3.75 \(\int \frac {\coth ^{-1}(a+b x)^2}{x^3} \, dx\)

Optimal. Leaf size=370 \[ \frac {a b^2 \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )}{\left (1-a^2\right )^2}-\frac {a b^2 \text {Li}_2\left (1-\frac {2 b x}{(1-a) (a+b x+1)}\right )}{\left (1-a^2\right )^2}+\frac {b^2 \log (x)}{\left (1-a^2\right )^2}-\frac {2 a b^2 \log \left (\frac {2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{\left (1-a^2\right )^2}+\frac {2 a b^2 \log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)}{\left (1-a^2\right )^2}-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}+\frac {b^2 \text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{4 (1-a)^2}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )}{4 (a+1)^2}-\frac {b^2 \log (-a-b x+1)}{2 (1-a)^2 (a+1)}-\frac {b^2 \log (a+b x+1)}{2 (1-a) (a+1)^2}+\frac {b^2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{2 (1-a)^2}-\frac {b^2 \log \left (\frac {2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{2 (a+1)^2}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.83, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.333, Rules used = {6110, 371, 710, 801, 6741, 6122, 6725, 5927, 706, 31, 633, 5921, 2402, 2315, 2447, 5919} \[ \frac {a b^2 \text {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{\left (1-a^2\right )^2}-\frac {a b^2 \text {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (a+b x+1)}\right )}{\left (1-a^2\right )^2}+\frac {b^2 \text {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{4 (1-a)^2}+\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{4 (a+1)^2}+\frac {b^2 \log (x)}{\left (1-a^2\right )^2}-\frac {2 a b^2 \log \left (\frac {2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{\left (1-a^2\right )^2}+\frac {2 a b^2 \log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)}{\left (1-a^2\right )^2}-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {b^2 \log (-a-b x+1)}{2 (1-a)^2 (a+1)}-\frac {b^2 \log (a+b x+1)}{2 (1-a) (a+1)^2}+\frac {b^2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{2 (1-a)^2}-\frac {b^2 \log \left (\frac {2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{2 (a+1)^2}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]^2/x^3,x]

[Out]

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

Rule 31

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

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 706

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

Rule 710

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 5921

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcCoth[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 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 6110

