\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}-\frac {x}{2}-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+1+\sqrt {3}\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x*(x^2 + 1)^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {atan}{\left (x \sqrt {x^{2} + 1} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(x*sqrt(x**2 + 1)), x)

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