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

Optimal. Leaf size=54 \[ -\frac{b \text{PolyLog}(2,-c-d x)}{2 d e}+\frac{b \text{PolyLog}(2,c+d x)}{2 d e}+\frac{a \log (c+d x)}{d e} \]

[Out]

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

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Rubi [A]  time = 0.0447492, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6107, 12, 5912} \[ -\frac{b \text{PolyLog}(2,-c-d x)}{2 d e}+\frac{b \text{PolyLog}(2,c+d x)}{2 d e}+\frac{a \log (c+d x)}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0247697, size = 54, normalized size = 1. \[ -\frac{b \text{PolyLog}(2,-c-d x)}{2 d e}+\frac{b \text{PolyLog}(2,c+d x)}{2 d e}+\frac{a \log (c+d x)}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 89, normalized size = 1.7 \begin{align*}{\frac{a\ln \left ( dx+c \right ) }{de}}+{\frac{b\ln \left ( dx+c \right ){\it Artanh} \left ( dx+c \right ) }{de}}-{\frac{b{\it dilog} \left ( dx+c \right ) }{2\,de}}-{\frac{b{\it dilog} \left ( dx+c+1 \right ) }{2\,de}}-{\frac{b\ln \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{2\,de}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a/e*ln(d*x+c)+1/d*b/e*ln(d*x+c)*arctanh(d*x+c)-1/2/d*b/e*dilog(d*x+c)-1/2/d*b/e*dilog(d*x+c+1)-1/2/d*b/e*l
n(d*x+c)*ln(d*x+c+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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