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

Rule 6122

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x
_)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(-(C/d^2) + (C*x^2)/d^2)^q*(a + b*Ar
cCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 - c^2) + 2*A*c*
d, 0] && EqQ[2*c*C - B*d, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a+b x)^2}{x^3} \, dx &=-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+b \int \frac {\coth ^{-1}(a+b x)}{x^2 \left (1-(a+b x)^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+b \int \frac {\coth ^{-1}(a+b x)}{x^2 \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\operatorname {Subst}\left (\int \left (-\frac {b^2 \coth ^{-1}(x)}{\left (-1+a^2\right ) (a-x)^2}-\frac {2 a b^2 \coth ^{-1}(x)}{\left (-1+a^2\right )^2 (a-x)}-\frac {b^2 \coth ^{-1}(x)}{2 (-1+a)^2 (-1+x)}+\frac {b^2 \coth ^{-1}(x)}{2 (1+a)^2 (1+x)}\right ) \, dx,x,a+b x\right )\\ &=-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{-1+x} \, dx,x,a+b x\right )}{2 (1-a)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{1+x} \, dx,x,a+b x\right )}{2 (1+a)^2}-\frac {\left (2 a b^2\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{a-x} \, dx,x,a+b x\right )}{\left (1-a^2\right )^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{(a-x)^2} \, dx,x,a+b x\right )}{1-a^2}\\ &=-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{2 (1-a)^2}-\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{2 (1+a)^2}-\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}+\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,a+b x\right )}{2 (1-a)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,a+b x\right )}{2 (1+a)^2}+\frac {\left (2 a b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,a+b x\right )}{\left (1-a^2\right )^2}-\frac {\left (2 a b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 (a-x)}{(-1+a) (1+x)}\right )}{1-x^2} \, dx,x,a+b x\right )}{\left (1-a^2\right )^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{(a-x) \left (1-x^2\right )} \, dx,x,a+b x\right )}{1-a^2}\\ &=-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{2 (1-a)^2}-\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{2 (1+a)^2}-\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}+\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}-\frac {a b^2 \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a-b x}\right )}{2 (1-a)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+a+b x}\right )}{2 (1+a)^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a-x} \, dx,x,a+b x\right )}{\left (1-a^2\right )^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {-a-x}{1-x^2} \, dx,x,a+b x\right )}{\left (1-a^2\right )^2}+\frac {\left (2 a b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+a+b x}\right )}{\left (1-a^2\right )^2}\\ &=-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\frac {b^2 \log (x)}{\left (1-a^2\right )^2}+\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{2 (1-a)^2}-\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{2 (1+a)^2}-\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}+\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-a-b x}\right )}{4 (1-a)^2}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{4 (1+a)^2}+\frac {a b^2 \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}-\frac {a b^2 \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,a+b x\right )}{2 (1-a) (1+a)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,a+b x\right )}{2 (1-a)^2 (1+a)}\\ &=-\frac {b \coth ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\coth ^{-1}(a+b x)^2}{2 x^2}+\frac {b^2 \log (x)}{\left (1-a^2\right )^2}+\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{2 (1-a)^2}-\frac {b^2 \log (1-a-b x)}{2 (1-a)^2 (1+a)}-\frac {b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{2 (1+a)^2}-\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}+\frac {2 a b^2 \coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}-\frac {b^2 \log (1+a+b x)}{2 (1-a) (1+a)^2}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-a-b x}\right )}{4 (1-a)^2}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{4 (1+a)^2}+\frac {a b^2 \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{\left (1-a^2\right )^2}-\frac {a b^2 \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{\left (1-a^2\right )^2}\\ \end {align*}

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Mathematica [C]  time = 2.35, size = 291, normalized size = 0.79 \[ \frac {2 b x \coth ^{-1}(a+b x) \left (a^2+a b x+i \pi a b x-2 a b x \tanh ^{-1}\left (\frac {1}{a}\right )+2 a b x \log \left (1-e^{2 \tanh ^{-1}\left (\frac {1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )-1\right )+\left (-a^4+a^2 \left (b^2 x^2 \left (2 \sqrt {1-\frac {1}{a^2}} e^{\tanh ^{-1}\left (\frac {1}{a}\right )}-1\right )+2\right )+b^2 x^2-1\right ) \coth ^{-1}(a+b x)^2-2 a b^2 x^2 \text {Li}_2\left (e^{2 \tanh ^{-1}\left (\frac {1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )+2 b^2 x^2 \left (i \pi a \log \left (\frac {1}{\sqrt {1-\frac {1}{(a+b x)^2}}}\right )+\log \left (-\frac {b x}{(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}\right )-i \pi a \log \left (e^{2 \coth ^{-1}(a+b x)}+1\right )-2 a \tanh ^{-1}\left (\frac {1}{a}\right ) \left (\log \left (1-e^{2 \tanh ^{-1}\left (\frac {1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )-\log \left (i \sinh \left (\coth ^{-1}(a+b x)-\tanh ^{-1}\left (\frac {1}{a}\right )\right )\right )\right )\right )}{2 \left (a^2-1\right )^2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a + b*x]^2/x^3,x]

[Out]

