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

Optimal. Leaf size=40 \[ a x+\frac{b \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b (c+d x) \coth ^{-1}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0245296, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6104, 5911, 260} \[ a x+\frac{b \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b (c+d x) \coth ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcCoth[c + d*x],x]

[Out]

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

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0149388, size = 48, normalized size = 1.2 \[ a x+\frac{b ((c+1) \log (c+d x+1)-(c-1) \log (-c-d x+1))}{2 d}+b x \coth ^{-1}(c+d x) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcCoth[c + d*x],x]

[Out]

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

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Maple [A]  time = 0.033, size = 42, normalized size = 1.1 \begin{align*} ax+b{\rm arccoth} \left (dx+c\right )x+{\frac{b{\rm arccoth} \left (dx+c\right )c}{d}}+{\frac{b\ln \left ( \left ( dx+c \right ) ^{2}-1 \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970909, size = 49, normalized size = 1.22 \begin{align*} a x + \frac{{\left (2 \,{\left (d x + c\right )} \operatorname{arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.91274, size = 157, normalized size = 3.92 \begin{align*} \frac{b d x \log \left (\frac{d x + c + 1}{d x + c - 1}\right ) + 2 \, a d x +{\left (b c + b\right )} \log \left (d x + c + 1\right ) -{\left (b c - b\right )} \log \left (d x + c - 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.761472, size = 46, normalized size = 1.15 \begin{align*} a x + b \left (\begin{cases} \frac{c \operatorname{acoth}{\left (c + d x \right )}}{d} + x \operatorname{acoth}{\left (c + d x \right )} + \frac{\log{\left (c + d x + 1 \right )}}{d} - \frac{\operatorname{acoth}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \operatorname{acoth}{\left (c \right )} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int b \operatorname{arcoth}\left (d x + c\right ) + a\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(b*arccoth(d*x + c) + a, x)