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

Optimal. Leaf size=106 \[ -\sqrt{\frac{1}{2} \left (1+\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{\frac{1}{2} \left (1+\sqrt{5}\right )} \sqrt{1-x^2}\right )+x \tan ^{-1}\left (x \sqrt{1-x^2}\right )+\sqrt{\frac{1}{2} \left (\sqrt{5}-1\right )} \tanh ^{-1}\left (\sqrt{\frac{1}{2} \left (\sqrt{5}-1\right )} \sqrt{1-x^2}\right ) \]

[Out]

-(Sqrt[(1 + Sqrt[5])/2]*ArcTan[Sqrt[(1 + Sqrt[5])/2]*Sqrt[1 - x^2]]) + x*ArcTan[x*Sqrt[1 - x^2]] + Sqrt[(-1 +
Sqrt[5])/2]*ArcTanh[Sqrt[(-1 + Sqrt[5])/2]*Sqrt[1 - x^2]]

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Rubi [A]  time = 0.109557, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5203, 1685, 826, 1166, 204, 206} \[ -\sqrt{\frac{2}{\sqrt{5}-1}} \tan ^{-1}\left (\sqrt{\frac{2}{\sqrt{5}-1}} \sqrt{1-x^2}\right )+x \tan ^{-1}\left (x \sqrt{1-x^2}\right )+\sqrt{\frac{2}{1+\sqrt{5}}} \tanh ^{-1}\left (\sqrt{\frac{2}{1+\sqrt{5}}} \sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x*Sqrt[1 - x^2]],x]

[Out]

-(Sqrt[2/(-1 + Sqrt[5])]*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*Sqrt[1 - x^2]]) + x*ArcTan[x*Sqrt[1 - x^2]] + Sqrt[2/(1
+ Sqrt[5])]*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*Sqrt[1 - 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]

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 1166

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 )+\operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}-\frac{\sqrt{5}}{2}-x^2} \, dx,x,\sqrt{1-x^2}\right )+\operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}+\frac{\sqrt{5}}{2}-x^2} \, dx,x,\sqrt{1-x^2}\right )\\ &=-\sqrt{\frac{2}{-1+\sqrt{5}}} \tan ^{-1}\left (\sqrt{\frac{2}{-1+\sqrt{5}}} \sqrt{1-x^2}\right )+x \tan ^{-1}\left (x \sqrt{1-x^2}\right )+\sqrt{\frac{2}{1+\sqrt{5}}} \tanh ^{-1}\left (\sqrt{\frac{2}{1+\sqrt{5}}} \sqrt{1-x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.178952, size = 106, normalized size = 1. \[ x \tan ^{-1}\left (x \sqrt{1-x^2}\right )-\frac{\sqrt{\frac{2}{\sqrt{5}-1}} \left (\left (1+\sqrt{5}\right ) \tan ^{-1}\left (\sqrt{\frac{2}{\sqrt{5}-1}} \sqrt{1-x^2}\right )-2 \tanh ^{-1}\left (\frac{\sqrt{2-2 x^2}}{\sqrt{1+\sqrt{5}}}\right )\right )}{1+\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x*Sqrt[1 - x^2]],x]

[Out]

x*ArcTan[x*Sqrt[1 - x^2]] - (Sqrt[2/(-1 + Sqrt[5])]*((1 + Sqrt[5])*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*Sqrt[1 - x^2]
] - 2*ArcTanh[Sqrt[2 - 2*x^2]/Sqrt[1 + Sqrt[5]]]))/(1 + Sqrt[5])

