3.289 \(\int x \sqrt {\frac {1-x^2}{1+x^2}} \, dx\)

Optimal. Leaf size=51 \[ \frac {1}{2} \sqrt {\frac {1-x^2}{x^2+1}} \left (x^2+1\right )-\tan ^{-1}\left (\sqrt {\frac {1-x^2}{x^2+1}}\right ) \]

[Out]

-arctan(((-x^2+1)/(x^2+1))^(1/2))+1/2*(x^2+1)*((-x^2+1)/(x^2+1))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1960, 288, 204} \[ \frac {1}{2} \sqrt {\frac {1-x^2}{x^2+1}} \left (x^2+1\right )-\tan ^{-1}\left (\sqrt {\frac {1-x^2}{x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[(1 - x^2)/(1 + x^2)],x]

[Out]

(Sqrt[(1 - x^2)/(1 + x^2)]*(1 + x^2))/2 - ArcTan[Sqrt[(1 - x^2)/(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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 86, normalized size = 1.69 \[ \frac {\sqrt {\frac {1-x^2}{x^2+1}} \sqrt {x^2+1} \left (\sqrt {x^2+1} \left (x^2-1\right )+2 \sqrt {1-x^2} \sin ^{-1}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right )\right )}{2 \left (x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[(1 - x^2)/(1 + x^2)],x]

[Out]

(Sqrt[(1 - x^2)/(1 + x^2)]*Sqrt[1 + x^2]*((-1 + x^2)*Sqrt[1 + x^2] + 2*Sqrt[1 - x^2]*ArcSin[Sqrt[1 - x^2]/Sqrt
[2]]))/(2*(-1 + x^2))

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fricas [A]  time = 0.42, size = 55, normalized size = 1.08 \[ \frac {1}{2} \, {\left (x^{2} + 1\right )} \sqrt {-\frac {x^{2} - 1}{x^{2} + 1}} - \arctan \left (\frac {{\left (x^{2} + 1\right )} \sqrt {-\frac {x^{2} - 1}{x^{2} + 1}} - 1}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((-x^2+1)/(x^2+1))^(1/2),x, algorithm="fricas")

[Out]

1/2*(x^2 + 1)*sqrt(-(x^2 - 1)/(x^2 + 1)) - arctan(((x^2 + 1)*sqrt(-(x^2 - 1)/(x^2 + 1)) - 1)/x^2)

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giac [A]  time = 0.28, size = 18, normalized size = 0.35 \[ \frac {1}{2} \, \sqrt {-x^{4} + 1} + \frac {1}{2} \, \arcsin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((-x^2+1)/(x^2+1))^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-x^4 + 1) + 1/2*arcsin(x^2)

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maple [A]  time = 0.02, size = 52, normalized size = 1.02 \[ \frac {\sqrt {-\frac {x^{2}-1}{x^{2}+1}}\, \left (x^{2}+1\right ) \left (\arcsin \left (x^{2}\right )+\sqrt {-x^{4}+1}\right )}{2 \sqrt {-\left (x^{2}-1\right ) \left (x^{2}+1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-(x^2-1)/(x^2+1))^(1/2)*(x^2+1)*((-x^4+1)^(1/2)+arcsin(x^2))/(-(x^2-1)*(x^2+1))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {-\frac {x^{2} - 1}{x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((-x^2+1)/(x^2+1))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.67, size = 55, normalized size = 1.08 \[ -\mathrm {atan}\left (\sqrt {-\frac {x^2-1}{x^2+1}}\right )-\frac {\sqrt {-\frac {x^2-1}{x^2+1}}}{\frac {x^2-1}{x^2+1}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-(x^2 - 1)/(x^2 + 1))^(1/2),x)

[Out]

- atan((-(x^2 - 1)/(x^2 + 1))^(1/2)) - (-(x^2 - 1)/(x^2 + 1))^(1/2)/((x^2 - 1)/(x^2 + 1) - 1)

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sympy [A]  time = 21.54, size = 39, normalized size = 0.76 \[ \begin {cases} \frac {\sqrt {1 - x^{2}} \sqrt {x^{2} + 1}}{2} - \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {1 - x^{2}}}{2} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(1 - x**2)*sqrt(x**2 + 1)/2 - asin(sqrt(2)*sqrt(1 - x**2)/2), (x > -1) & (x < 1)))

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