((-1 - a^4 + b^2*x^2 + a^2*(2 + b^2*(-1 + 2*Sqrt[1 - a^(-2)]*E^ArcTanh[a^(-1)])*x^2))*ArcCoth[a + b*x]^2 + 2*b
*x*ArcCoth[a + b*x]*(-1 + a^2 + a*b*x + I*a*b*Pi*x - 2*a*b*x*ArcTanh[a^(-1)] + 2*a*b*x*Log[1 - E^(-2*ArcCoth[a
+ b*x] + 2*ArcTanh[a^(-1)])]) + 2*b^2*x^2*((-I)*a*Pi*Log[1 + E^(2*ArcCoth[a + b*x])] + I*a*Pi*Log[1/Sqrt[1 -
(a + b*x)^(-2)]] + Log[-((b*x)/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]))] - 2*a*ArcTanh[a^(-1)]*(Log[1 - E^(-2*Arc
Coth[a + b*x] + 2*ArcTanh[a^(-1)])] - Log[I*Sinh[ArcCoth[a + b*x] - ArcTanh[a^(-1)]]])) - 2*a*b^2*x^2*PolyLog[
2, E^(-2*ArcCoth[a + b*x] + 2*ArcTanh[a^(-1)])])/(2*(-1 + a^2)^2*x^2)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (b x + a\right )^{2}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^3,x, algorithm="fricas")

[Out]

integral(arccoth(b*x + a)^2/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (b x + a\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^3,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)^2/x^3, x)

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maple [A]  time = 0.08, size = 467, normalized size = 1.26 \[ -\frac {\mathrm {arccoth}\left (b x +a \right )^{2}}{2 x^{2}}-\frac {b^{2} \mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a -1\right )}{2 \left (a -1\right )^{2}}+\frac {b^{2} \mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a +1\right )}{2 \left (1+a \right )^{2}}+\frac {b \,\mathrm {arccoth}\left (b x +a \right )}{\left (a -1\right ) \left (1+a \right ) x}+\frac {2 b^{2} \mathrm {arccoth}\left (b x +a \right ) a \ln \left (b x \right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}}-\frac {b^{2} \ln \left (b x +a -1\right )^{2}}{8 \left (a -1\right )^{2}}+\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 \left (a -1\right )^{2}}+\frac {b^{2} \ln \left (b x +a -1\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 \left (a -1\right )^{2}}-\frac {b^{2} \ln \left (b x +a +1\right )^{2}}{8 \left (1+a \right )^{2}}-\frac {b^{2} \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 \left (1+a \right )^{2}}+\frac {b^{2} \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (b x +a +1\right )}{4 \left (1+a \right )^{2}}-\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 \left (1+a \right )^{2}}-\frac {b^{2} \ln \left (b x +a -1\right )}{\left (a -1\right ) \left (1+a \right ) \left (2 a -2\right )}+\frac {b^{2} \ln \left (b x +a +1\right )}{\left (a -1\right ) \left (1+a \right ) \left (2+2 a \right )}+\frac {b^{2} \ln \left (b x \right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}}-\frac {b^{2} a \dilog \left (\frac {b x +a +1}{1+a}\right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}}-\frac {b^{2} a \ln \left (b x \right ) \ln \left (\frac {b x +a +1}{1+a}\right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}}+\frac {b^{2} a \dilog \left (\frac {b x +a -1}{a -1}\right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}}+\frac {b^{2} a \ln \left (b x \right ) \ln \left (\frac {b x +a -1}{a -1}\right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)^2/x^3,x)

[Out]

-1/2*arccoth(b*x+a)^2/x^2-1/2*b^2*arccoth(b*x+a)/(a-1)^2*ln(b*x+a-1)+1/2*b^2*arccoth(b*x+a)/(1+a)^2*ln(b*x+a+1
)+b*arccoth(b*x+a)/(a-1)/(1+a)/x+2*b^2*arccoth(b*x+a)*a/(a-1)^2/(1+a)^2*ln(b*x)-1/8*b^2/(a-1)^2*ln(b*x+a-1)^2+
1/4*b^2/(a-1)^2*dilog(1/2+1/2*b*x+1/2*a)+1/4*b^2/(a-1)^2*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)-1/8*b^2/(1+a)^2*ln(
b*x+a+1)^2-1/4*b^2/(1+a)^2*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)+1/4*b^2/(1+a)^2*ln(-1/2*b*x-1/2*a+1/2)
*ln(b*x+a+1)-1/4*b^2/(1+a)^2*dilog(1/2+1/2*b*x+1/2*a)-b^2/(a-1)/(1+a)/(2*a-2)*ln(b*x+a-1)+b^2/(a-1)/(1+a)/(2+2
*a)*ln(b*x+a+1)+b^2/(a-1)^2/(1+a)^2*ln(b*x)-b^2*a/(a-1)^2/(1+a)^2*dilog((b*x+a+1)/(1+a))-b^2*a/(a-1)^2/(1+a)^2
*ln(b*x)*ln((b*x+a+1)/(1+a))+b^2*a/(a-1)^2/(1+a)^2*dilog((b*x+a-1)/(a-1))+b^2*a/(a-1)^2/(1+a)^2*ln(b*x)*ln((b*
x+a-1)/(a-1))

