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

Optimal. Leaf size=32 \[ \frac{\text{PolyLog}(2,a+b x)}{2 d}-\frac{\text{PolyLog}(2,-a-b x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0310902, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {6107, 12, 5912} \[ \frac{\text{PolyLog}(2,a+b x)}{2 d}-\frac{\text{PolyLog}(2,-a-b x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0056341, size = 52, normalized size = 1.62 \[ b \left (\frac{\text{PolyLog}\left (2,\frac{a d+b d x}{d}\right )}{2 b d}-\frac{\text{PolyLog}\left (2,-\frac{a d+b d x}{d}\right )}{2 b d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.043, size = 59, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( bx+a \right ){\it Artanh} \left ( bx+a \right ) }{d}}-{\frac{{\it dilog} \left ( bx+a \right ) }{2\,d}}-{\frac{{\it dilog} \left ( bx+a+1 \right ) }{2\,d}}-{\frac{\ln \left ( bx+a \right ) \ln \left ( bx+a+1 \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.978404, size = 178, normalized size = 5.56 \begin{align*} -\frac{1}{2} \, b{\left (\frac{\log \left (b x + a\right ) \log \left (b x + a - 1\right ) +{\rm Li}_2\left (-b x - a + 1\right )}{b d} - \frac{\log \left (b x + a + 1\right ) \log \left (-b x - a\right ) +{\rm Li}_2\left (b x + a + 1\right )}{b d}\right )} - \frac{b{\left (\frac{\log \left (b x + a + 1\right )}{b} - \frac{\log \left (b x + a - 1\right )}{b}\right )} \log \left (d x + \frac{a d}{b}\right )}{2 \, d} + \frac{\operatorname{artanh}\left (b x + a\right ) \log \left (d x + \frac{a d}{b}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*((log(b*x + a)*log(b*x + a - 1) + dilog(-b*x - a + 1))/(b*d) - (log(b*x + a + 1)*log(-b*x - a) + dilog(
b*x + a + 1))/(b*d)) - 1/2*b*(log(b*x + a + 1)/b - log(b*x + a - 1)/b)*log(d*x + a*d/b)/d + arctanh(b*x + a)*l
og(d*x + a*d/b)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arctanh(b*x + a)/(b*d*x + a*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Integral(atanh(a + b*x)/(a + b*x), x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(b*x + a)/(d*x + a*d/b), x)