3.48 \(\int \sqrt{a-a x^2} \coth ^{-1}(x) \, dx\)

Optimal. Leaf size=186 \[ -\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \coth ^{-1}(x)-\frac{a \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \coth ^{-1}(x)}{\sqrt{a-a x^2}} \]

[Out]

Sqrt[a - a*x^2]/2 + (x*Sqrt[a - a*x^2]*ArcCoth[x])/2 - (a*Sqrt[1 - x^2]*ArcCoth[x]*ArcTan[Sqrt[1 - x]/Sqrt[1 +
x]])/Sqrt[a - a*x^2] - ((I/2)*a*Sqrt[1 - x^2]*PolyLog[2, ((-I)*Sqrt[1 - x])/Sqrt[1 + x]])/Sqrt[a - a*x^2] + (
(I/2)*a*Sqrt[1 - x^2]*PolyLog[2, (I*Sqrt[1 - x])/Sqrt[1 + x]])/Sqrt[a - a*x^2]

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Rubi [A]  time = 0.0831836, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5943, 5955, 5951} \[ -\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{2 \sqrt{a-a x^2}}+\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \coth ^{-1}(x)-\frac{a \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \coth ^{-1}(x)}{\sqrt{a-a x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*x^2]*ArcCoth[x],x]

[Out]

Sqrt[a - a*x^2]/2 + (x*Sqrt[a - a*x^2]*ArcCoth[x])/2 - (a*Sqrt[1 - x^2]*ArcCoth[x]*ArcTan[Sqrt[1 - x]/Sqrt[1 +
x]])/Sqrt[a - a*x^2] - ((I/2)*a*Sqrt[1 - x^2]*PolyLog[2, ((-I)*Sqrt[1 - x])/Sqrt[1 + x]])/Sqrt[a - a*x^2] + (
(I/2)*a*Sqrt[1 - x^2]*PolyLog[2, (I*Sqrt[1 - x])/Sqrt[1 + x]])/Sqrt[a - a*x^2]

Rule 5943

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcCoth[c*x]), x], x] + Simp[(x*(d
+ e*x^2)^q*(a + b*ArcCoth[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 0]

Rule 5955

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcCoth[c*x])^p/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d
+ e, 0] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 5951

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*(a + b*ArcCoth[c*x])*
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(c*Sqrt[d]), x] + (-Simp[(I*b*PolyLog[2, -((I*Sqrt[1 - c*x])/Sqrt[1 + c*x
])])/(c*Sqrt[d]), x] + Simp[(I*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]])/(c*Sqrt[d]), x]) /; FreeQ[{a, b,
c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]

Rubi steps

\begin{align*} \int \sqrt{a-a x^2} \coth ^{-1}(x) \, dx &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \coth ^{-1}(x)+\frac{1}{2} a \int \frac{\coth ^{-1}(x)}{\sqrt{a-a x^2}} \, dx\\ &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \coth ^{-1}(x)+\frac{\left (a \sqrt{1-x^2}\right ) \int \frac{\coth ^{-1}(x)}{\sqrt{1-x^2}} \, dx}{2 \sqrt{a-a x^2}}\\ &=\frac{1}{2} \sqrt{a-a x^2}+\frac{1}{2} x \sqrt{a-a x^2} \coth ^{-1}(x)-\frac{a \sqrt{1-x^2} \coth ^{-1}(x) \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{1+x}}\right )}{\sqrt{a-a x^2}}-\frac{i a \sqrt{1-x^2} \text{Li}_2\left (-\frac{i \sqrt{1-x}}{\sqrt{1+x}}\right )}{2 \sqrt{a-a x^2}}+\frac{i a \sqrt{1-x^2} \text{Li}_2\left (\frac{i \sqrt{1-x}}{\sqrt{1+x}}\right )}{2 \sqrt{a-a x^2}}\\ \end{align*}

