3.102 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{x^4 (d+c d x)} \, dx\)

Optimal. Leaf size=334 \[ \frac{b c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{4 b^2 c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{3 d}+\frac{b^2 c^3 \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}+\frac{8 b c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 d}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d}-\frac{b^2 c^2}{3 d x}-\frac{b^2 c^3 \log (x)}{d}+\frac{b^2 c^3 \tanh ^{-1}(c x)}{3 d} \]

[Out]

-(b^2*c^2)/(3*d*x) + (b^2*c^3*ArcTanh[c*x])/(3*d) - (b*c*(a + b*ArcTanh[c*x]))/(3*d*x^2) + (b*c^2*(a + b*ArcTa
nh[c*x]))/(d*x) + (5*c^3*(a + b*ArcTanh[c*x])^2)/(6*d) - (a + b*ArcTanh[c*x])^2/(3*d*x^3) + (c*(a + b*ArcTanh[
c*x])^2)/(2*d*x^2) - (c^2*(a + b*ArcTanh[c*x])^2)/(d*x) - (b^2*c^3*Log[x])/d + (b^2*c^3*Log[1 - c^2*x^2])/(2*d
) + (8*b*c^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/(3*d) - (c^3*(a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x
)])/d - (4*b^2*c^3*PolyLog[2, -1 + 2/(1 + c*x)])/(3*d) + (b*c^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 + c*
x)])/d + (b^2*c^3*PolyLog[3, -1 + 2/(1 + c*x)])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.97563, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 15, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.682, Rules used = {5934, 5916, 5982, 325, 206, 5988, 5932, 2447, 266, 36, 29, 31, 5948, 6056, 6610} \[ \frac{b c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{4 b^2 c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{3 d}+\frac{b^2 c^3 \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac{c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}+\frac{8 b c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 d}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d}-\frac{b^2 c^2}{3 d x}-\frac{b^2 c^3 \log (x)}{d}+\frac{b^2 c^3 \tanh ^{-1}(c x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x^4*(d + c*d*x)),x]

[Out]

-(b^2*c^2)/(3*d*x) + (b^2*c^3*ArcTanh[c*x])/(3*d) - (b*c*(a + b*ArcTanh[c*x]))/(3*d*x^2) + (b*c^2*(a + b*ArcTa
nh[c*x]))/(d*x) + (5*c^3*(a + b*ArcTanh[c*x])^2)/(6*d) - (a + b*ArcTanh[c*x])^2/(3*d*x^3) + (c*(a + b*ArcTanh[
c*x])^2)/(2*d*x^2) - (c^2*(a + b*ArcTanh[c*x])^2)/(d*x) - (b^2*c^3*Log[x])/d + (b^2*c^3*Log[1 - c^2*x^2])/(2*d
) + (8*b*c^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/(3*d) - (c^3*(a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x
)])/d - (4*b^2*c^3*PolyLog[2, -1 + 2/(1 + c*x)])/(3*d) + (b*c^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 + c*
x)])/d + (b^2*c^3*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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 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 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^4 (d+c d x)} \, dx &=-\left (c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3 (d+c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx}{d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+c^2 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2 (d+c d x)} \, dx-\frac{c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx}{d}+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-c^3 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+c d x)} \, dx+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx}{3 d}+\frac{c^2 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx}{d}-\frac{\left (b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (2 b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{\left (b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}+\frac{\left (b^2 c^2\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx}{3 d}+\frac{\left (2 b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{3 d}+\frac{\left (2 b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d}-\frac{\left (b c^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{d}+\frac{\left (2 b c^4\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^2 c^2}{3 d x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{3 d}-\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{b c^3 \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^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d}-\frac{\left (b^2 c^3\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (b^2 c^4\right ) \int \frac{1}{1-c^2 x^2} \, dx}{3 d}-\frac{\left (2 b^2 c^4\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{3 d}-\frac{\left (b^2 c^4\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b^2 c^2}{3 d x}+\frac{b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{3 d}-\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b^2 c^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{3 d}+\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b^2 c^3 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}-\frac{\left (b^2 c^3\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^4\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac{b^2 c^2}{3 d x}+\frac{b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{3 d}-\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{4 b^2 c^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{3 d}+\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b^2 c^3 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}-\frac{\left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 d}-\frac{\left (b^2 c^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b^2 c^2}{3 d x}+\frac{b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac{b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac{5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{b^2 c^3 \log (x)}{d}+\frac{b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d}+\frac{8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{3 d}-\frac{c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{4 b^2 c^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{3 d}+\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b^2 c^3 \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}\\ \end{align*}

