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

Optimal. Leaf size=204 \[ -\frac{\left (3 a^2+1\right ) \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{3 b^3}+\frac{a \left (a^2+3\right ) \coth ^{-1}(a+b x)^2}{3 b^3}+\frac{\left (3 a^2+1\right ) \coth ^{-1}(a+b x)^2}{3 b^3}-\frac{2 \left (3 a^2+1\right ) \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{3 b^3}-\frac{a \log \left (1-(a+b x)^2\right )}{b^3}-\frac{\tanh ^{-1}(a+b x)}{3 b^3}+\frac{(a+b x)^2 \coth ^{-1}(a+b x)}{3 b^3}-\frac{2 a (a+b x) \coth ^{-1}(a+b x)}{b^3}+\frac{1}{3} x^3 \coth ^{-1}(a+b x)^2+\frac{x}{3 b^2} \]

[Out]

x/(3*b^2) - (2*a*(a + b*x)*ArcCoth[a + b*x])/b^3 + ((a + b*x)^2*ArcCoth[a + b*x])/(3*b^3) + (a*(3 + a^2)*ArcCo
th[a + b*x]^2)/(3*b^3) + ((1 + 3*a^2)*ArcCoth[a + b*x]^2)/(3*b^3) + (x^3*ArcCoth[a + b*x]^2)/3 - ArcTanh[a + b
*x]/(3*b^3) - (2*(1 + 3*a^2)*ArcCoth[a + b*x]*Log[2/(1 - a - b*x)])/(3*b^3) - (a*Log[1 - (a + b*x)^2])/b^3 - (
(1 + 3*a^2)*PolyLog[2, -((1 + a + b*x)/(1 - a - b*x))])/(3*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.278272, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.083, Rules used = {6112, 5929, 5911, 260, 5917, 321, 206, 6049, 5949, 5985, 5919, 2402, 2315} \[ -\frac{\left (3 a^2+1\right ) \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{3 b^3}+\frac{a \left (a^2+3\right ) \coth ^{-1}(a+b x)^2}{3 b^3}+\frac{\left (3 a^2+1\right ) \coth ^{-1}(a+b x)^2}{3 b^3}-\frac{2 \left (3 a^2+1\right ) \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{3 b^3}-\frac{a \log \left (1-(a+b x)^2\right )}{b^3}-\frac{\tanh ^{-1}(a+b x)}{3 b^3}+\frac{(a+b x)^2 \coth ^{-1}(a+b x)}{3 b^3}-\frac{2 a (a+b x) \coth ^{-1}(a+b x)}{b^3}+\frac{1}{3} x^3 \coth ^{-1}(a+b x)^2+\frac{x}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[a + b*x]^2,x]

[Out]

x/(3*b^2) - (2*a*(a + b*x)*ArcCoth[a + b*x])/b^3 + ((a + b*x)^2*ArcCoth[a + b*x])/(3*b^3) + (a*(3 + a^2)*ArcCo
th[a + b*x]^2)/(3*b^3) + ((1 + 3*a^2)*ArcCoth[a + b*x]^2)/(3*b^3) + (x^3*ArcCoth[a + b*x]^2)/3 - ArcTanh[a + b
*x]/(3*b^3) - (2*(1 + 3*a^2)*ArcCoth[a + b*x]*Log[2/(1 - a - b*x)])/(3*b^3) - (a*Log[1 - (a + b*x)^2])/b^3 - (
(1 + 3*a^2)*PolyLog[2, -((1 + a + b*x)/(1 - a - b*x))])/(3*b^3)

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 5929

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

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 260

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

Rule 5917

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6049

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcCoth[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

Rule 5949

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

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 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 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 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]

