3.101 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{x^3 (d+c d x)} \, dx\)

Optimal. Leaf size=250 \[ -\frac{b c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b^2 c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}+\frac{c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{2 b c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

[Out]

-((b*c*(a + b*ArcTanh[c*x]))/(d*x)) - (c^2*(a + b*ArcTanh[c*x])^2)/(2*d) - (a + b*ArcTanh[c*x])^2/(2*d*x^2) +
(c*(a + b*ArcTanh[c*x])^2)/(d*x) + (b^2*c^2*Log[x])/d - (b^2*c^2*Log[1 - c^2*x^2])/(2*d) - (2*b*c^2*(a + b*Arc
Tanh[c*x])*Log[2 - 2/(1 + c*x)])/d + (c^2*(a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x)])/d + (b^2*c^2*PolyLog[2,
-1 + 2/(1 + c*x)])/d - (b*c^2*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 + c*x)])/d - (b^2*c^2*PolyLog[3, -1 +
2/(1 + c*x)])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.634345, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.591, Rules used = {5934, 5916, 5982, 266, 36, 29, 31, 5948, 5988, 5932, 2447, 6056, 6610} \[ -\frac{b c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b^2 c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}+\frac{c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{2 b c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x^3*(d + c*d*x)),x]

[Out]

-((b*c*(a + b*ArcTanh[c*x]))/(d*x)) - (c^2*(a + b*ArcTanh[c*x])^2)/(2*d) - (a + b*ArcTanh[c*x])^2/(2*d*x^2) +
(c*(a + b*ArcTanh[c*x])^2)/(d*x) + (b^2*c^2*Log[x])/d - (b^2*c^2*Log[1 - c^2*x^2])/(2*d) - (2*b*c^2*(a + b*Arc
Tanh[c*x])*Log[2 - 2/(1 + c*x)])/d + (c^2*(a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x)])/d + (b^2*c^2*PolyLog[2,
-1 + 2/(1 + c*x)])/d - (b*c^2*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 + c*x)])/d - (b^2*c^2*PolyLog[3, -1 +
2/(1 + c*x)])/(2*d)

Rule 5934

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[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 5982

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[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 5988

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 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 6056

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3 (d+c d x)} \, dx &=-\left (c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2 (d+c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx}{d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+c^2 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+c d x)} \, dx-\frac{c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx}{d}+\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx}{d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}-\frac{\left (2 b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{d}-\frac{\left (2 b c^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{\left (2 b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d}+\frac{\left (b^2 c^2\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (b^2 c^3\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{2 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}+\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac{\left (2 b^2 c^3\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{2 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{b^2 c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}+\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 d}+\frac{\left (b^2 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{b^2 c^2 \log (x)}{d}-\frac{b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d}-\frac{2 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{b^2 c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}\\ \end{align*}

Mathematica [C]  time = 1.0157, size = 317, normalized size = 1.27 \[ \frac{\frac{2 a b \left (-c^2 x^2 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-c x \left (2 c x \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+1\right )+\tanh ^{-1}(c x) \left (c^2 x^2+2 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+2 c x-1\right )\right )}{x^2}+2 b^2 c^2 \left (\tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )+\log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )-\frac{\tanh ^{-1}(c x)^2}{2 c^2 x^2}-\frac{2}{3} \tanh ^{-1}(c x)^3+\frac{\tanh ^{-1}(c x)^2}{c x}-\frac{1}{2} \tanh ^{-1}(c x)^2-\frac{\tanh ^{-1}(c x)}{c x}+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+\frac{i \pi ^3}{24}\right )+2 a^2 c^2 \log (x)-2 a^2 c^2 \log (c x+1)+\frac{2 a^2 c}{x}-\frac{a^2}{x^2}}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x^3*(d + c*d*x)),x]

[Out]

(-(a^2/x^2) + (2*a^2*c)/x + 2*a^2*c^2*Log[x] - 2*a^2*c^2*Log[1 + c*x] + (2*a*b*(ArcTanh[c*x]*(-1 + 2*c*x + c^2
*x^2 + 2*c^2*x^2*Log[1 - E^(-2*ArcTanh[c*x])]) - c*x*(1 + 2*c*x*Log[(c*x)/Sqrt[1 - c^2*x^2]]) - c^2*x^2*PolyLo
g[2, E^(-2*ArcTanh[c*x])]))/x^2 + 2*b^2*c^2*((I/24)*Pi^3 - ArcTanh[c*x]/(c*x) - ArcTanh[c*x]^2/2 - ArcTanh[c*x
]^2/(2*c^2*x^2) + ArcTanh[c*x]^2/(c*x) - (2*ArcTanh[c*x]^3)/3 - 2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] +
ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + Log[(c*x)/Sqrt[1 - c^2*x^2]] + PolyLog[2, E^(-2*ArcTanh[c*x])] +
ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - PolyLog[3, E^(2*ArcTanh[c*x])]/2))/(2*d)

