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

Optimal. Leaf size=211 \[ -\frac{\text{PolyLog}\left (3,\frac{b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac{\text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )}{2 d^2 \log ^2(f)}+\frac{x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac{1}{4} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac{1}{4} x^2 \log \left (a+b f^{c+d x}+1\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (\frac{b f^{c+d x}}{a+1}+1\right ) \]

[Out]

-(x^2*Log[1 - a - b*f^(c + d*x)])/4 + (x^2*Log[1 + a + b*f^(c + d*x)])/4 + (x^2*Log[1 - (b*f^(c + d*x))/(1 - a
)])/4 - (x^2*Log[1 + (b*f^(c + d*x))/(1 + a)])/4 + (x*PolyLog[2, (b*f^(c + d*x))/(1 - a)])/(2*d*Log[f]) - (x*P
olyLog[2, -((b*f^(c + d*x))/(1 + a))])/(2*d*Log[f]) - PolyLog[3, (b*f^(c + d*x))/(1 - a)]/(2*d^2*Log[f]^2) + P
olyLog[3, -((b*f^(c + d*x))/(1 + a))]/(2*d^2*Log[f]^2)

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Rubi [A]  time = 0.150448, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6213, 2532, 2531, 2282, 6589} \[ -\frac{\text{PolyLog}\left (3,\frac{b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac{\text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )}{2 d^2 \log ^2(f)}+\frac{x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac{1}{4} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac{1}{4} x^2 \log \left (a+b f^{c+d x}+1\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (\frac{b f^{c+d x}}{a+1}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[a + b*f^(c + d*x)],x]

[Out]

-(x^2*Log[1 - a - b*f^(c + d*x)])/4 + (x^2*Log[1 + a + b*f^(c + d*x)])/4 + (x^2*Log[1 - (b*f^(c + d*x))/(1 - a
)])/4 - (x^2*Log[1 + (b*f^(c + d*x))/(1 + a)])/4 + (x*PolyLog[2, (b*f^(c + d*x))/(1 - a)])/(2*d*Log[f]) - (x*P
olyLog[2, -((b*f^(c + d*x))/(1 + a))])/(2*d*Log[f]) - PolyLog[3, (b*f^(c + d*x))/(1 - a)]/(2*d^2*Log[f]^2) + P
olyLog[3, -((b*f^(c + d*x))/(1 + a))]/(2*d^2*Log[f]^2)

Rule 6213

Int[ArcTanh[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + a + b*
f^(c + d*x)], x], x] - Dist[1/2, Int[x^m*Log[1 - a - b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] && IG
tQ[m, 0]

Rule 2532

Int[Log[(d_) + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[
((f + g*x)^(m + 1)*Log[d + e*(F^(c*(a + b*x)))^n])/(g*(m + 1)), x] + (Int[(f + g*x)^m*Log[1 + (e*(F^(c*(a + b*
x)))^n)/d], x] - Simp[((f + g*x)^(m + 1)*Log[1 + (e*(F^(c*(a + b*x)))^n)/d])/(g*(m + 1)), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[d, 1]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x \tanh ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=-\left (\frac{1}{2} \int x \log \left (1-a-b f^{c+d x}\right ) \, dx\right )+\frac{1}{2} \int x \log \left (1+a+b f^{c+d x}\right ) \, dx\\ &=-\frac{1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (1+\frac{b f^{c+d x}}{1+a}\right )-\frac{1}{2} \int x \log \left (1-\frac{b f^{c+d x}}{1-a}\right ) \, dx+\frac{1}{2} \int x \log \left (1+\frac{b f^{c+d x}}{1+a}\right ) \, dx\\ &=-\frac{1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (1+\frac{b f^{c+d x}}{1+a}\right )+\frac{x \text{Li}_2\left (\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x \text{Li}_2\left (-\frac{b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac{\int \text{Li}_2\left (\frac{b f^{c+d x}}{1-a}\right ) \, dx}{2 d \log (f)}+\frac{\int \text{Li}_2\left (-\frac{b f^{c+d x}}{1+a}\right ) \, dx}{2 d \log (f)}\\ &=-\frac{1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (1+\frac{b f^{c+d x}}{1+a}\right )+\frac{x \text{Li}_2\left (\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x \text{Li}_2\left (-\frac{b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{1-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{1+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}\\ &=-\frac{1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac{1}{4} x^2 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{4} x^2 \log \left (1+\frac{b f^{c+d x}}{1+a}\right )+\frac{x \text{Li}_2\left (\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x \text{Li}_2\left (-\frac{b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac{\text{Li}_3\left (\frac{b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac{\text{Li}_3\left (-\frac{b f^{c+d x}}{1+a}\right )}{2 d^2 \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.118409, size = 177, normalized size = 0.84 \[ \frac{-2 \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a-1}\right )+2 \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )+2 d x \log (f) \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a-1}\right )-2 d x \log (f) \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )+d^2 x^2 \log ^2(f) \log \left (\frac{b f^{c+d x}}{a-1}+1\right )-d^2 x^2 \log ^2(f) \log \left (\frac{b f^{c+d x}}{a+1}+1\right )+2 d^2 x^2 \log ^2(f) \tanh ^{-1}\left (a+b f^{c+d x}\right )}{4 d^2 \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*x^2*ArcTanh[a + b*f^(c + d*x)]*Log[f]^2 + d^2*x^2*Log[f]^2*Log[1 + (b*f^(c + d*x))/(-1 + a)] - d^2*x^2*
Log[f]^2*Log[1 + (b*f^(c + d*x))/(1 + a)] + 2*d*x*Log[f]*PolyLog[2, -((b*f^(c + d*x))/(-1 + a))] - 2*d*x*Log[f
]*PolyLog[2, -((b*f^(c + d*x))/(1 + a))] - 2*PolyLog[3, -((b*f^(c + d*x))/(-1 + a))] + 2*PolyLog[3, -((b*f^(c
+ d*x))/(1 + a))])/(4*d^2*Log[f]^2)

