\(\int \coth ^{-1}(a+b x)^2 \, dx\) [72]

Optimal result

Integrand size = 8, antiderivative size = 81 \[ \int \coth ^{-1}(a+b x)^2 \, dx=\frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-\frac {1+a+b x}{1-a-b x}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6239, 6022, 6132, 6056, 2449, 2352} \[ \int \coth ^{-1}(a+b x)^2 \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}+\frac {\coth ^{-1}(a+b x)^2}{b}-\frac {2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b} \]

[In]

Int[ArcCoth[a + b*x]^2,x]

[Out]

ArcCoth[a + b*x]^2/b + ((a + b*x)*ArcCoth[a + b*x]^2)/b - (2*ArcCoth[a + b*x]*Log[2/(1 - a - b*x)])/b - PolyLo
g[2, -((1 + a + b*x)/(1 - a - b*x))]/b

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 6022

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6056

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

Rule 6132

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

Rule 6239

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \coth ^{-1}(a+b x)^2 \, dx=\frac {\coth ^{-1}(a+b x) \left ((-1+a+b x) \coth ^{-1}(a+b x)-2 \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,e^{-2 \coth ^{-1}(a+b x)}\right )}{b} \]

[In]

Integrate[ArcCoth[a + b*x]^2,x]

[Out]

(ArcCoth[a + b*x]*((-1 + a + b*x)*ArcCoth[a + b*x] - 2*Log[1 - E^(-2*ArcCoth[a + b*x])]) + PolyLog[2, E^(-2*Ar
cCoth[a + b*x])])/b

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.64

method	result	size
derivativedivides	\(\frac {\operatorname {arccoth}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arccoth}\left (b x +a \right )^{2}-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1-\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, \frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1+\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, -\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}\)	\(133\)
default	\(\frac {\operatorname {arccoth}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arccoth}\left (b x +a \right )^{2}-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1-\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, \frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1+\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, -\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}\)	\(133\)
risch	\(\frac {\left (b x +a +1\right ) \ln \left (b x +a +1\right )^{2}}{4 b}+\left (-\frac {x \ln \left (b x +a -1\right )}{2}+\frac {-\ln \left (b x +a -1\right ) a +\ln \left (b x +a -1\right )}{2 b}\right ) \ln \left (b x +a +1\right )+\frac {x \ln \left (b x +a -1\right )^{2}}{4}+\frac {\ln \left (b x +a -1\right )^{2} a}{4 b}-\frac {\ln \left (b x +a -1\right )^{2}}{4 b}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{b}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{b}\)	\(142\)

[In]

int(arccoth(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(arccoth(b*x+a)^2*(b*x+a-1)+2*arccoth(b*x+a)^2-2*arccoth(b*x+a)*ln(1-1/((b*x+a-1)/(b*x+a+1))^(1/2))-2*poly
log(2,1/((b*x+a-1)/(b*x+a+1))^(1/2))-2*arccoth(b*x+a)*ln(1+1/((b*x+a-1)/(b*x+a+1))^(1/2))-2*polylog(2,-1/((b*x
+a-1)/(b*x+a+1))^(1/2)))

Fricas [F]

\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcoth}\left (b x + a\right )^{2} \,d x } \]

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arccoth(b*x + a)^2, x)

Sympy [F]

\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int \operatorname {acoth}^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate(acoth(b*x+a)**2,x)

[Out]

Integral(acoth(a + b*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.72 \[ \int \coth ^{-1}(a+b x)^2 \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a - 1\right )} \log \left (b x + a - 1\right )^{2}}{b^{3}} + \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{b^{3}}\right )} + b {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} \operatorname {arcoth}\left (b x + a\right ) + x \operatorname {arcoth}\left (b x + a\right )^{2} \]

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*b^2*(((a + 1)*log(b*x + a + 1)^2 - 2*(a + 1)*log(b*x + a + 1)*log(b*x + a - 1) + (a - 1)*log(b*x + a - 1)
^2)/b^3 + 4*(log(b*x + a - 1)*log(1/2*b*x + 1/2*a + 1/2) + dilog(-1/2*b*x - 1/2*a + 1/2))/b^3) + b*((a + 1)*lo
g(b*x + a + 1)/b^2 - (a - 1)*log(b*x + a - 1)/b^2)*arccoth(b*x + a) + x*arccoth(b*x + a)^2

Giac [F]

\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcoth}\left (b x + a\right )^{2} \,d x } \]

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \coth ^{-1}(a+b x)^2 \, dx=\int {\mathrm {acoth}\left (a+b\,x\right )}^2 \,d x \]

[In]

int(acoth(a + b*x)^2,x)

[Out]

int(acoth(a + b*x)^2, x)