3.3.89 \(\int \coth ^{-1}(a+b f^{c+d x}) \, dx\) [289]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2320, 6247, 6058, 2449, 2352, 2497} \begin {gather*} \frac {\text {Li}_2\left (1-\frac {2}{b f^{c+d x}+a+1}\right )}{2 d \log (f)}-\frac {\text {Li}_2\left (1-\frac {2 b f^{c+d x}}{(1-a) \left (b f^{c+d x}+a+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)} \end {gather*}

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 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 2352

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

Rule 2449

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 2497

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]]

Rule 6058

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 6247

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]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 108, normalized size = 0.64 \begin {gather*} \frac {d x \log (f) \left (2 \coth ^{-1}\left (a+b f^{c+d x}\right )+\log \left (\frac {-1+a+b f^{c+d x}}{-1+a}\right )-\log \left (\frac {1+a+b f^{c+d x}}{1+a}\right )\right )+\text {PolyLog}\left (2,-\frac {b f^{c+d x}}{-1+a}\right )-\text {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )}{2 d \log (f)} \end {gather*}

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 = 0.52, size = 160, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\ln \left (-b \,f^{d x +c}\right ) \mathrm {arccoth}\left (a +b \,f^{d x +c}\right )-\frac {\dilog \left (\frac {-b \,f^{d x +c}-a -1}{-a -1}\right )}{2}-\frac {\ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {-b \,f^{d x +c}-a -1}{-a -1}\right )}{2}+\frac {\dilog \left (\frac {-b \,f^{d x +c}-a +1}{1-a}\right )}{2}+\frac {\ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {-b \,f^{d x +c}-a +1}{1-a}\right )}{2}}{d \ln \left (f \right )}\)	\(160\)
default	\(\frac {\ln \left (-b \,f^{d x +c}\right ) \mathrm {arccoth}\left (a +b \,f^{d x +c}\right )-\frac {\dilog \left (\frac {-b \,f^{d x +c}-a -1}{-a -1}\right )}{2}-\frac {\ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {-b \,f^{d x +c}-a -1}{-a -1}\right )}{2}+\frac {\dilog \left (\frac {-b \,f^{d x +c}-a +1}{1-a}\right )}{2}+\frac {\ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {-b \,f^{d x +c}-a +1}{1-a}\right )}{2}}{d \ln \left (f \right )}\)	\(160\)
risch	\(-\frac {x \ln \left (a +b \,f^{d x +c}-1\right )}{2}+\frac {\dilog \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right )}{2 \ln \left (f \right ) d}+\frac {\ln \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right ) x}{2}+\frac {\ln \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right ) c}{2 d}-\frac {c \ln \left (b \,f^{d x} f^{c}+a -1\right )}{2 d}+\frac {\ln \left (1+a +b \,f^{d x +c}\right ) \ln \left (\frac {b \,f^{d x +c}}{-a -1}\right )}{2 d \ln \left (f \right )}+\frac {\dilog \left (\frac {b \,f^{d x +c}}{-a -1}\right )}{2 d \ln \left (f \right )}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 202, normalized size = 1.20 \begin {gather*} \frac {{\left (d x + c\right )} \operatorname {arcoth}\left (b f^{d x + c} + a\right )}{d} - \frac {{\left (d x + c\right )} 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\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)*arccoth(b*f^(d*x + c) + a)/d - 1/2*((d*x + c)*b*(log(b*f^(d*x + c) + a + 1)/b - log(b*f^(d*x + c) +
a - 1)/b)*log(f) - 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 = 0.36, size = 283, normalized size = 1.68 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {acoth}{\left (a + b f^{c + d x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
0.55Unable to divide, perhaps due to rounding error%%%{1,[0,1,2,0,0,0]%%%}+%%%{2,[0,1,1,1,1,0]%%%}+%%%{-2,[0,1
,1,0,0,0]%%%}+

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {acoth}\left (a+b\,f^{c+d\,x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*f^(c + d*x)),x)

[Out]

int(acoth(a + b*f^(c + d*x)), x)

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