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Maple [B]  time = 0.131, size = 596, normalized size = 2.8 \begin{align*}{\frac{{x}^{2}\ln \left ( 1+a+b{f}^{dx+c} \right ) }{4}}-{\frac{{x}^{2}\ln \left ( 1-a-b{f}^{dx+c} \right ) }{4}}+{\frac{{x}^{2}}{4}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }+{\frac{cx}{2\,d}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }+{\frac{{c}^{2}}{4\,{d}^{2}}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }+{\frac{x}{2\,d\ln \left ( f \right ) }{\it polylog} \left ( 2,{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }+{\frac{c}{2\,\ln \left ( f \right ){d}^{2}}{\it polylog} \left ( 2,{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }-{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}{\it polylog} \left ( 3,{\frac{b{f}^{dx}{f}^{c}}{1-a}} \right ) }+{\frac{{c}^{2}\ln \left ( 1-a-b{f}^{dx}{f}^{c} \right ) }{4\,{d}^{2}}}-{\frac{c}{2\,\ln \left ( f \right ){d}^{2}}{\it dilog} \left ({\frac{b{f}^{dx}{f}^{c}+a-1}{a-1}} \right ) }-{\frac{cx}{2\,d}\ln \left ({\frac{b{f}^{dx}{f}^{c}+a-1}{a-1}} \right ) }-{\frac{{c}^{2}}{2\,{d}^{2}}\ln \left ({\frac{b{f}^{dx}{f}^{c}+a-1}{a-1}} \right ) }-{\frac{{x}^{2}}{4}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }-{\frac{cx}{2\,d}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }-{\frac{{c}^{2}}{4\,{d}^{2}}\ln \left ( 1-{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }-{\frac{x}{2\,d\ln \left ( f \right ) }{\it polylog} \left ( 2,{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }-{\frac{c}{2\,\ln \left ( f \right ){d}^{2}}{\it polylog} \left ( 2,{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }+{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}{\it polylog} \left ( 3,{\frac{b{f}^{dx}{f}^{c}}{-1-a}} \right ) }-{\frac{{c}^{2}\ln \left ( 1+a+b{f}^{dx}{f}^{c} \right ) }{4\,{d}^{2}}}+{\frac{c}{2\,\ln \left ( f \right ){d}^{2}}{\it dilog} \left ({\frac{1+a+b{f}^{dx}{f}^{c}}{1+a}} \right ) }+{\frac{cx}{2\,d}\ln \left ({\frac{1+a+b{f}^{dx}{f}^{c}}{1+a}} \right ) }+{\frac{{c}^{2}}{2\,{d}^{2}}\ln \left ({\frac{1+a+b{f}^{dx}{f}^{c}}{1+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2*ln(1+a+b*f^(d*x+c))-1/4*x^2*ln(1-a-b*f^(d*x+c))+1/4*ln(1-b*f^(d*x)*f^c/(1-a))*x^2+1/2/d*ln(1-b*f^(d*x)
*f^c/(1-a))*x*c+1/4/d^2*ln(1-b*f^(d*x)*f^c/(1-a))*c^2+1/2/ln(f)/d*polylog(2,b*f^(d*x)*f^c/(1-a))*x+1/2/ln(f)/d
^2*polylog(2,b*f^(d*x)*f^c/(1-a))*c-1/2/ln(f)^2/d^2*polylog(3,b*f^(d*x)*f^c/(1-a))+1/4/d^2*c^2*ln(1-a-b*f^(d*x
)*f^c)-1/2/ln(f)/d^2*c*dilog((b*f^(d*x)*f^c+a-1)/(a-1))-1/2/d*c*ln((b*f^(d*x)*f^c+a-1)/(a-1))*x-1/2/d^2*c^2*ln
((b*f^(d*x)*f^c+a-1)/(a-1))-1/4*ln(1-b*f^(d*x)*f^c/(-1-a))*x^2-1/2/d*ln(1-b*f^(d*x)*f^c/(-1-a))*x*c-1/4/d^2*ln
(1-b*f^(d*x)*f^c/(-1-a))*c^2-1/2/ln(f)/d*polylog(2,b*f^(d*x)*f^c/(-1-a))*x-1/2/ln(f)/d^2*polylog(2,b*f^(d*x)*f
^c/(-1-a))*c+1/2/ln(f)^2/d^2*polylog(3,b*f^(d*x)*f^c/(-1-a))-1/4/d^2*c^2*ln(1+a+b*f^(d*x)*f^c)+1/2/ln(f)/d^2*c
*dilog((1+a+b*f^(d*x)*f^c)/(1+a))+1/2/d*c*ln((1+a+b*f^(d*x)*f^c)/(1+a))*x+1/2/d^2*c^2*ln((1+a+b*f^(d*x)*f^c)/(
1+a))