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maxima [A]  time = 0.34, size = 360, normalized size = 0.97 \[ \frac {1}{8} \, {\left (\frac {8 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )} a}{a^{4} - 2 \, a^{2} + 1} - \frac {8 \, {\left (\log \left (\frac {b x}{a + 1} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a + 1}\right )\right )} a}{a^{4} - 2 \, a^{2} + 1} + \frac {8 \, {\left (\log \left (\frac {b x}{a - 1} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a - 1}\right )\right )} a}{a^{4} - 2 \, a^{2} + 1} - \frac {{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a - 1\right )^{2}}{a^{4} - 2 \, a^{2} + 1} + \frac {4 \, \log \left (b x + a + 1\right )}{a^{3} + a^{2} - a - 1} - \frac {4 \, \log \left (b x + a - 1\right )}{a^{3} - a^{2} - a + 1} + \frac {8 \, \log \relax (x)}{a^{4} - 2 \, a^{2} + 1}\right )} b^{2} + \frac {1}{2} \, {\left (\frac {4 \, a b \log \relax (x)}{a^{4} - 2 \, a^{2} + 1} + \frac {b \log \left (b x + a + 1\right )}{a^{2} + 2 \, a + 1} - \frac {b \log \left (b x + a - 1\right )}{a^{2} - 2 \, a + 1} + \frac {2}{{\left (a^{2} - 1\right )} x}\right )} b \operatorname {arcoth}\left (b x + a\right ) - \frac {\operatorname {arcoth}\left (b x + a\right )^{2}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^3,x, algorithm="maxima")

[Out]

1/8*(8*(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))*a/(a^4 - 2*a^2 + 1) - 8*(
log(b*x/(a + 1) + 1)*log(x) + dilog(-b*x/(a + 1)))*a/(a^4 - 2*a^2 + 1) + 8*(log(b*x/(a - 1) + 1)*log(x) + dilo
g(-b*x/(a - 1)))*a/(a^4 - 2*a^2 + 1) - ((a^2 - 2*a + 1)*log(b*x + a + 1)^2 - 2*(a^2 - 2*a + 1)*log(b*x + a + 1
)*log(b*x + a - 1) + (a^2 + 2*a + 1)*log(b*x + a - 1)^2)/(a^4 - 2*a^2 + 1) + 4*log(b*x + a + 1)/(a^3 + a^2 - a
- 1) - 4*log(b*x + a - 1)/(a^3 - a^2 - a + 1) + 8*log(x)/(a^4 - 2*a^2 + 1))*b^2 + 1/2*(4*a*b*log(x)/(a^4 - 2*
a^2 + 1) + b*log(b*x + a + 1)/(a^2 + 2*a + 1) - b*log(b*x + a - 1)/(a^2 - 2*a + 1) + 2/((a^2 - 1)*x))*b*arccot
h(b*x + a) - 1/2*arccoth(b*x + a)^2/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {acoth}\left (a+b\,x\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)^2/x^3,x)

[Out]

int(acoth(a + b*x)^2/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}^{2}{\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(b*x+a)**2/x**3,x)

[Out]

Integral(acoth(a + b*x)**2/x**3, x)

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