3.139 \(\int \frac{a+b \tanh ^{-1}(\frac{c}{x})}{x} \, dx\)

Optimal. Leaf size=30 \[ \frac{1}{2} b \text{PolyLog}\left (2,-\frac{c}{x}\right )-\frac{1}{2} b \text{PolyLog}\left (2,\frac{c}{x}\right )+a \log (x) \]

[Out]

a*Log[x] + (b*PolyLog[2, -(c/x)])/2 - (b*PolyLog[2, c/x])/2

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Rubi [A]  time = 0.031678, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6095, 5912} \[ \frac{1}{2} b \text{PolyLog}\left (2,-\frac{c}{x}\right )-\frac{1}{2} b \text{PolyLog}\left (2,\frac{c}{x}\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x])/x,x]

[Out]

a*Log[x] + (b*PolyLog[2, -(c/x)])/2 - (b*PolyLog[2, c/x])/2

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 5912

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

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (\frac{c}{x}\right )}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx,x,\frac{1}{x}\right )\\ &=a \log (x)+\frac{1}{2} b \text{Li}_2\left (-\frac{c}{x}\right )-\frac{1}{2} b \text{Li}_2\left (\frac{c}{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.012366, size = 28, normalized size = 0.93 \[ \frac{1}{2} b \left (\text{PolyLog}\left (2,-\frac{c}{x}\right )-\text{PolyLog}\left (2,\frac{c}{x}\right )\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x])/x,x]

[Out]

a*Log[x] + (b*(PolyLog[2, -(c/x)] - PolyLog[2, c/x]))/2

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Maple [B]  time = 0.014, size = 63, normalized size = 2.1 \begin{align*} -a\ln \left ({\frac{c}{x}} \right ) -b\ln \left ({\frac{c}{x}} \right ){\it Artanh} \left ({\frac{c}{x}} \right ) +{\frac{b}{2}{\it dilog} \left ({\frac{c}{x}} \right ) }+{\frac{b}{2}{\it dilog} \left ( 1+{\frac{c}{x}} \right ) }+{\frac{b}{2}\ln \left ({\frac{c}{x}} \right ) \ln \left ( 1+{\frac{c}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x))/x,x)

[Out]

-a*ln(c/x)-b*ln(c/x)*arctanh(c/x)+1/2*b*dilog(c/x)+1/2*b*dilog(1+c/x)+1/2*b*ln(c/x)*ln(1+c/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, b \int \frac{\log \left (\frac{c}{x} + 1\right ) - \log \left (-\frac{c}{x} + 1\right )}{x}\,{d x} + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x))/x,x, algorithm="maxima")

[Out]

1/2*b*integrate((log(c/x + 1) - log(-c/x + 1))/x, x) + a*log(x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (\frac{c}{x}\right ) + a}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x))/x,x, algorithm="fricas")

[Out]

integral((b*arctanh(c/x) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c/x))/x,x)

[Out]

Integral((a + b*atanh(c/x))/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (\frac{c}{x}\right ) + a}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x))/x,x, algorithm="giac")

[Out]

integrate((b*arctanh(c/x) + a)/x, x)