Mathematica [C]  time = 1.42146, size = 388, normalized size = 1.16 \[ \frac{-\frac{8 a b \left (-3 c^3 x^3 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-c x \left (c^2 x^2+8 c^2 x^2 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+3 c x-1\right )+\tanh ^{-1}(c x) \left (3 c^3 x^3+6 c^2 x^2+6 c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-3 c x+2\right )\right )}{x^3}+b^2 c^3 \left (-24 \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-32 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+12 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-24 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+\frac{12 \tanh ^{-1}(c x)^2}{c^2 x^2}-\frac{8 \tanh ^{-1}(c x)^2}{c^3 x^3}-\frac{8 \tanh ^{-1}(c x)}{c^2 x^2}-\frac{8}{c x}+16 \tanh ^{-1}(c x)^3-\frac{24 \tanh ^{-1}(c x)^2}{c x}+20 \tanh ^{-1}(c x)^2+\frac{24 \tanh ^{-1}(c x)}{c x}+8 \tanh ^{-1}(c x)-24 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+64 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-i \pi ^3\right )-\frac{24 a^2 c^2}{x}-24 a^2 c^3 \log (x)+24 a^2 c^3 \log (c x+1)+\frac{12 a^2 c}{x^2}-\frac{8 a^2}{x^3}}{24 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x^4*(d + c*d*x)),x]

[Out]

((-8*a^2)/x^3 + (12*a^2*c)/x^2 - (24*a^2*c^2)/x - 24*a^2*c^3*Log[x] + 24*a^2*c^3*Log[1 + c*x] - (8*a*b*(ArcTan
h[c*x]*(2 - 3*c*x + 6*c^2*x^2 + 3*c^3*x^3 + 6*c^3*x^3*Log[1 - E^(-2*ArcTanh[c*x])]) - c*x*(-1 + 3*c*x + c^2*x^
2 + 8*c^2*x^2*Log[(c*x)/Sqrt[1 - c^2*x^2]]) - 3*c^3*x^3*PolyLog[2, E^(-2*ArcTanh[c*x])]))/x^3 + b^2*c^3*((-I)*
Pi^3 - 8/(c*x) + 8*ArcTanh[c*x] - (8*ArcTanh[c*x])/(c^2*x^2) + (24*ArcTanh[c*x])/(c*x) + 20*ArcTanh[c*x]^2 - (
8*ArcTanh[c*x]^2)/(c^3*x^3) + (12*ArcTanh[c*x]^2)/(c^2*x^2) - (24*ArcTanh[c*x]^2)/(c*x) + 16*ArcTanh[c*x]^3 +
64*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - 24*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - 24*Log[(c*x)/Sq
rt[1 - c^2*x^2]] - 32*PolyLog[2, E^(-2*ArcTanh[c*x])] - 24*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + 12*Po
lyLog[3, E^(2*ArcTanh[c*x])]))/(24*d)

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Maple [C]  time = 1.181, size = 2010, normalized size = 6. \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^4/(c*d*x+d),x)

[Out]

