3.289 \(\int \coth ^{-1}(a+b f^{c+d x}) \, dx\)

Optimal. Leaf size=168 \[ \frac{\text{PolyLog}\left (2,1-\frac{2}{a+b f^{c+d x}+1}\right )}{2 d \log (f)}-\frac{\text{PolyLog}\left (2,1-\frac{2 b f^{c+d x}}{(1-a) \left (a+b f^{c+d x}+1\right )}\right )}{2 d \log (f)}-\frac{\log \left (\frac{2}{a+b f^{c+d x}+1}\right ) \coth ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}+\frac{\log \left (\frac{2 b f^{c+d x}}{(1-a) \left (a+b f^{c+d x}+1\right )}\right ) \coth ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)} \]

[Out]

-((ArcCoth[a + b*f^(c + d*x)]*Log[2/(1 + a + b*f^(c + d*x))])/(d*Log[f])) + (ArcCoth[a + b*f^(c + d*x)]*Log[(2
*b*f^(c + d*x))/((1 - a)*(1 + a + b*f^(c + d*x)))])/(d*Log[f]) + PolyLog[2, 1 - 2/(1 + a + b*f^(c + d*x))]/(2*
d*Log[f]) - PolyLog[2, 1 - (2*b*f^(c + d*x))/((1 - a)*(1 + a + b*f^(c + d*x)))]/(2*d*Log[f])

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Rubi [A]  time = 0.132084, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2282, 6112, 5921, 2402, 2315, 2447} \[ \frac{\text{PolyLog}\left (2,1-\frac{2}{a+b f^{c+d x}+1}\right )}{2 d \log (f)}-\frac{\text{PolyLog}\left (2,1-\frac{2 b f^{c+d x}}{(1-a) \left (a+b f^{c+d x}+1\right )}\right )}{2 d \log (f)}-\frac{\log \left (\frac{2}{a+b f^{c+d x}+1}\right ) \coth ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}+\frac{\log \left (\frac{2 b f^{c+d x}}{(1-a) \left (a+b f^{c+d x}+1\right )}\right ) \coth ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*f^(c + d*x)],x]

[Out]

-((ArcCoth[a + b*f^(c + d*x)]*Log[2/(1 + a + b*f^(c + d*x))])/(d*Log[f])) + (ArcCoth[a + b*f^(c + d*x)]*Log[(2
*b*f^(c + d*x))/((1 - a)*(1 + a + b*f^(c + d*x)))])/(d*Log[f]) + PolyLog[2, 1 - 2/(1 + a + b*f^(c + d*x))]/(2*
d*Log[f]) - PolyLog[2, 1 - (2*b*f^(c + d*x))/((1 - a)*(1 + a + b*f^(c + d*x)))]/(2*d*Log[f])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6112

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

Rule 5921

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcCoth[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [A]  time = 0.0750964, size = 108, normalized size = 0.64 \[ \frac{\text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a-1}\right )-\text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )+d x \log (f) \left (\log \left (\frac{a+b f^{c+d x}-1}{a-1}\right )-\log \left (\frac{a+b f^{c+d x}+1}{a+1}\right )+2 \coth ^{-1}\left (a+b f^{c+d x}\right )\right )}{2 d \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a + b*f^(c + d*x)],x]

[Out]

(d*x*Log[f]*(2*ArcCoth[a + b*f^(c + d*x)] + Log[(-1 + a + b*f^(c + d*x))/(-1 + a)] - Log[(1 + a + b*f^(c + d*x
))/(1 + a)]) + PolyLog[2, -((b*f^(c + d*x))/(-1 + a))] - PolyLog[2, -((b*f^(c + d*x))/(1 + a))])/(2*d*Log[f])

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Maple [A]  time = 0.199, size = 164, normalized size = 1. \begin{align*}{\frac{\ln \left ( b{f}^{dx+c} \right ){\rm arccoth} \left (a+b{f}^{dx+c}\right )}{d\ln \left ( f \right ) }}-{\frac{1}{2\,d\ln \left ( f \right ) }{\it dilog} \left ({\frac{1+a+b{f}^{dx+c}}{1+a}} \right ) }-{\frac{\ln \left ( b{f}^{dx+c} \right ) }{2\,d\ln \left ( f \right ) }\ln \left ({\frac{1+a+b{f}^{dx+c}}{1+a}} \right ) }+{\frac{1}{2\,d\ln \left ( f \right ) }{\it dilog} \left ({\frac{b{f}^{dx+c}+a-1}{a-1}} \right ) }+{\frac{\ln \left ( b{f}^{dx+c} \right ) }{2\,d\ln \left ( f \right ) }\ln \left ({\frac{b{f}^{dx+c}+a-1}{a-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a+b*f^(d*x+c)),x)

