3.72 \(\int \coth ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=81 \[ -\frac {\text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}+\frac {\coth ^{-1}(a+b x)^2}{b}-\frac {2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6104, 5911, 5985, 5919, 2402, 2315} \[ -\frac {\text {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}+\frac {\coth ^{-1}(a+b x)^2}{b}-\frac {2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]^2,x]

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 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 5919

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

Rule 5985

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 55, normalized size = 0.68 \[ \frac {\text {Li}_2\left (e^{-2 \coth ^{-1}(a+b x)}\right )+\coth ^{-1}(a+b x) \left ((a+b x-1) \coth ^{-1}(a+b x)-2 \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a + b*x]^2,x]

[Out]

(ArcCoth[a + b*x]*((-1 + a + b*x)*ArcCoth[a + b*x] - 2*Log[1 - E^(-2*ArcCoth[a + b*x])]) + PolyLog[2, E^(-2*Ar
cCoth[a + b*x])])/b

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arcoth}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arccoth(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)^2, x)

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maple [A]  time = 0.30, size = 151, normalized size = 1.86 \[ x \mathrm {arccoth}\left (b x +a \right )^{2}+\frac {\mathrm {arccoth}\left (b x +a \right )^{2} a}{b}+\frac {\mathrm {arccoth}\left (b x +a \right )^{2}}{b}-\frac {2 \,\mathrm {arccoth}\left (b x +a \right ) \ln \left (1-\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}-\frac {2 \,\mathrm {arccoth}\left (b x +a \right ) \ln \left (1+\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}-\frac {2 \polylog \left (2, -\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}-\frac {2 \polylog \left (2, \frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)^2,x)

[Out]

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

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maxima [A]  time = 0.33, size = 139, normalized size = 1.72 \[ -\frac {1}{4} \, b^{2} {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a - 1\right )} \log \left (b x + a - 1\right )^{2}}{b^{3}} + \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{b^{3}}\right )} + b {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} \operatorname {arcoth}\left (b x + a\right ) + x \operatorname {arcoth}\left (b x + a\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {acoth}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)^2,x)

[Out]

int(acoth(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {acoth}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(b*x+a)**2,x)

[Out]

Integral(acoth(a + b*x)**2, x)

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