-1/2*I*c^3*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^
3*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*arcta
nh(c*x)^2+1/2*I*c^3*b^2/d*arctanh(c*x)^2*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+1/2*I*c^3*b^2/d*arctanh(c*x)^2*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+1/2*I*c^3*b^2/d*Pi*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))^2*arctanh(c*x)^2-I*c^3*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+4/3*b^2*c^3*arctanh(c*x)/d-1/3*a^2/d/x^3-2*c^2*a*b/d*a
rctanh(c*x)/x+c*a*b/d*arctanh(c*x)/x^2+c^3*a*b/d*ln(c*x)*ln(c*x+1)-2*c^3*a*b/d*arctanh(c*x)*ln(c*x)+2*c^3*a*b/
d*arctanh(c*x)*ln(c*x+1)+c^3*a*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)-c^3*a*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/2*c
*a^2/d/x^2-c^2*a^2/d/x-1/3*b^2/d*arctanh(c*x)^2/x^3+8/3*c^3*b^2/d*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-8/3*c^3*
b^2/d*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^3*b^2/d*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^3*b^2/d*polylog(
3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+c^3*a^2/d*ln(c*x+1)+2/3*c^3*b^2/d*arctanh(c*x)^3-11/6*c^3*b^2/d*arctanh(c*x)^2-
c^3*a^2/d*ln(c*x)-c^3*b^2/d*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)-c^3*b^2/d*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-1/2*I*
c^3*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^3*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^3*b^2/d*Pi*csgn(I*(c*x+1)^2
/(c^2*x^2-1))^3*arctanh(c*x)^2+c^3*a*b/d*dilog(c*x)+c^3*a*b/d*dilog(c*x+1)+8/3*c^3*a*b/d*ln(c*x)-c^3*b^2/d*arc
tanh(c*x)^2*ln(2)-c^3*a*b/d*dilog(1/2+1/2*c*x)-1/2*c^3*a*b/d*ln(c*x+1)^2-5/6*c^3*a*b/d*ln(c*x-1)-11/6*c^3*a*b/
d*ln(c*x+1)-c^3*b^2/d*arctanh(c*x)^2*ln(c*x)+8/3*c^3*b^2/d*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-c^3*b
^2/d*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^3*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^
(1/2))-c^3*b^2/d*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^3*b^2/d*arctanh(c*x)*polylog(2,(c*x+1)/(-
c^2*x^2+1)^(1/2))+c^3*b^2/d*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)-2*c^3*b^2/d*arctanh(c*x)^2*ln((c*x+1)/
(-c^2*x^2+1)^(1/2))+c^3*b^2/d*arctanh(c*x)^2*ln(c*x+1)+1/3*c^3*b^2/d/((-c^2*x^2+1)^(1/2)+c*x+1)*(-c^2*x^2+1)^(
1/2)-1/3*c^3*b^2/d/(c*x+1-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)-1/3*c*a*b/d/x^2+c^2*a*b/d/x-2/3*a*b/d*arctanh
(c*x)/x^3-c^2*b^2/d*arctanh(c*x)^2/x+1/2*c*b^2/d*arctanh(c*x)^2/x^2+c^2*b^2/d*arctanh(c*x)/x-1/3*c*b^2/d*arcta
nh(c*x)/x^2+1/2*I*c^3*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)^2-1/2*I*c^3*b^2/d*arctanh(c*x)^2*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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \,{\left (\frac{6 \, c^{3} \log \left (c x + 1\right )}{d} - \frac{6 \, c^{3} \log \left (x\right )}{d} - \frac{6 \, c^{2} x^{2} - 3 \, c x + 2}{d x^{3}}\right )} a^{2} + \frac{{\left (6 \, b^{2} c^{3} x^{3} \log \left (c x + 1\right ) - 6 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x - 2 \, b^{2}\right )} \log \left (-c x + 1\right )^{2}}{24 \, d x^{3}} - \int -\frac{3 \,{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 12 \,{\left (a b c x - a b\right )} \log \left (c x + 1\right ) +{\left (6 \, b^{2} c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} - b^{2} c^{2} x^{2} + 12 \, a b - 2 \,{\left (6 \, a b c - b^{2} c\right )} x - 6 \,{\left (b^{2} c^{5} x^{5} + b^{2} c^{4} x^{4} + b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{12 \,{\left (c^{2} d x^{6} - d x^{4}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4/(c*d*x+d),x, algorithm="maxima")

[Out]

1/6*(6*c^3*log(c*x + 1)/d - 6*c^3*log(x)/d - (6*c^2*x^2 - 3*c*x + 2)/(d*x^3))*a^2 + 1/24*(6*b^2*c^3*x^3*log(c*
x + 1) - 6*b^2*c^2*x^2 + 3*b^2*c*x - 2*b^2)*log(-c*x + 1)^2/(d*x^3) - integrate(-1/12*(3*(b^2*c*x - b^2)*log(c
*x + 1)^2 + 12*(a*b*c*x - a*b)*log(c*x + 1) + (6*b^2*c^4*x^4 + 3*b^2*c^3*x^3 - b^2*c^2*x^2 + 12*a*b - 2*(6*a*b
*c - b^2*c)*x - 6*(b^2*c^5*x^5 + b^2*c^4*x^4 + b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1))/(c^2*d*x^6 - d*x^4)
, x)

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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^{5} + d x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4/(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^5 + d*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c x^{5} + x^{4}}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c x^{5} + x^{4}}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{c x^{5} + x^{4}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c*x**5 + x**4), x) + Integral(b**2*atanh(c*x)**2/(c*x**5 + x**4), x) + Integral(2*a*b*atanh(c*
x)/(c*x**5 + x**4), x))/d

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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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4/(c*d*x+d),x, algorithm="giac")

[Out]

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