3.696 \(\int \tan ^{-1}(\sqrt{\frac{-a+x}{a+x}}) \, dx\)

Optimal. Leaf size=40 \[ x \tan ^{-1}\left (\sqrt{-\frac{a-x}{a+x}}\right )-a \tanh ^{-1}\left (\sqrt{-\frac{a-x}{a+x}}\right ) \]

[Out]

x*ArcTan[Sqrt[-((a - x)/(a + x))]] - a*ArcTanh[Sqrt[-((a - x)/(a + x))]]

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Rubi [A]  time = 0.0425593, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5203, 12, 1961, 208} \[ x \tan ^{-1}\left (\sqrt{-\frac{a-x}{a+x}}\right )-a \tanh ^{-1}\left (\sqrt{-\frac{a-x}{a+x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[Sqrt[(-a + x)/(a + x)]],x]

[Out]

x*ArcTan[Sqrt[-((a - x)/(a + x))]] - a*ArcTanh[Sqrt[-((a - x)/(a + x))]]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1961

Int[(u_)^(r_.)*(((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[SimplifyIntegrand[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(1/n -
1)*(u /. x -> (-(a*e) + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r)/(b*e - d*x^q)^(1/n + 1), x], x], x, ((e*(a + b*x
^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && PolynomialQ[u, x] && FractionQ[p] && IntegerQ[1/
n] && IntegerQ[r]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0488343, size = 71, normalized size = 1.78 \[ x \tan ^{-1}\left (\sqrt{\frac{x-a}{a+x}}\right )-\frac{a \sqrt{x-a} \tanh ^{-1}\left (\frac{\sqrt{x-a}}{\sqrt{a+x}}\right )}{\sqrt{\frac{x-a}{a+x}} \sqrt{a+x}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Sqrt[(-a + x)/(a + x)]],x]

[Out]

x*ArcTan[Sqrt[(-a + x)/(a + x)]] - (a*Sqrt[-a + x]*ArcTanh[Sqrt[-a + x]/Sqrt[a + x]])/(Sqrt[(-a + x)/(a + x)]*
Sqrt[a + x])

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Maple [A]  time = 0.013, size = 64, normalized size = 1.6 \begin{align*} x\arctan \left ( \sqrt{{\frac{-a+x}{a+x}}} \right ) -{\frac{ \left ( -a+x \right ) a}{2}\ln \left ( x+\sqrt{-{a}^{2}+{x}^{2}} \right ){\frac{1}{\sqrt{{\frac{-a+x}{a+x}}}}}{\frac{1}{\sqrt{ \left ( a+x \right ) \left ( -a+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(((-a+x)/(a+x))^(1/2)),x)

[Out]

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

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Maxima [B]  time = 1.41736, size = 120, normalized size = 3. \begin{align*} \frac{1}{2} \, a{\left (\frac{4 \, \arctan \left (\sqrt{-\frac{a - x}{a + x}}\right )}{\frac{a - x}{a + x} + 1} - 2 \, \arctan \left (\sqrt{-\frac{a - x}{a + x}}\right ) - \log \left (\sqrt{-\frac{a - x}{a + x}} + 1\right ) + \log \left (\sqrt{-\frac{a - x}{a + x}} - 1\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(4*arctan(sqrt(-(a - x)/(a + x)))/((a - x)/(a + x) + 1) - 2*arctan(sqrt(-(a - x)/(a + x))) - log(sqrt(-(
a - x)/(a + x)) + 1) + log(sqrt(-(a - x)/(a + x)) - 1))

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Fricas [A]  time = 2.88018, size = 154, normalized size = 3.85 \begin{align*} x \arctan \left (\sqrt{-\frac{a - x}{a + x}}\right ) - \frac{1}{2} \, a \log \left (\sqrt{-\frac{a - x}{a + x}} + 1\right ) + \frac{1}{2} \, a \log \left (\sqrt{-\frac{a - x}{a + x}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(sqrt(-(a - x)/(a + x))) - 1/2*a*log(sqrt(-(a - x)/(a + x)) + 1) + 1/2*a*log(sqrt(-(a - x)/(a + x)) -
1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{atan}{\left (\sqrt{\frac{- a + x}{a + x}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(sqrt((-a + x)/(a + x))), x)

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Giac [A]  time = 1.11585, size = 66, normalized size = 1.65 \begin{align*} \frac{1}{2} \, a \log \left ({\left | -x + \sqrt{-a^{2} + x^{2}} \right |}\right ) \mathrm{sgn}\left (a + x\right ) + x \arctan \left (\frac{\sqrt{-a^{2} + x^{2}} \mathrm{sgn}\left (a + x\right )}{a + x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*log(abs(-x + sqrt(-a^2 + x^2)))*sgn(a + x) + x*arctan(sqrt(-a^2 + x^2)*sgn(a + x)/(a + x))