Mathematica [A]  time = 1.03914, size = 125, normalized size = 0.67 \[ -\frac{\sqrt{a-a x^2} \left (-4 \text{PolyLog}\left (2,-e^{-\coth ^{-1}(x)}\right )+4 \text{PolyLog}\left (2,e^{-\coth ^{-1}(x)}\right )-2 \coth \left (\frac{1}{2} \coth ^{-1}(x)\right )-4 \coth ^{-1}(x) \log \left (1-e^{-\coth ^{-1}(x)}\right )+4 \coth ^{-1}(x) \log \left (e^{-\coth ^{-1}(x)}+1\right )+2 \tanh \left (\frac{1}{2} \coth ^{-1}(x)\right )-\coth ^{-1}(x) \text{csch}^2\left (\frac{1}{2} \coth ^{-1}(x)\right )-\coth ^{-1}(x) \text{sech}^2\left (\frac{1}{2} \coth ^{-1}(x)\right )\right )}{8 \sqrt{1-\frac{1}{x^2}} x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a - a*x^2]*ArcCoth[x],x]

[Out]

-(Sqrt[a - a*x^2]*(-2*Coth[ArcCoth[x]/2] - ArcCoth[x]*Csch[ArcCoth[x]/2]^2 - 4*ArcCoth[x]*Log[1 - E^(-ArcCoth[
x])] + 4*ArcCoth[x]*Log[1 + E^(-ArcCoth[x])] - 4*PolyLog[2, -E^(-ArcCoth[x])] + 4*PolyLog[2, E^(-ArcCoth[x])]
- ArcCoth[x]*Sech[ArcCoth[x]/2]^2 + 2*Tanh[ArcCoth[x]/2]))/(8*Sqrt[1 - x^(-2)]*x)

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Maple [A]  time = 0.448, size = 199, normalized size = 1.1 \begin{align*}{\frac{{\rm arccoth} \left (x\right )x+1}{2}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}}-{\frac{{\rm arccoth} \left (x\right )}{-2+2\,x}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{{\frac{-1+x}{1+x}}}\ln \left ( 1+{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) }-{\frac{1}{-2+2\,x}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{{\frac{-1+x}{1+x}}}{\it polylog} \left ( 2,-{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) }+{\frac{{\rm arccoth} \left (x\right )}{-2+2\,x}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{{\frac{-1+x}{1+x}}}\ln \left ( 1-{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) }+{\frac{1}{-2+2\,x}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}\sqrt{{\frac{-1+x}{1+x}}}{\it polylog} \left ( 2,{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(arccoth(x)*x+1)*(-(-1+x)*(1+x)*a)^(1/2)-1/2*(-(-1+x)*(1+x)*a)^(1/2)*((-1+x)/(1+x))^(1/2)/(-1+x)*arccoth(x
)*ln(1+1/((-1+x)/(1+x))^(1/2))-1/2*(-(-1+x)*(1+x)*a)^(1/2)*((-1+x)/(1+x))^(1/2)/(-1+x)*polylog(2,-1/((-1+x)/(1
+x))^(1/2))+1/2*(-(-1+x)*(1+x)*a)^(1/2)*((-1+x)/(1+x))^(1/2)/(-1+x)*arccoth(x)*ln(1-1/((-1+x)/(1+x))^(1/2))+1/
2*(-(-1+x)*(1+x)*a)^(1/2)*((-1+x)/(1+x))^(1/2)/(-1+x)*polylog(2,1/((-1+x)/(1+x))^(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((-a*x^2+a)^(1/2)*arccoth(x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-a x^{2} + a} \operatorname{arcoth}\left (x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a*x^2 + a)*arccoth(x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- a \left (x - 1\right ) \left (x + 1\right )} \operatorname{acoth}{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-a*(x - 1)*(x + 1))*acoth(x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a x^{2} + a} \operatorname{arcoth}\left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-a*x^2 + a)*arccoth(x), x)