Rubi steps

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

Mathematica [B]  time = 4.52669, size = 607, normalized size = 2.98 \[ -\frac{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \left (1-(a+b x)^2\right ) \left (\frac{4 \left (3 a^2+1\right ) \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(a+b x)}\right )}{(a+b x)^3 \left (1-\frac{1}{(a+b x)^2}\right )^{3/2}}+\frac{9 a^2 \coth ^{-1}(a+b x)^2}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}+\frac{-3 \left (a^2-1\right ) \coth ^{-1}(a+b x)^2+6 a \coth ^{-1}(a+b x)-1}{\sqrt{1-\frac{1}{(a+b x)^2}}}+3 a^2 \coth ^{-1}(a+b x)^2 \cosh \left (3 \coth ^{-1}(a+b x)\right )+\frac{18 a^2 \coth ^{-1}(a+b x) \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}-3 a^2 \coth ^{-1}(a+b x)^2 \sinh \left (3 \coth ^{-1}(a+b x)\right )-6 a^2 \coth ^{-1}(a+b x) \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right ) \sinh \left (3 \coth ^{-1}(a+b x)\right )-\frac{18 a \log \left (\frac{1}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}\right )}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}-\frac{12 a \coth ^{-1}(a+b x)^2}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}+\frac{3 \coth ^{-1}(a+b x)^2}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}+\frac{4 \coth ^{-1}(a+b x)}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}-6 a \coth ^{-1}(a+b x) \cosh \left (3 \coth ^{-1}(a+b x)\right )+\coth ^{-1}(a+b x)^2 \cosh \left (3 \coth ^{-1}(a+b x)\right )+\cosh \left (3 \coth ^{-1}(a+b x)\right )+\frac{6 \coth ^{-1}(a+b x) \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}-\coth ^{-1}(a+b x)^2 \sinh \left (3 \coth ^{-1}(a+b x)\right )+6 a \log \left (\frac{1}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}\right ) \sinh \left (3 \coth ^{-1}(a+b x)\right )-2 \coth ^{-1}(a+b x) \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right ) \sinh \left (3 \coth ^{-1}(a+b x)\right )\right )}{12 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*ArcCoth[a + b*x]^2,x]

[Out]

-((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*(1 - (a + b*x)^2)*((4*ArcCoth[a + b*x])/((a + b*x)*Sqrt[1 - (a + b*x)^(-2
)]) + (3*ArcCoth[a + b*x]^2)/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]) - (12*a*ArcCoth[a + b*x]^2)/((a + b*x)*Sqrt[
1 - (a + b*x)^(-2)]) + (9*a^2*ArcCoth[a + b*x]^2)/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]) + (-1 + 6*a*ArcCoth[a +
b*x] - 3*(-1 + a^2)*ArcCoth[a + b*x]^2)/Sqrt[1 - (a + b*x)^(-2)] + Cosh[3*ArcCoth[a + b*x]] - 6*a*ArcCoth[a +
b*x]*Cosh[3*ArcCoth[a + b*x]] + ArcCoth[a + b*x]^2*Cosh[3*ArcCoth[a + b*x]] + 3*a^2*ArcCoth[a + b*x]^2*Cosh[3
*ArcCoth[a + b*x]] + (6*ArcCoth[a + b*x]*Log[1 - E^(-2*ArcCoth[a + b*x])])/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]
) + (18*a^2*ArcCoth[a + b*x]*Log[1 - E^(-2*ArcCoth[a + b*x])])/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]) - (18*a*Lo
g[1/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)])])/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]) + (4*(1 + 3*a^2)*PolyLog[2, E^
(-2*ArcCoth[a + b*x])])/((a + b*x)^3*(1 - (a + b*x)^(-2))^(3/2)) - ArcCoth[a + b*x]^2*Sinh[3*ArcCoth[a + b*x]]
- 3*a^2*ArcCoth[a + b*x]^2*Sinh[3*ArcCoth[a + b*x]] - 2*ArcCoth[a + b*x]*Log[1 - E^(-2*ArcCoth[a + b*x])]*Sin
h[3*ArcCoth[a + b*x]] - 6*a^2*ArcCoth[a + b*x]*Log[1 - E^(-2*ArcCoth[a + b*x])]*Sinh[3*ArcCoth[a + b*x]] + 6*a
*Log[1/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)])]*Sinh[3*ArcCoth[a + b*x]]))/(12*b^3)

