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

Optimal. Leaf size=58 \[ \sqrt{c+d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c+d}}\right )-\sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c-d}}\right ) \]

[Out]

-(Sqrt[c - d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c - d]]) + Sqrt[c + d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c + d]]

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Rubi [A]  time = 0.0580653, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {700, 1130, 206} \[ \sqrt{c+d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c+d}}\right )-\sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c-d}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(1 - x^2),x]

[Out]

-(Sqrt[c - d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c - d]]) + Sqrt[c + d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c + d]]

Rule 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1130

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

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 \frac{\sqrt{c+d x}}{1-x^2} \, dx &=(2 d) \operatorname{Subst}\left (\int \frac{x^2}{-c^2+d^2+2 c x^2-x^4} \, dx,x,\sqrt{c+d x}\right )\\ &=-\left ((c-d) \operatorname{Subst}\left (\int \frac{1}{c-d-x^2} \, dx,x,\sqrt{c+d x}\right )\right )+(c+d) \operatorname{Subst}\left (\int \frac{1}{c+d-x^2} \, dx,x,\sqrt{c+d x}\right )\\ &=-\sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c-d}}\right )+\sqrt{c+d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c+d}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0346304, size = 58, normalized size = 1. \[ \sqrt{c+d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c+d}}\right )-\sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c-d}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]/(1 - x^2),x]

[Out]

-(Sqrt[c - d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c - d]]) + Sqrt[c + d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c + d]]

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Maple [A]  time = 0.131, size = 47, normalized size = 0.8 \begin{align*}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c+d}}}} \right ) \sqrt{c+d}-\sqrt{-c+d}\arctan \left ({\sqrt{dx+c}{\frac{1}{\sqrt{-c+d}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh((d*x+c)^(1/2)/(c+d)^(1/2))*(c+d)^(1/2)-(-c+d)^(1/2)*arctan((d*x+c)^(1/2)/(-c+d)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.40727, size = 741, normalized size = 12.78 \begin{align*} \left [\frac{1}{2} \, \sqrt{c - d} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c - d} + 2 \, c - d}{x + 1}\right ) + \frac{1}{2} \, \sqrt{c + d} \log \left (\frac{d x + 2 \, \sqrt{d x + c} \sqrt{c + d} + 2 \, c + d}{x - 1}\right ), -\sqrt{-c + d} \arctan \left (-\frac{\sqrt{d x + c} \sqrt{-c + d}}{c - d}\right ) + \frac{1}{2} \, \sqrt{c + d} \log \left (\frac{d x + 2 \, \sqrt{d x + c} \sqrt{c + d} + 2 \, c + d}{x - 1}\right ), -\sqrt{-c - d} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c - d}}{c + d}\right ) + \frac{1}{2} \, \sqrt{c - d} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c - d} + 2 \, c - d}{x + 1}\right ), -\sqrt{-c + d} \arctan \left (-\frac{\sqrt{d x + c} \sqrt{-c + d}}{c - d}\right ) - \sqrt{-c - d} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c - d}}{c + d}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(c - d)*log((d*x - 2*sqrt(d*x + c)*sqrt(c - d) + 2*c - d)/(x + 1)) + 1/2*sqrt(c + d)*log((d*x + 2*sqr
t(d*x + c)*sqrt(c + d) + 2*c + d)/(x - 1)), -sqrt(-c + d)*arctan(-sqrt(d*x + c)*sqrt(-c + d)/(c - d)) + 1/2*sq
rt(c + d)*log((d*x + 2*sqrt(d*x + c)*sqrt(c + d) + 2*c + d)/(x - 1)), -sqrt(-c - d)*arctan(sqrt(d*x + c)*sqrt(
-c - d)/(c + d)) + 1/2*sqrt(c - d)*log((d*x - 2*sqrt(d*x + c)*sqrt(c - d) + 2*c - d)/(x + 1)), -sqrt(-c + d)*a
rctan(-sqrt(d*x + c)*sqrt(-c + d)/(c - d)) - sqrt(-c - d)*arctan(sqrt(d*x + c)*sqrt(-c - d)/(c + d))]

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Sympy [A]  time = 3.65656, size = 61, normalized size = 1.05 \begin{align*} \frac{2 \left (\frac{d \left (c - d\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{- c + d}} \right )}}{2 \sqrt{- c + d}} - \frac{d \left (c + d\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{- c - d}} \right )}}{2 \sqrt{- c - d}}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*(c - d)*atan(sqrt(c + d*x)/sqrt(-c + d))/(2*sqrt(-c + d)) - d*(c + d)*atan(sqrt(c + d*x)/sqrt(-c - d))/(2
*sqrt(-c - d)))/d

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Giac [A]  time = 1.36741, size = 84, normalized size = 1.45 \begin{align*} \frac{{\left (c - d\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c + d}}\right )}{\sqrt{-c + d}} - \frac{{\left (c + d\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c - d}}\right )}{\sqrt{-c - d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c - d)*arctan(sqrt(d*x + c)/sqrt(-c + d))/sqrt(-c + d) - (c + d)*arctan(sqrt(d*x + c)/sqrt(-c - d))/sqrt(-c -
d)