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]

x*arctan(x*(-x^2+1)^(1/2))+1/2*arctanh(1/2*(-x^2+1)^(1/2)*(-2+2*5^(1/2))^(1/2))*(-2+2*5^(1/2))^(1/2)-1/2*arcta
n(1/2*(-x^2+1)^(1/2)*(2+2*5^(1/2))^(1/2))*(2+2*5^(1/2))^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 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])

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 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 )+\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*}

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Mathematica [A]  time = 0.18, size = 106, normalized size = 1.00 \[ x \tan ^{-1}\left (x \sqrt {1-x^2}\right )-\frac {\sqrt {\frac {2}{\sqrt {5}-1}} \left (\left (1+\sqrt {5}\right ) \tan ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} \sqrt {1-x^2}\right )-2 \tanh ^{-1}\left (\frac {\sqrt {2-2 x^2}}{\sqrt {1+\sqrt {5}}}\right )\right )}{1+\sqrt {5}} \]

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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fricas [B]  time = 0.51, size = 164, normalized size = 1.55 \[ x \arctan \left (\sqrt {-x^{2} + 1} x\right ) + \sqrt {2} \sqrt {\sqrt {5} + 1} \arctan \left (-\frac {1}{2} \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {\sqrt {5} + 1} + \frac {1}{8} \, \sqrt {2} \sqrt {-16 \, x^{2} + 8 \, \sqrt {5} + 8} \sqrt {\sqrt {5} + 1}\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {5} - 1} \log \left ({\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 1} + 4 \, \sqrt {-x^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {5} - 1} \log \left (-{\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 1} + 4 \, \sqrt {-x^{2} + 1}\right ) \]

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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giac [A]  time = 1.46, size = 111, normalized size = 1.05 \[ x \arctan \left (\sqrt {-x^{2} + 1} x\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {5} + 2} \arctan \left (\frac {\sqrt {-x^{2} + 1}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (\sqrt {-x^{2} + 1} + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left ({\left | \sqrt {-x^{2} + 1} - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) \]

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)))

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maple [B]  time = 0.09, size = 198, normalized size = 1.87 \[ x \arctan \left (\sqrt {-x^{2}+1}\, x \right )+\frac {\sqrt {5}\, \arctanh \left (\frac {\frac {2 \left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}+4+2 \sqrt {5}}{4 \sqrt {2+\sqrt {5}}}\right )}{2 \sqrt {2+\sqrt {5}}}+\frac {\arctanh \left (\frac {\frac {2 \left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}+4+2 \sqrt {5}}{4 \sqrt {2+\sqrt {5}}}\right )}{2 \sqrt {2+\sqrt {5}}}+\frac {\sqrt {5}\, \arctan \left (\frac {\frac {2 \left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-2 \sqrt {5}+4}{4 \sqrt {-2+\sqrt {5}}}\right )}{2 \sqrt {-2+\sqrt {5}}}-\frac {\arctan \left (\frac {\frac {2 \left (\sqrt {-x^{2}+1}-1\right )^{2}}{x^{2}}-2 \sqrt {5}+4}{4 \sqrt {-2+\sqrt {5}}}\right )}{2 \sqrt {-2+\sqrt {5}}} \]

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

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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mupad [B]  time = 1.16, size = 455, normalized size = 4.29 \[ x\,\mathrm {atan}\left (x\,\sqrt {1-x^2}\right )+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}-1\right )\,1{}\mathrm {i}}{\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\,\left (\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}-2\,{\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )}^{3/2}\right )}{\left (2\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}-4\,{\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}-1\right )\,1{}\mathrm {i}}{\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )\,\left (\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}-2\,{\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right )}^{3/2}\right )}{\left (2\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}-4\,{\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right )}^{3/2}\right )\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}}+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}+1\right )\,1{}\mathrm {i}}{\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\,\left (\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}-2\,{\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )}^{3/2}\right )}{\left (2\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}-4\,{\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}+1\right )\,1{}\mathrm {i}}{\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )\,\left (\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}-2\,{\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right )}^{3/2}\right )}{\left (2\,\sqrt {\frac {1}{2}-\frac {\sqrt {5}}{2}}-4\,{\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right )}^{3/2}\right )\,\sqrt {\frac {\sqrt {5}}{2}+\frac {1}{2}}} \]

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)) + (log((((x*(5^(1/2)/2 + 1/2)^(1/2) - 1)*1i)/(1/2 - 5^(1/2)/2)^(1/2) - (1 - x^2)^(1/
2)*1i)/(x - (5^(1/2)/2 + 1/2)^(1/2)))*((5^(1/2)/2 + 1/2)^(1/2) - 2*(5^(1/2)/2 + 1/2)^(3/2)))/((2*(5^(1/2)/2 +
1/2)^(1/2) - 4*(5^(1/2)/2 + 1/2)^(3/2))*(1/2 - 5^(1/2)/2)^(1/2)) + (log((((x*(1/2 - 5^(1/2)/2)^(1/2) - 1)*1i)/
(5^(1/2)/2 + 1/2)^(1/2) - (1 - x^2)^(1/2)*1i)/(x - (1/2 - 5^(1/2)/2)^(1/2)))*((1/2 - 5^(1/2)/2)^(1/2) - 2*(1/2
- 5^(1/2)/2)^(3/2)))/((2*(1/2 - 5^(1/2)/2)^(1/2) - 4*(1/2 - 5^(1/2)/2)^(3/2))*(5^(1/2)/2 + 1/2)^(1/2)) + (log
((((x*(5^(1/2)/2 + 1/2)^(1/2) + 1)*1i)/(1/2 - 5^(1/2)/2)^(1/2) + (1 - x^2)^(1/2)*1i)/(x + (5^(1/2)/2 + 1/2)^(1
/2)))*((5^(1/2)/2 + 1/2)^(1/2) - 2*(5^(1/2)/2 + 1/2)^(3/2)))/((2*(5^(1/2)/2 + 1/2)^(1/2) - 4*(5^(1/2)/2 + 1/2)
^(3/2))*(1/2 - 5^(1/2)/2)^(1/2)) + (log((((x*(1/2 - 5^(1/2)/2)^(1/2) + 1)*1i)/(5^(1/2)/2 + 1/2)^(1/2) + (1 - x
^2)^(1/2)*1i)/(x + (1/2 - 5^(1/2)/2)^(1/2)))*((1/2 - 5^(1/2)/2)^(1/2) - 2*(1/2 - 5^(1/2)/2)^(3/2)))/((2*(1/2 -
5^(1/2)/2)^(1/2) - 4*(1/2 - 5^(1/2)/2)^(3/2))*(5^(1/2)/2 + 1/2)^(1/2))

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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

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