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

Optimal. Leaf size=35 \[ \frac {\text {Li}_2\left (-\frac {1}{a+b x}\right )}{2 b}-\frac {\text {Li}_2\left (\frac {1}{a+b x}\right )}{2 b} \]

[Out]

1/2*polylog(2,-1/(b*x+a))/b-1/2*polylog(2,1/(b*x+a))/b

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6108, 5913} \[ \frac {\text {PolyLog}\left (2,-\frac {1}{a+b x}\right )}{2 b}-\frac {\text {PolyLog}\left (2,\frac {1}{a+b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/(a + b*x),x]

[Out]

PolyLog[2, -(a + b*x)^(-1)]/(2*b) - PolyLog[2, (a + b*x)^(-1)]/(2*b)

Rule 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

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]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a+b x)}{a+b x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{x} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Li}_2\left (-\frac {1}{a+b x}\right )}{2 b}-\frac {\text {Li}_2\left (\frac {1}{a+b x}\right )}{2 b}\\ \end {align*}

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Mathematica [B]  time = 0.03, size = 286, normalized size = 8.17 \[ -\frac {\text {Li}_2(-a-b x)}{2 b}+\frac {\text {Li}_2(a+b x)}{2 b}-\frac {\log ^2\left (\frac {a b-(a-1) b}{b (a+b x)}\right )}{4 b}+\frac {\log ^2\left (\frac {a b-(a+1) b}{b (a+b x)}\right )}{4 b}-\frac {\log \left (\frac {b (a+b x-1)}{(a-1) b-a b}\right ) \log \left (\frac {a b-(a-1) b}{b (a+b x)}\right )}{2 b}+\frac {\log \left (\frac {a+b x-1}{a+b x}\right ) \log \left (\frac {a b-(a-1) b}{b (a+b x)}\right )}{2 b}+\frac {\log \left (\frac {b (-a-b x-1)}{(-a-1) b+a b}\right ) \log \left (\frac {a b-(a+1) b}{b (a+b x)}\right )}{2 b}-\frac {\log \left (\frac {a b-(a+1) b}{b (a+b x)}\right ) \log \left (\frac {a+b x+1}{a+b x}\right )}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (b x + a\right )}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(b*x + a)/(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 59, normalized size = 1.69 \[ \frac {\ln \left (b x +a \right ) \mathrm {arccoth}\left (b x +a \right )}{b}-\frac {\dilog \left (b x +a \right )}{2 b}-\frac {\dilog \left (b x +a +1\right )}{2 b}-\frac {\ln \left (b x +a \right ) \ln \left (b x +a +1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 112, normalized size = 3.20 \[ -\frac {1}{2} \, b {\left (\frac {\log \left (b x + a\right ) \log \left (b x + a - 1\right ) + {\rm Li}_2\left (-b x - a + 1\right )}{b^{2}} - \frac {\log \left (b x + a + 1\right ) \log \left (-b x - a\right ) + {\rm Li}_2\left (b x + a + 1\right )}{b^{2}}\right )} - \frac {1}{2} \, {\left (\frac {\log \left (b x + a + 1\right )}{b} - \frac {\log \left (b x + a - 1\right )}{b}\right )} \log \left (b x + a\right ) + \frac {\operatorname {arcoth}\left (b x + a\right ) \log \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {acoth}\left (a+b\,x\right )}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/(a + b*x),x)

[Out]

int(acoth(a + b*x)/(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}{\left (a + b x \right )}}{a + b x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a + b*x)/(a + b*x), x)

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