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

Optimal. Leaf size=29 \[ -b \text{PolyLog}\left (2,-c \sqrt{x}\right )+b \text{PolyLog}\left (2,c \sqrt{x}\right )+a \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0315208, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6095, 5912} \[ -b \text{PolyLog}\left (2,-c \sqrt{x}\right )+b \text{PolyLog}\left (2,c \sqrt{x}\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/x,x]

[Out]

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

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

Mathematica [A]  time = 0.0114324, size = 29, normalized size = 1. \[ -b \text{PolyLog}\left (2,-c \sqrt{x}\right )+b \text{PolyLog}\left (2,c \sqrt{x}\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/x,x]

[Out]

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

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Maple [B]  time = 0.036, size = 63, normalized size = 2.2 \begin{align*} 2\,a\ln \left ( c\sqrt{x} \right ) +2\,b\ln \left ( c\sqrt{x} \right ){\it Artanh} \left ( c\sqrt{x} \right ) -b{\it dilog} \left ( c\sqrt{x} \right ) -b{\it dilog} \left ( 1+c\sqrt{x} \right ) -b\ln \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x,x)

[Out]

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

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Maxima [B]  time = 1.49312, size = 82, normalized size = 2.83 \begin{align*} -{\left (\log \left (c \sqrt{x}\right ) \log \left (-c \sqrt{x} + 1\right ) +{\rm Li}_2\left (-c \sqrt{x} + 1\right )\right )} b +{\left (\log \left (c \sqrt{x} + 1\right ) \log \left (-c \sqrt{x}\right ) +{\rm Li}_2\left (c \sqrt{x} + 1\right )\right )} b + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x,x, algorithm="maxima")

[Out]

-(log(c*sqrt(x))*log(-c*sqrt(x) + 1) + dilog(-c*sqrt(x) + 1))*b + (log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(
c*sqrt(x) + 1))*b + 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 (c \sqrt{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^(1/2)))/x,x, algorithm="fricas")

[Out]

integral((b*arctanh(c*sqrt(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 (c \sqrt{x} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*sqrt(x)))/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c \sqrt{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^(1/2)))/x,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)/x, x)