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

Optimal. Leaf size=39 \[ -\frac{\tanh ^{-1}(a+b x)}{2 b}+\frac{(a+b x)^2 \coth ^{-1}(a+b x)}{2 b}+\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0224787, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6108, 5917, 321, 206} \[ -\frac{\tanh ^{-1}(a+b x)}{2 b}+\frac{(a+b x)^2 \coth ^{-1}(a+b x)}{2 b}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*ArcCoth[a + b*x],x]

[Out]

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

Rule 6108

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

Rule 5917

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

Rule 321

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0320044, size = 66, normalized size = 1.69 \[ \frac{a^2 \log (a+b x+1)-\left (a^2-1\right ) \log (-a-b x+1)-\log (a+b x+1)+2 b x (2 a+b x) \coth ^{-1}(a+b x)+2 b x}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*ArcCoth[a + b*x],x]

[Out]

(2*b*x + 2*b*x*(2*a + b*x)*ArcCoth[a + b*x] - (-1 + a^2)*Log[1 - a - b*x] - Log[1 + a + b*x] + a^2*Log[1 + a +
b*x])/(4*b)

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Maple [B]  time = 0.036, size = 70, normalized size = 1.8 \begin{align*}{\frac{b{\rm arccoth} \left (bx+a\right ){x}^{2}}{2}}+{\rm arccoth} \left (bx+a\right )xa+{\frac{{\rm arccoth} \left (bx+a\right ){a}^{2}}{2\,b}}+{\frac{x}{2}}+{\frac{a}{2\,b}}+{\frac{\ln \left ( bx+a-1 \right ) }{4\,b}}-{\frac{\ln \left ( bx+a+1 \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*arccoth(b*x+a)*x^2+arccoth(b*x+a)*x*a+1/2/b*arccoth(b*x+a)*a^2+1/2*x+1/2*a/b+1/4/b*ln(b*x+a-1)-1/4/b*ln(
b*x+a+1)

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Maxima [A]  time = 0.966219, size = 84, normalized size = 2.15 \begin{align*} \frac{1}{4} \, b{\left (\frac{2 \, x}{b} + \frac{{\left (a^{2} - 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac{{\left (a^{2} - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} + \frac{1}{2} \,{\left (b x^{2} + 2 \, a x\right )} \operatorname{arcoth}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*(2*x/b + (a^2 - 1)*log(b*x + a + 1)/b^2 - (a^2 - 1)*log(b*x + a - 1)/b^2) + 1/2*(b*x^2 + 2*a*x)*arccoth(
b*x + a)

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Fricas [A]  time = 1.92371, size = 108, normalized size = 2.77 \begin{align*} \frac{2 \, b x +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \log \left (\frac{b x + a + 1}{b x + a - 1}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*x + (b^2*x^2 + 2*a*b*x + a^2 - 1)*log((b*x + a + 1)/(b*x + a - 1)))/b

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Sympy [A]  time = 1.73783, size = 56, normalized size = 1.44 \begin{align*} \begin{cases} \frac{a^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 b} + a x \operatorname{acoth}{\left (a + b x \right )} + \frac{b x^{2} \operatorname{acoth}{\left (a + b x \right )}}{2} + \frac{x}{2} - \frac{\operatorname{acoth}{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\a x \operatorname{acoth}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*acoth(a + b*x)/(2*b) + a*x*acoth(a + b*x) + b*x**2*acoth(a + b*x)/2 + x/2 - acoth(a + b*x)/(2*
b), Ne(b, 0)), (a*x*acoth(a), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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