\(\int \coth ^{-1}(a+b x) \, dx\) [65]

Optimal result

Integrand size = 6, antiderivative size = 35 \[ \int \coth ^{-1}(a+b x) \, dx=\frac {(a+b x) \coth ^{-1}(a+b x)}{b}+\frac {\log \left (1-(a+b x)^2\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6239, 6022, 266} \[ \int \coth ^{-1}(a+b x) \, dx=\frac {\log \left (1-(a+b x)^2\right )}{2 b}+\frac {(a+b x) \coth ^{-1}(a+b x)}{b} \]

[In]

Int[ArcCoth[a + b*x],x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \coth ^{-1}(a+b x) \, dx=x \coth ^{-1}(a+b x)+\frac {-((-1+a) \log (1-a-b x))+(1+a) \log (1+a+b x)}{2 b} \]

[In]

Integrate[ArcCoth[a + b*x],x]

[Out]

x*ArcCoth[a + b*x] + (-((-1 + a)*Log[1 - a - b*x]) + (1 + a)*Log[1 + a + b*x])/(2*b)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\left (b x +a \right ) \operatorname {arccoth}\left (b x +a \right )+\frac {\ln \left (\left (b x +a \right )^{2}-1\right )}{2}}{b}\)	\(30\)
default	\(\frac {\left (b x +a \right ) \operatorname {arccoth}\left (b x +a \right )+\frac {\ln \left (\left (b x +a \right )^{2}-1\right )}{2}}{b}\)	\(30\)
parts	\(x \,\operatorname {arccoth}\left (b x +a \right )+b \left (\frac {\left (1-a \right ) \ln \left (b x +a -1\right )}{2 b^{2}}+\frac {\left (1+a \right ) \ln \left (b x +a +1\right )}{2 b^{2}}\right )\)	\(45\)
parallelrisch	\(-\frac {-\operatorname {arccoth}\left (b x +a \right ) b^{2} x -\operatorname {arccoth}\left (b x +a \right ) a b -b \ln \left (b x +a -1\right )-\operatorname {arccoth}\left (b x +a \right ) b}{b^{2}}\)	\(48\)
risch	\(\frac {x \ln \left (b x +a +1\right )}{2}-\frac {\ln \left (b x +a -1\right ) x}{2}-\frac {\ln \left (b x +a -1\right ) a}{2 b}+\frac {\ln \left (-b x -a -1\right ) a}{2 b}+\frac {\ln \left (b x +a -1\right )}{2 b}+\frac {\ln \left (-b x -a -1\right )}{2 b}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \coth ^{-1}(a+b x) \, dx=\frac {b x \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + {\left (a + 1\right )} \log \left (b x + a + 1\right ) - {\left (a - 1\right )} \log \left (b x + a - 1\right )}{2 \, b} \]

[In]

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

[Out]

1/2*(b*x*log((b*x + a + 1)/(b*x + a - 1)) + (a + 1)*log(b*x + a + 1) - (a - 1)*log(b*x + a - 1))/b

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \coth ^{-1}(a+b x) \, dx=\begin {cases} \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b} + x \operatorname {acoth}{\left (a + b x \right )} + \frac {\log {\left (a + b x + 1 \right )}}{b} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \operatorname {acoth}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*acoth(a + b*x)/b + x*acoth(a + b*x) + log(a + b*x + 1)/b - acoth(a + b*x)/b, Ne(b, 0)), (x*acoth(
a), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \coth ^{-1}(a+b x) \, dx=\frac {2 \, {\left (b x + a\right )} \operatorname {arcoth}\left (b x + a\right ) + \log \left (-{\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

[In]

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

[Out]

1/2*(2*(b*x + a)*arccoth(b*x + a) + log(-(b*x + a)^2 + 1))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (33) = 66\).

Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 5.63 \[ \int \coth ^{-1}(a+b x) \, dx=\frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{b^{2}} + \frac {\log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}}\right )} \]

[In]

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

[Out]

1/2*((a + 1)*b - (a - 1)*b)*(log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^2 - log(abs((b*x + a + 1)/(b*x + a - 1)
- 1))/b^2 + log(-(1/(a - ((b*x + a + 1)*(a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/(b*x + a - 1) - b))
+ 1)/(1/(a - ((b*x + a + 1)*(a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/(b*x + a - 1) - b)) - 1))/(b^2*(
(b*x + a + 1)/(b*x + a - 1) - 1)))

Mupad [B] (verification not implemented)

Time = 4.64 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \coth ^{-1}(a+b x) \, dx=\frac {\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2}+a\,\mathrm {acoth}\left (a+b\,x\right )}{b}+x\,\mathrm {acoth}\left (a+b\,x\right ) \]

[In]

int(acoth(a + b*x),x)

[Out]

(log(a^2 + b^2*x^2 + 2*a*b*x - 1)/2 + a*acoth(a + b*x))/b + x*acoth(a + b*x)