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

Optimal. Leaf size=144 \[ -\frac{i \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{\sqrt{a-a x^2}}+\frac{i \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{\sqrt{a-a x^2}}-\frac{2 \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \coth ^{-1}(x)}{\sqrt{a-a x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0495992, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {5955, 5951} \[ -\frac{i \sqrt{1-x^2} \text{PolyLog}\left (2,-\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{\sqrt{a-a x^2}}+\frac{i \sqrt{1-x^2} \text{PolyLog}\left (2,\frac{i \sqrt{1-x}}{\sqrt{x+1}}\right )}{\sqrt{a-a x^2}}-\frac{2 \sqrt{1-x^2} \tan ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{x+1}}\right ) \coth ^{-1}(x)}{\sqrt{a-a x^2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/Sqrt[a - a*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.112466, size = 77, normalized size = 0.53 \[ \frac{\sqrt{a-a x^2} \left (\text{PolyLog}\left (2,-e^{-\coth ^{-1}(x)}\right )-\text{PolyLog}\left (2,e^{-\coth ^{-1}(x)}\right )+\coth ^{-1}(x) \left (\log \left (1-e^{-\coth ^{-1}(x)}\right )-\log \left (e^{-\coth ^{-1}(x)}+1\right )\right )\right )}{a \sqrt{1-\frac{1}{x^2}} x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.368, size = 190, normalized size = 1.3 \begin{align*} -{\frac{{\rm arccoth} \left (x\right )}{ \left ( -1+x \right ) a}\ln \left ( 1+{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) \sqrt{{\frac{-1+x}{1+x}}}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}}-{\frac{1}{ \left ( -1+x \right ) a}{\it polylog} \left ( 2,-{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) \sqrt{{\frac{-1+x}{1+x}}}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}}+{\frac{{\rm arccoth} \left (x\right )}{ \left ( -1+x \right ) a}\ln \left ( 1-{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) \sqrt{{\frac{-1+x}{1+x}}}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}}+{\frac{1}{ \left ( -1+x \right ) a}{\it polylog} \left ( 2,{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}} \right ) \sqrt{{\frac{-1+x}{1+x}}}\sqrt{- \left ( -1+x \right ) \left ( 1+x \right ) a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(arccoth(x)/(-a*x^2+a)^(1/2),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 (-\frac{\sqrt{-a x^{2} + a} \operatorname{arcoth}\left (x\right )}{a x^{2} - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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