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

Optimal. Leaf size=35 \[ \frac {\log \left (1-(a+b x)^2\right )}{2 b}+\frac {(a+b x) \coth ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6104, 5911, 260} \[ \frac {\log \left (1-(a+b x)^2\right )}{2 b}+\frac {(a+b x) \coth ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[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 260

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 5911

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

Rule 6104

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*} \int \coth ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b}-\frac {\operatorname {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*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.23 \[ \frac {(a+1) \log (a+b x+1)-(a-1) \log (-a-b x+1)}{2 b}+x \coth ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.67, size = 48, normalized size = 1.37 \[ \frac {b x \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + {\left (a + 1\right )} \log \left (b x + a + 1\right ) - {\left (a - 1\right )} \log \left (b x + a - 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(b*x + a), x)

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maple [A]  time = 0.03, size = 36, normalized size = 1.03 \[ x \,\mathrm {arccoth}\left (b x +a \right )+\frac {\mathrm {arccoth}\left (b x +a \right ) a}{b}+\frac {\ln \left (\left (b x +a \right )^{2}-1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a),x)

[Out]

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

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maxima [A]  time = 0.31, size = 31, normalized size = 0.89 \[ \frac {2 \, {\left (b x + a\right )} \operatorname {arcoth}\left (b x + a\right ) + \log \left (-{\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.71, size = 42, normalized size = 1.20 \[ \frac {\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2}+a\,\mathrm {acoth}\left (a+b\,x\right )}{b}+x\,\mathrm {acoth}\left (a+b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.51, size = 41, normalized size = 1.17 \[ \begin {cases} \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b} + x \operatorname {acoth}{\left (a + b x \right )} + \frac {\log {\left (a + b x + 1 \right )}}{b} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \operatorname {acoth}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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