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Maple [B]  time = 0.052, size = 198, normalized size = 1.9 \begin{align*} x\arctan \left ( x\sqrt{-{x}^{2}+1} \right ) +{\frac{\sqrt{5}}{2\,\sqrt{2+\sqrt{5}}}{\it Artanh} \left ({\frac{1}{4\,\sqrt{2+\sqrt{5}}} \left ( 2\,{\frac{ \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}{{x}^{2}}}+4+2\,\sqrt{5} \right ) } \right ) }+{\frac{\sqrt{5}}{2\,\sqrt{-2+\sqrt{5}}}\arctan \left ({\frac{1}{4\,\sqrt{-2+\sqrt{5}}} \left ( 2\,{\frac{ \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}{{x}^{2}}}-2\,\sqrt{5}+4 \right ) } \right ) }+{\frac{1}{2\,\sqrt{2+\sqrt{5}}}{\it Artanh} \left ({\frac{1}{4\,\sqrt{2+\sqrt{5}}} \left ( 2\,{\frac{ \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}{{x}^{2}}}+4+2\,\sqrt{5} \right ) } \right ) }-{\frac{1}{2\,\sqrt{-2+\sqrt{5}}}\arctan \left ({\frac{1}{4\,\sqrt{-2+\sqrt{5}}} \left ( 2\,{\frac{ \left ( \sqrt{-{x}^{2}+1}-1 \right ) ^{2}}{{x}^{2}}}-2\,\sqrt{5}+4 \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))+1/2*5^(1/2)/(2+5^(1/2))^(1/2)*arctanh(1/4*(2*((-x^2+1)^(1/2)-1)^2/x^2+4+2*5^(1/2))/
(2+5^(1/2))^(1/2))+1/2*5^(1/2)/(-2+5^(1/2))^(1/2)*arctan(1/4*(2*((-x^2+1)^(1/2)-1)^2/x^2-2*5^(1/2)+4)/(-2+5^(1
/2))^(1/2))+1/2/(2+5^(1/2))^(1/2)*arctanh(1/4*(2*((-x^2+1)^(1/2)-1)^2/x^2+4+2*5^(1/2))/(2+5^(1/2))^(1/2))-1/2/
(-2+5^(1/2))^(1/2)*arctan(1/4*(2*((-x^2+1)^(1/2)-1)^2/x^2-2*5^(1/2)+4)/(-2+5^(1/2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \arctan \left (\sqrt{x + 1} x \sqrt{-x + 1}\right ) - \int \frac{{\left (2 \, x^{3} - x\right )} e^{\left (\frac{1}{2} \, \log \left (x + 1\right ) + \frac{1}{2} \, \log \left (-x + 1\right )\right )}}{x^{2} +{\left (x^{4} - x^{2}\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(arctan(x*(-x^2+1)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.19223, size = 509, normalized size = 4.8 \begin{align*} x \arctan \left (\sqrt{-x^{2} + 1} x\right ) + \sqrt{2} \sqrt{\sqrt{5} + 1} \arctan \left (-\frac{1}{2} \, \sqrt{2} \sqrt{-x^{2} + 1} \sqrt{\sqrt{5} + 1} + \frac{1}{8} \, \sqrt{2} \sqrt{-16 \, x^{2} + 8 \, \sqrt{5} + 8} \sqrt{\sqrt{5} + 1}\right ) + \frac{1}{4} \, \sqrt{2} \sqrt{\sqrt{5} - 1} \log \left ({\left (\sqrt{5} \sqrt{2} + \sqrt{2}\right )} \sqrt{\sqrt{5} - 1} + 4 \, \sqrt{-x^{2} + 1}\right ) - \frac{1}{4} \, \sqrt{2} \sqrt{\sqrt{5} - 1} \log \left (-{\left (\sqrt{5} \sqrt{2} + \sqrt{2}\right )} \sqrt{\sqrt{5} - 1} + 4 \, \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) + sqrt(2)*sqrt(sqrt(5) + 1)*arctan(-1/2*sqrt(2)*sqrt(-x^2 + 1)*sqrt(sqrt(5) + 1) +
1/8*sqrt(2)*sqrt(-16*x^2 + 8*sqrt(5) + 8)*sqrt(sqrt(5) + 1)) + 1/4*sqrt(2)*sqrt(sqrt(5) - 1)*log((sqrt(5)*sqrt
(2) + sqrt(2))*sqrt(sqrt(5) - 1) + 4*sqrt(-x^2 + 1)) - 1/4*sqrt(2)*sqrt(sqrt(5) - 1)*log(-(sqrt(5)*sqrt(2) + s
qrt(2))*sqrt(sqrt(5) - 1) + 4*sqrt(-x^2 + 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 [A]  time = 1.13074, size = 150, normalized size = 1.42 \begin{align*} x \arctan \left (\sqrt{-x^{2} + 1} x\right ) - \frac{1}{2} \, \sqrt{2 \, \sqrt{5} + 2} \arctan \left (\frac{\sqrt{-x^{2} + 1}}{\sqrt{\frac{1}{2} \, \sqrt{5} - \frac{1}{2}}}\right ) + \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2} \log \left (\sqrt{-x^{2} + 1} + \sqrt{\frac{1}{2} \, \sqrt{5} + \frac{1}{2}}\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2} \log \left ({\left | \sqrt{-x^{2} + 1} - \sqrt{\frac{1}{2} \, \sqrt{5} + \frac{1}{2}} \right |}\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(sqrt(-x^2 + 1)*x) - 1/2*sqrt(2*sqrt(5) + 2)*arctan(sqrt(-x^2 + 1)/sqrt(1/2*sqrt(5) - 1/2)) + 1/4*sqrt
(2*sqrt(5) - 2)*log(sqrt(-x^2 + 1) + sqrt(1/2*sqrt(5) + 1/2)) - 1/4*sqrt(2*sqrt(5) - 2)*log(abs(sqrt(-x^2 + 1)
- sqrt(1/2*sqrt(5) + 1/2)))