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

Optimal. Leaf size=65 \[ \frac{(1-a)^2 \log (-a-b x+1)}{4 b^2}-\frac{(a+1)^2 \log (a+b x+1)}{4 b^2}+\frac{1}{2} x^2 \coth ^{-1}(a+b x)+\frac{x}{2 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0716315, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6112, 5927, 702, 633, 31} \[ \frac{(1-a)^2 \log (-a-b x+1)}{4 b^2}-\frac{(a+1)^2 \log (a+b x+1)}{4 b^2}+\frac{1}{2} x^2 \coth ^{-1}(a+b x)+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[a + b*x],x]

[Out]

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

Rule 6112

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

Rule 5927

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

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

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

Rubi steps

\begin{align*} \int x \coth ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right ) \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{2} x^2 \coth ^{-1}(a+b x)-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^2}{1-x^2} \, dx,x,a+b x\right )\\ &=\frac{1}{2} x^2 \coth ^{-1}(a+b x)-\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{1}{b^2}+\frac{1+a^2-2 a x}{b^2 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac{x}{2 b}+\frac{1}{2} x^2 \coth ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \frac{1+a^2-2 a x}{1-x^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{x}{2 b}+\frac{1}{2} x^2 \coth ^{-1}(a+b x)-\frac{(1-a)^2 \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,a+b x\right )}{4 b^2}+\frac{(1+a)^2 \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,a+b x\right )}{4 b^2}\\ &=\frac{x}{2 b}+\frac{1}{2} x^2 \coth ^{-1}(a+b x)+\frac{(1-a)^2 \log (1-a-b x)}{4 b^2}-\frac{(1+a)^2 \log (1+a+b x)}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0219382, size = 56, normalized size = 0.86 \[ \frac{2 b^2 x^2 \coth ^{-1}(a+b x)+(a-1)^2 \log (-a-b x+1)-(a+1)^2 \log (a+b x+1)+2 b x}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 89, normalized size = 1.4 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left (bx+a\right )}{2}}-{\frac{{\rm arccoth} \left (bx+a\right ){a}^{2}}{2\,{b}^{2}}}+{\frac{x}{2\,b}}+{\frac{a}{2\,{b}^{2}}}-{\frac{\ln \left ( bx+a-1 \right ) a}{2\,{b}^{2}}}+{\frac{\ln \left ( bx+a-1 \right ) }{4\,{b}^{2}}}-{\frac{\ln \left ( bx+a+1 \right ) a}{2\,{b}^{2}}}-{\frac{\ln \left ( bx+a+1 \right ) }{4\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.971898, size = 82, normalized size = 1.26 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (b x + a\right ) + \frac{1}{4} \, b{\left (\frac{2 \, x}{b^{2}} - \frac{{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} + \frac{{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.59959, size = 176, normalized size = 2.71 \begin{align*} \frac{b^{2} x^{2} \log \left (\frac{b x + a + 1}{b x + a - 1}\right ) + 2 \, b x -{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right ) +{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.60945, size = 76, normalized size = 1.17 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 b^{2}} - \frac{a \log{\left (a + b x + 1 \right )}}{b^{2}} + \frac{a \operatorname{acoth}{\left (a + b x \right )}}{b^{2}} + \frac{x^{2} \operatorname{acoth}{\left (a + b x \right )}}{2} + \frac{x}{2 b} - \frac{\operatorname{acoth}{\left (a + b x \right )}}{2 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{acoth}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*acoth(a + b*x)/(2*b**2) - a*log(a + b*x + 1)/b**2 + a*acoth(a + b*x)/b**2 + x**2*acoth(a + b*
x)/2 + x/(2*b) - acoth(a + b*x)/(2*b**2), Ne(b, 0)), (x**2*acoth(a)/2, True))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(b*x + a), x)