________________________________________________________________________________________

Maple [B]  time = 0.055, size = 729, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^3*ln(b*x+a-1)*a-1/b^3*ln(b*x+a+1)*a+1/3*x^3*arccoth(b*x+a)^2+1/6/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)+1
/3/b^3*arccoth(b*x+a)*ln(b*x+a+1)-1/6/b^3*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)-1/b^3*dilog(1/2+1/2*b*x+1/2*a)*a^2
-1/12/b^3*ln(b*x+a+1)^2*a^3-1/4/b^3*ln(b*x+a+1)^2*a^2-1/4/b^3*ln(b*x+a+1)^2*a-1/12/b^3*ln(b*x+a-1)^2*a^3+1/4/b
^3*ln(b*x+a-1)^2*a^2-1/4/b^3*ln(b*x+a-1)^2*a-5/3/b^3*arccoth(b*x+a)*a^2+1/3/b*arccoth(b*x+a)*x^2+1/3/b^3*arcco
th(b*x+a)*ln(b*x+a-1)-1/6/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)-1/2/b^3*ln(1/2+1/2*b*x+1/2*a)*ln(b*
x+a-1)*a^2+1/6/b^3*ln(1/2+1/2*b*x+1/2*a)*ln(b*x+a-1)*a^3+1/3/b^3*arccoth(b*x+a)*ln(b*x+a+1)*a^3-1/6/b^3*ln(-1/
2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a^3-1/2/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a^2-1/b^3*arcc
oth(b*x+a)*ln(b*x+a-1)*a-4/3/b^2*arccoth(b*x+a)*x*a-1/2/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a+1/6
/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)*a^3+1/2/b^3*ln(1/2+1/2*b*x+1/2*a)*ln(b*x+a-1)*a+1/2/b^3*ln(-1/2*b*x-1/
2*a+1/2)*ln(b*x+a+1)*a^2+1/2/b^3*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)*a+1/b^3*arccoth(b*x+a)*ln(b*x+a+1)*a^2+1/b
^3*arccoth(b*x+a)*ln(b*x+a+1)*a-1/3/b^3*arccoth(b*x+a)*ln(b*x+a-1)*a^3+1/b^3*arccoth(b*x+a)*ln(b*x+a-1)*a^2+1/
6/b^3*ln(b*x+a-1)-1/6/b^3*ln(b*x+a+1)-1/12/b^3*ln(b*x+a+1)^2-1/3/b^3*dilog(1/2+1/2*b*x+1/2*a)+1/3/b^3*a+1/12/b
^3*ln(b*x+a-1)^2+1/3*x/b^2

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Maxima [A]  time = 1.00212, size = 350, normalized size = 1.72 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (b x + a\right )^{2} - \frac{1}{12} \, b^{2}{\left (\frac{4 \,{\left (3 \, a^{2} + 1\right )}{\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^{5}} + \frac{2 \,{\left (5 \, a^{2} + 6 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac{{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \,{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) +{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )^{2} - 4 \, b x - 2 \,{\left (5 \, a^{2} - 6 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} + \frac{1}{3} \, b{\left (\frac{b x^{2} - 4 \, a x}{b^{3}} + \frac{{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} - \frac{{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} \operatorname{arcoth}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(b*x + a)^2 - 1/12*b^2*(4*(3*a^2 + 1)*(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^5 + 2*(5*a^2 + 6*a + 1)*log(b*x + a + 1)/b^5 + ((a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1
)^2 - 2*(a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1)*log(b*x + a - 1) + (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1)^2
- 4*b*x - 2*(5*a^2 - 6*a + 1)*log(b*x + a - 1))/b^5) + 1/3*b*((b*x^2 - 4*a*x)/b^3 + (a^3 + 3*a^2 + 3*a + 1)*l
og(b*x + a + 1)/b^4 - (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1)/b^4)*arccoth(b*x + a)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{arcoth}\left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arccoth(b*x + a)^2, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccoth(b*x + a)^2, x)