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]

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

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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)

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fricas [A]  time = 0.84, size = 66, normalized size = 1.02 \[ \frac {b^{2} x^{2} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + 2 \, b x - {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right ) + {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{4 \, b^{2}} \]

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

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

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)

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maple [A]  time = 0.03, size = 89, normalized size = 1.37 \[ \frac {x^{2} \mathrm {arccoth}\left (b x +a \right )}{2}-\frac {\mathrm {arccoth}\left (b x +a \right ) a^{2}}{2 b^{2}}+\frac {x}{2 b}+\frac {a}{2 b^{2}}+\frac {\ln \left (b x +a -1\right )}{4 b^{2}}-\frac {\ln \left (b x +a -1\right ) a}{2 b^{2}}-\frac {\ln \left (b x +a +1\right )}{4 b^{2}}-\frac {\ln \left (b x +a +1\right ) a}{2 b^{2}} \]

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

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maxima [A]  time = 0.30, size = 61, normalized size = 0.94 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{4} \, b {\left (\frac {2 \, x}{b^{2}} - \frac {{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} + \frac {{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{3}}\right )} \]

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)

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mupad [B]  time = 2.00, size = 62, normalized size = 0.95 \[ \frac {x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {\frac {\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {b\,x}{2}+\frac {a^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}+\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(a + b*x),x)

[Out]

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

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sympy [A]  time = 0.81, size = 76, normalized size = 1.17 \[ \begin {cases} - \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b^{2}} - \frac {a \log {\left (a + b x + 1 \right )}}{b^{2}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{2}} + \frac {x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2} + \frac {x}{2 b} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acoth}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

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))

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