\(\int x \text {arctanh}(a+b f^{c+d x}) \, dx\) [353]

Optimal result

Integrand size = 14, antiderivative size = 211 \[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac {1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac {1}{4} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{4} x^2 \log \left (1+\frac {b f^{c+d x}}{1+a}\right )+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac {\operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac {\operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{2 d^2 \log ^2(f)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 d^2 x^2 \text {arctanh}\left (a+b f^{c+d x}\right ) \log ^2(f)+d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{-1+a}\right )-d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{1+a}\right )+2 d x \log (f) \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{-1+a}\right )-2 d x \log (f) \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )-2 \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{-1+a}\right )+2 \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{4 d^2 \log ^2(f)} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6767, 3012, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx\)

\(\Big \downarrow \) 6767

\(\displaystyle \frac {1}{2} \int x \log \left (b f^{c+d x}+a+1\right )dx-\frac {1}{2} \int x \log \left (-b f^{c+d x}-a+1\right )dx\)

\(\Big \downarrow \) 3012

\(\displaystyle \frac {1}{2} \left (-\int x \log \left (1-\frac {b f^{c+d x}}{1-a}\right )dx-\frac {1}{2} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{2} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )\right )+\frac {1}{2} \left (\int x \log \left (\frac {b f^{c+d x}}{a+1}+1\right )dx+\frac {1}{2} x^2 \log \left (a+b f^{c+d x}+1\right )-\frac {1}{2} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right )\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {1}{2} \left (-\frac {\int \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )dx}{d \log (f)}+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{d \log (f)}-\frac {1}{2} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{2} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )\right )+\frac {1}{2} \left (\frac {\int \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )dx}{d \log (f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )}{d \log (f)}+\frac {1}{2} x^2 \log \left (a+b f^{c+d x}+1\right )-\frac {1}{2} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right )\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{2} \left (-\frac {\int f^{-c-d x} \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )df^{c+d x}}{d^2 \log ^2(f)}+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{d \log (f)}-\frac {1}{2} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{2} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )\right )+\frac {1}{2} \left (\frac {\int f^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )}{d \log (f)}+\frac {1}{2} x^2 \log \left (a+b f^{c+d x}+1\right )-\frac {1}{2} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right )\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{2} \left (-\frac {\operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{1-a}\right )}{d^2 \log ^2(f)}+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{d \log (f)}-\frac {1}{2} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{2} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )\right )+\frac {1}{2} \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+1}\right )}{d^2 \log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )}{d \log (f)}+\frac {1}{2} x^2 \log \left (a+b f^{c+d x}+1\right )-\frac {1}{2} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3012

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/d)*(F^(c*(a + b*x) 
))^n], x] - Simp[(f + g*x)^(m + 1)*(Log[1 + (e/d)*(F^(c*(a + b*x)))^n]/(g*( 
m + 1))), x]) /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[ 
d, 1]
 

rule 6767

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

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(595\) vs. \(2(195)=390\).

Time = 0.86 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.82

method	result	size
risch	\(\frac {x^{2} \ln \left (1+a +b \,f^{d x +c}\right )}{4}-\frac {x^{2} \ln \left (1-a -b \,f^{d x +c}\right )}{4}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x^{2}}{4}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c x}{2 d}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c^{2}}{4 d^{2}}+\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) x}{2 \ln \left (f \right ) d}+\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) c}{2 \ln \left (f \right ) d^{2}}-\frac {\operatorname {polylog}\left (3, \frac {b \,f^{d x} f^{c}}{1-a}\right )}{2 \ln \left (f \right )^{2} d^{2}}+\frac {c^{2} \ln \left (1-a -b \,f^{d x} f^{c}\right )}{4 d^{2}}-\frac {c \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -1}{a -1}\right )}{2 \ln \left (f \right ) d^{2}}-\frac {c \ln \left (\frac {b \,f^{d x} f^{c}+a -1}{a -1}\right ) x}{2 d}-\frac {c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -1}{a -1}\right )}{2 d^{2}}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) x^{2}}{4}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) c x}{2 d}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) c^{2}}{4 d^{2}}-\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{-a -1}\right ) x}{2 \ln \left (f \right ) d}-\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{-a -1}\right ) c}{2 \ln \left (f \right ) d^{2}}+\frac {\operatorname {polylog}\left (3, \frac {b \,f^{d x} f^{c}}{-a -1}\right )}{2 \ln \left (f \right )^{2} d^{2}}-\frac {c^{2} \ln \left (1+a +b \,f^{d x} f^{c}\right )}{4 d^{2}}+\frac {c \operatorname {dilog}\left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right )}{2 \ln \left (f \right ) d^{2}}+\frac {c \ln \left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right ) x}{2 d}+\frac {c^{2} \ln \left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right )}{2 d^{2}}\)	\(596\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (188) = 376\).

Time = 0.10 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.88 \[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\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}} \] Input:

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

Output:

1/4*(d^2*x^2*log(f)^2*log(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*lo 
g(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)*lo 
g(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)
 

Sympy [F]

\[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\int x \operatorname {atanh}{\left (a + b f^{c + d x} \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92 \[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{4} \, b d {\left (\frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a + 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a + 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a + 1})}{b d^{3} \log \left (f\right )^{3}} - \frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a - 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a - 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a - 1})}{b d^{3} \log \left (f\right )^{3}}\right )} \log \left (f\right ) + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (b f^{d x + c} + a\right ) \] Input:

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

Output:

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

Giac [F]

\[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\int { x \operatorname {artanh}\left (b f^{d x + c} + a\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\int x\,\mathrm {atanh}\left (a+b\,f^{c+d\,x}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x \text {arctanh}\left (a+b f^{c+d x}\right ) \, dx=\int \mathit {atanh} \left (f^{d x +c} b +a \right ) x d x \] Input:

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

Output:

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