3.7 \(\int \frac {\coth ^{-1}(a x)}{x} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{2} \text {Li}_2\left (-\frac {1}{a x}\right )-\frac {1}{2} \text {Li}_2\left (\frac {1}{a x}\right ) \]

[Out]

1/2*polylog(2,-1/a/x)-1/2*polylog(2,1/a/x)

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5913} \[ \frac {1}{2} \text {PolyLog}\left (2,-\frac {1}{a x}\right )-\frac {1}{2} \text {PolyLog}\left (2,\frac {1}{a x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/x,x]

[Out]

PolyLog[2, -(1/(a*x))]/2 - PolyLog[2, 1/(a*x)]/2

Rule 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.93 \[ \frac {1}{2} \left (\text {Li}_2\left (-\frac {1}{a x}\right )-\text {Li}_2\left (\frac {1}{a x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]/x,x]

[Out]

(PolyLog[2, -(1/(a*x))] - PolyLog[2, 1/(a*x)])/2

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x,x, algorithm="fricas")

[Out]

integral(arccoth(a*x)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (a x\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)/x, x)

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maple [A]  time = 0.05, size = 37, normalized size = 1.32 \[ \ln \left (a x \right ) \mathrm {arccoth}\left (a x \right )-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)/x,x)

[Out]

ln(a*x)*arccoth(a*x)-1/2*dilog(a*x)-1/2*dilog(a*x+1)-1/2*ln(a*x)*ln(a*x+1)

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maxima [B]  time = 0.30, size = 86, normalized size = 3.07 \[ -\frac {1}{2} \, a {\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \log \relax (x) - \frac {1}{2} \, a {\left (\frac {\log \left (a x - 1\right ) \log \left (a x\right ) + {\rm Li}_2\left (-a x + 1\right )}{a} - \frac {\log \left (a x + 1\right ) \log \left (-a x\right ) + {\rm Li}_2\left (a x + 1\right )}{a}\right )} + \operatorname {arcoth}\left (a x\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x,x, algorithm="maxima")

[Out]

-1/2*a*(log(a*x + 1)/a - log(a*x - 1)/a)*log(x) - 1/2*a*((log(a*x - 1)*log(a*x) + dilog(-a*x + 1))/a - (log(a*
x + 1)*log(-a*x) + dilog(a*x + 1))/a) + arccoth(a*x)*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {acoth}\left (a\,x\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)/x,x)

[Out]

int(acoth(a*x)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}{\left (a x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)/x,x)

[Out]

Integral(acoth(a*x)/x, x)

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