3.1.49 \(\int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx\) [49]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 144, normalized size of antiderivative = 1.00, 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 = {6102, 6098} \begin {gather*} -\frac {2 \sqrt {1-x^2} \text {ArcTan}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \coth ^{-1}(x)}{\sqrt {a-a x^2}}-\frac {i \sqrt {1-x^2} \text {Li}_2\left (-\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \text {Li}_2\left (\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}} \end {gather*}

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 6098

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]

Rule 6102

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]

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*}

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Mathematica [A]
time = 0.08, size = 77, normalized size = 0.53 \begin {gather*} \frac {\sqrt {a-a x^2} \left (\coth ^{-1}(x) \left (\log \left (1-e^{-\coth ^{-1}(x)}\right )-\log \left (1+e^{-\coth ^{-1}(x)}\right )\right )+\text {PolyLog}\left (2,-e^{-\coth ^{-1}(x)}\right )-\text {PolyLog}\left (2,e^{-\coth ^{-1}(x)}\right )\right )}{a \sqrt {1-\frac {1}{x^2}} x} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.41, size = 190, normalized size = 1.32

method	result	size
default	\(-\frac {\ln \left (\frac {1}{\sqrt {\frac {-1+x}{1+x}}}+1\right ) \mathrm {arccoth}\left (x \right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {-a \left (1+x \right ) \left (-1+x \right )}}{\left (-1+x \right ) a}-\frac {\polylog \left (2, -\frac {1}{\sqrt {\frac {-1+x}{1+x}}}\right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {-a \left (1+x \right ) \left (-1+x \right )}}{\left (-1+x \right ) a}+\frac {\ln \left (1-\frac {1}{\sqrt {\frac {-1+x}{1+x}}}\right ) \mathrm {arccoth}\left (x \right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {-a \left (1+x \right ) \left (-1+x \right )}}{\left (-1+x \right ) a}+\frac {\polylog \left (2, \frac {1}{\sqrt {\frac {-1+x}{1+x}}}\right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {-a \left (1+x \right ) \left (-1+x \right )}}{\left (-1+x \right ) a}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acoth}{\left (x \right )}}{\sqrt {- a \left (x - 1\right ) \left (x + 1\right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acoth}\left (x\right )}{\sqrt {a-a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(x)/(a - a*x^2)^(1/2),x)

[Out]

int(acoth(x)/(a - a*x^2)^(1/2), x)

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