\(\int x \coth ^{-1}(a+b f^{c+d x}) \, dx\) [290]

Optimal result

Integrand size = 14, antiderivative size = 216 \[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {1}{4} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{4} x^2 \log \left (1+\frac {b f^{c+d x}}{1+a}\right )-\frac {1}{4} x^2 \log \left (1-\frac {1}{a+b f^{c+d x}}\right )+\frac {1}{4} x^2 \log \left (1+\frac {1}{a+b f^{c+d x}}\right )+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac {\operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac {\operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{2 d^2 \log ^2(f)} \]

[Out]

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

Rubi [A] (verified)

Time = 1.91 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6349, 2631, 12, 6874, 2221, 2611, 2320, 6724} \[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {\operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac {\operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+1}\right )}{2 d^2 \log ^2(f)}+\frac {x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+1}\right )}{2 d \log (f)}+\frac {1}{4} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{4} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right )-\frac {1}{4} x^2 \log \left (1-\frac {1}{a+b f^{c+d x}}\right )+\frac {1}{4} x^2 \log \left (\frac {1}{a+b f^{c+d x}}+1\right ) \]

[In]

Int[x*ArcCoth[a + b*f^(c + d*x)],x]

[Out]

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

Rule 12

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

Rule 2221

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

Rule 2320

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 2611

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 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6349

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

Rule 6724

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.82 \[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 d^2 x^2 \coth ^{-1}\left (a+b f^{c+d x}\right ) \log ^2(f)+d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{-1+a}\right )-d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{1+a}\right )+2 d x \log (f) \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{-1+a}\right )-2 d x \log (f) \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )-2 \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{-1+a}\right )+2 \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{4 d^2 \log ^2(f)} \]

[In]

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

[Out]

(2*d^2*x^2*ArcCoth[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(589\) vs. \(2(200)=400\).

Time = 0.64 (sec) , antiderivative size = 590, normalized size of antiderivative = 2.73

method	result	size
risch	\(-\frac {x^{2} \ln \left (b \,f^{d x +c}+a -1\right )}{4}+\frac {x^{2} \ln \left (1+a +b \,f^{d x +c}\right )}{4}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) x^{2}}{4}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) c x}{2 d}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) c^{2}}{4 d^{2}}-\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{-1-a}\right ) x}{2 \ln \left (f \right ) d}-\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{-1-a}\right ) c}{2 \ln \left (f \right ) d^{2}}+\frac {\operatorname {polylog}\left (3, \frac {b \,f^{d x} f^{c}}{-1-a}\right )}{2 \ln \left (f \right )^{2} d^{2}}-\frac {c^{2} \ln \left (1+a +f^{d x} f^{c} b \right )}{4 d^{2}}+\frac {c \operatorname {dilog}\left (\frac {1+a +f^{d x} f^{c} b}{1+a}\right )}{2 \ln \left (f \right ) d^{2}}+\frac {c \ln \left (\frac {1+a +f^{d x} f^{c} b}{1+a}\right ) x}{2 d}+\frac {c^{2} \ln \left (\frac {1+a +f^{d x} f^{c} b}{1+a}\right )}{2 d^{2}}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x^{2}}{4}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c x}{2 d}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c^{2}}{4 d^{2}}+\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) x}{2 \ln \left (f \right ) d}+\frac {\operatorname {polylog}\left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) c}{2 \ln \left (f \right ) d^{2}}-\frac {\operatorname {polylog}\left (3, \frac {b \,f^{d x} f^{c}}{1-a}\right )}{2 \ln \left (f \right )^{2} d^{2}}+\frac {c^{2} \ln \left (f^{d x} f^{c} b +a -1\right )}{4 d^{2}}-\frac {c \operatorname {dilog}\left (\frac {f^{d x} f^{c} b +a -1}{-1+a}\right )}{2 \ln \left (f \right ) d^{2}}-\frac {c \ln \left (\frac {f^{d x} f^{c} b +a -1}{-1+a}\right ) x}{2 d}-\frac {c^{2} \ln \left (\frac {f^{d x} f^{c} b +a -1}{-1+a}\right )}{2 d^{2}}\)	\(590\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (193) = 386\).

Time = 0.26 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.83 \[ \int x \coth ^{-1}\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}} \]

[In]

integrate(x*arccoth(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*c
osh((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)

Sympy [F]

\[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x \operatorname {acoth}{\left (a + b f^{c + d x} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.90 \[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{4} \, b d {\left (\frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a + 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a + 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a + 1})}{b d^{3} \log \left (f\right )^{3}} - \frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a - 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a - 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a - 1})}{b d^{3} \log \left (f\right )^{3}}\right )} \log \left (f\right ) + \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (b f^{d x + c} + a\right ) \]

[In]

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

[Out]

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

Giac [F]

\[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { x \operatorname {arcoth}\left (b f^{d x + c} + a\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \coth ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x\,\mathrm {acoth}\left (a+b\,f^{c+d\,x}\right ) \,d x \]

[In]

int(x*acoth(a + b*f^(c + d*x)),x)

[Out]

int(x*acoth(a + b*f^(c + d*x)), x)