[Out]

1/d/ln(f)*ln(b*f^(d*x+c))*arccoth(a+b*f^(d*x+c))-1/2/d/ln(f)*dilog((1+a+b*f^(d*x+c))/(1+a))-1/2/d/ln(f)*ln(b*f
^(d*x+c))*ln((1+a+b*f^(d*x+c))/(1+a))+1/2/d/ln(f)*dilog((b*f^(d*x+c)+a-1)/(a-1))+1/2/d/ln(f)*ln(b*f^(d*x+c))*l
n((b*f^(d*x+c)+a-1)/(a-1))

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Maxima [A]  time = 1.08897, size = 284, normalized size = 1.69 \begin{align*} \frac{\operatorname{arcoth}\left (b f^{d x + c} + a\right ) \log \left (f^{d x + c}\right )}{d \log \left (f\right )} - \frac{b{\left (\frac{\log \left (b f^{d x + c} + a + 1\right )}{b} - \frac{\log \left (b f^{d x + c} + a - 1\right )}{b}\right )} \log \left (f^{d x + c}\right ) - b{\left (\frac{\log \left (b f^{d x + c} + a + 1\right ) \log \left (-\frac{b f^{d x + c} + a + 1}{a + 1} + 1\right ) +{\rm Li}_2\left (\frac{b f^{d x + c} + a + 1}{a + 1}\right )}{b} - \frac{\log \left (b f^{d x + c} + a - 1\right ) \log \left (-\frac{b f^{d x + c} + a - 1}{a - 1} + 1\right ) +{\rm Li}_2\left (\frac{b f^{d x + c} + a - 1}{a - 1}\right )}{b}\right )}}{2 \, d \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a+b*f^(d*x+c)),x, algorithm="maxima")

[Out]

arccoth(b*f^(d*x + c) + a)*log(f^(d*x + c))/(d*log(f)) - 1/2*(b*(log(b*f^(d*x + c) + a + 1)/b - log(b*f^(d*x +
c) + a - 1)/b)*log(f^(d*x + c)) - b*((log(b*f^(d*x + c) + a + 1)*log(-(b*f^(d*x + c) + a + 1)/(a + 1) + 1) +
dilog((b*f^(d*x + c) + a + 1)/(a + 1)))/b - (log(b*f^(d*x + c) + a - 1)*log(-(b*f^(d*x + c) + a - 1)/(a - 1) +
1) + dilog((b*f^(d*x + c) + a - 1)/(a - 1)))/b))/(d*log(f))

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Fricas [A]  time = 1.60135, size = 899, normalized size = 5.35 \begin{align*} \frac{d x \log \left (f\right ) \log \left (\frac{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}\right ) + c \log \left (b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1\right ) \log \left (f\right ) - c \log \left (b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1\right ) \log \left (f\right ) -{\left (d x + c\right )} \log \left (f\right ) \log \left (\frac{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{a + 1}\right ) +{\left (d x + c\right )} \log \left (f\right ) \log \left (\frac{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}{a - 1}\right ) -{\rm Li}_2\left (-\frac{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{a + 1} + 1\right ) +{\rm Li}_2\left (-\frac{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}{a - 1} + 1\right )}{2 \, d \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a+b*f^(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(d*x*log(f)*log((b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(b*cosh((d*x + c)*log(f)) +
b*sinh((d*x + c)*log(f)) + a - 1)) + c*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)*log(f)
- c*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)*log(f) - (d*x + c)*log(f)*log((b*cosh((d
*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(a + 1)) + (d*x + c)*log(f)*log((b*cosh((d*x + c)*log(f))
+ b*sinh((d*x + c)*log(f)) + a - 1)/(a - 1)) - dilog(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a
+ 1)/(a + 1) + 1) + dilog(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)/(a - 1) + 1))/(d*log
(f))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a+b*f**(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a+b*f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(arccoth(b*f^(d*x + c) + a), x)