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]

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

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Rubi [A]  time = 0.130676, antiderivative size = 120, normalized size of antiderivative = 1., 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 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 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 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 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]

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*}

Mathematica [C]  time = 0.307407, 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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Maple [B]  time = 0.035, size = 510, normalized size = 4.3 \begin{align*} \text{result too large to display} \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))+1/3*2^(1/2)/(((-1+x)^2/(-1-x)^2+1)/(1+(-1+x)/(-1-x))^2)^(1/2)/(1+(-1+x)/(-1-x))*(2*(
-1+x)^2/(-1-x)^2+2)^(1/2)*3^(1/2)*arctanh(1/2*(2*(-1+x)^2/(-1-x)^2+2)^(1/2)*3^(1/2))+1/3*2^(1/2)/(((1+x)^2/(1-
x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))*(2*(1+x)^2/(1-x)^2+2)^(1/2)*3^(1/2)*arctanh(1/2*(2*(1+x)^2/(1
-x)^2+2)^(1/2)*3^(1/2))+1/12*2^(1/2)*(2*(-1+x)^2/(-1-x)^2+2)^(1/2)*(-3^(1/2)*arctanh(1/2*(2*(-1+x)^2/(-1-x)^2+
2)^(1/2)*3^(1/2))+3*arctan(1/((-1+x)^2/(-1-x)^2+1)*(2*(-1+x)^2/(-1-x)^2+2)^(1/2)*(-1+x)/(-1-x)))/(((-1+x)^2/(-
1-x)^2+1)/(1+(-1+x)/(-1-x))^2)^(1/2)/(1+(-1+x)/(-1-x))+1/12*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)*(-3^(1/2)*arct
anh(1/2*(2*(1+x)^2/(1-x)^2+2)^(1/2)*3^(1/2))+3*arctan(1/((1+x)^2/(1-x)^2+1)*(2*(1+x)^2/(1-x)^2+2)^(1/2)*(1+x)/
(1-x)))/(((1+x)^2/(1-x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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} \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(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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Fricas [B]  time = 2.6853, size = 782, normalized size = 6.52 \begin{align*} 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 ) \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(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{atan}{\left (x \sqrt{x^{2} + 1} \right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left (\sqrt{x^{2} + 1} x\right )\,{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="giac")

[Out]

integrate(arctan(sqrt(x^2 + 1)*x), x)