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Maxima [A]  time = 1.07484, size = 262, normalized size = 1.24 \begin{align*} -\frac{1}{4} \, b d{\left (\frac{\log \left (\frac{b f^{d x} f^{c}}{a + 1} + 1\right ) \log \left (f^{d x}\right )^{2} + 2 \,{\rm Li}_2\left (-\frac{b f^{d x} f^{c}}{a + 1}\right ) \log \left (f^{d x}\right ) - 2 \,{\rm Li}_{3}(-\frac{b f^{d x} f^{c}}{a + 1})}{b d^{3} \log \left (f\right )^{3}} - \frac{\log \left (\frac{b f^{d x} f^{c}}{a - 1} + 1\right ) \log \left (f^{d x}\right )^{2} + 2 \,{\rm Li}_2\left (-\frac{b f^{d x} f^{c}}{a - 1}\right ) \log \left (f^{d x}\right ) - 2 \,{\rm Li}_{3}(-\frac{b f^{d x} f^{c}}{a - 1})}{b d^{3} \log \left (f\right )^{3}}\right )} \log \left (f\right ) + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (b f^{d x + c} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*d*((log(b*f^(d*x)*f^c/(a + 1) + 1)*log(f^(d*x))^2 + 2*dilog(-b*f^(d*x)*f^c/(a + 1))*log(f^(d*x)) - 2*po
lylog(3, -b*f^(d*x)*f^c/(a + 1)))/(b*d^3*log(f)^3) - (log(b*f^(d*x)*f^c/(a - 1) + 1)*log(f^(d*x))^2 + 2*dilog(
-b*f^(d*x)*f^c/(a - 1))*log(f^(d*x)) - 2*polylog(3, -b*f^(d*x)*f^c/(a - 1)))/(b*d^3*log(f)^3))*log(f) + 1/2*x^
2*arctanh(b*f^(d*x + c) + a)

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Fricas [C]  time = 1.74833, size = 1195, normalized size = 5.66 \begin{align*} \frac{d^{2} x^{2} \log \left (f\right )^{2} \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^{2} \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 )^{2} + c^{2} \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 )^{2} - 2 \, d x{\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 ) \log \left (f\right ) + 2 \, d x{\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 ) \log \left (f\right ) -{\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} \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^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} \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 ) + 2 \,{\rm polylog}\left (3, -\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}\right ) - 2 \,{\rm polylog}\left (3, -\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}\right )}{4 \, d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(d^2*x^2*log(f)^2*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^2*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a +
1)*log(f)^2 + c^2*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)*log(f)^2 - 2*d*x*dilog(-(b*
cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(a + 1) + 1)*log(f) + 2*d*x*dilog(-(b*cosh((d*x + c
)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)/(a - 1) + 1)*log(f) - (d^2*x^2 - c^2)*log(f)^2*log((b*cosh((d*x
+ c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(a + 1)) + (d^2*x^2 - c^2)*log(f)^2*log((b*cosh((d*x + c)*log
(f)) + b*sinh((d*x + c)*log(f)) + a - 1)/(a - 1)) + 2*polylog(3, -(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)
*log(f)))/(a + 1)) - 2*polylog(3, -(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)))/(a - 1)))/(d^2*log(f)
^2)

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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(x*atanh(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 x \operatorname{artanh}\left (b f^{d x + c} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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