________________________________________________________________________________________

Maple [C]  time = 1.205, size = 1841, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x^3/(c*d*x+d),x)

[Out]

I*c^2*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2+1/2*I*c^2*b^2
/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2-1/2*I*c^2*b^2/d*Pi*csg
n(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-1/2*I*c^2
*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*a
rctanh(c*x)^2-1/2*I*c^2*b^2/d*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)
^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+1/2*I*c^2*b^2/d*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c
^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+c^2*a*b/d*dilog(1/2+1/2*c*x)+1/2*c^2*a*b/d*ln(c*x+1)^2+
1/2*c^2*a*b/d*ln(c*x-1)+3/2*c^2*a*b/d*ln(c*x+1)+c^2*b^2/d*arctanh(c*x)^2*ln(c*x)-c*b^2/d*arctanh(c*x)/x+c*b^2/
d*arctanh(c*x)^2/x-a*b/d*arctanh(c*x)/x^2-2*c^2*b^2/d*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+c^2*b^2/d*
arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2)
)+c^2*b^2/d*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*b^2/d*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x
^2+1)^(1/2))+1/2*I*c^2*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2
+1/2*I*c^2*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2+1/2*I*c^2*b^2/d*
Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-1/2*a^2/d/x^2+1/2*I*c^2*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2
+1)-1))*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arcta
nh(c*x)^2-1/2*I*c^2*b^2/d*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2
/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2+2*c*a*b/d*arctanh(c*x)/x-c^2*a*b/d*ln(c*x)*ln(c*x+1)+2
*c^2*a*b/d*arctanh(c*x)*ln(c*x)-2*c^2*a*b/d*arctanh(c*x)*ln(c*x+1)-c^2*a*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)+c^2*a*
b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-c^2*b^2/d*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+c*a^2/d/x-1/2*b^2/d
*arctanh(c*x)^2/x^2+c^2*b^2/d*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)+c^2*b^2/d*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^
2*b^2/d*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*b^2/d*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^2*b^2/d*polylog(
3,(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^2*b^2/d*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))-c^2*a^2/d*ln(c*x+1)-2/3*c^2*b
^2/d*arctanh(c*x)^3-c^2*b^2/d*arctanh(c*x)+3/2*c^2*b^2/d*arctanh(c*x)^2+c^2*a^2/d*ln(c*x)+2*c^2*b^2/d*arctanh(
c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-c^2*b^2/d*arctanh(c*x)^2*ln(c*x+1)-c^2*a*b/d*dilog(c*x)-c^2*a*b/d*dilog(
c*x+1)-2*c^2*a*b/d*ln(c*x)-c*a*b/d/x+c^2*b^2/d*arctanh(c*x)^2*ln(2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left (\frac{2 \, c^{2} \log \left (c x + 1\right )}{d} - \frac{2 \, c^{2} \log \left (x\right )}{d} - \frac{2 \, c x - 1}{d x^{2}}\right )} a^{2} - \frac{{\left (2 \, b^{2} c^{2} x^{2} \log \left (c x + 1\right ) - 2 \, b^{2} c x + b^{2}\right )} \log \left (-c x + 1\right )^{2}}{8 \, d x^{2}} + \int \frac{{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \,{\left (a b c x - a b\right )} \log \left (c x + 1\right ) -{\left (2 \, b^{2} c^{3} x^{3} + b^{2} c^{2} x^{2} - 4 \, a b +{\left (4 \, a b c - b^{2} c\right )} x - 2 \,{\left (b^{2} c^{4} x^{4} + b^{2} c^{3} x^{3} - b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c^{2} d x^{5} - d x^{3}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*c^2*log(c*x + 1)/d - 2*c^2*log(x)/d - (2*c*x - 1)/(d*x^2))*a^2 - 1/8*(2*b^2*c^2*x^2*log(c*x + 1) - 2*b
^2*c*x + b^2)*log(-c*x + 1)^2/(d*x^2) + integrate(1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(
c*x + 1) - (2*b^2*c^3*x^3 + b^2*c^2*x^2 - 4*a*b + (4*a*b*c - b^2*c)*x - 2*(b^2*c^4*x^4 + b^2*c^3*x^3 - b^2*c*x
+ b^2)*log(c*x + 1))*log(-c*x + 1))/(c^2*d*x^5 - d*x^3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x\right ) + a^{2}}{c d x^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(c*d*x^4 + d*x^3), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c x^{4} + x^{3}}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c x^{4} + x^{3}}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{c x^{4} + x^{3}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x**3/(c*d*x+d),x)

[Out]

(Integral(a**2/(c*x**4 + x**3), x) + Integral(b**2*atanh(c*x)**2/(c*x**4 + x**3), x) + Integral(2*a*b*atanh(c*
x)/(c*x**4 + x**3), x))/d

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(c*x) + a)^2/((c*d*x + d)*x^3), x)