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

Optimal. Leaf size=361 \[ -3 b c d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+3 b c d^3 \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-3 b^2 c d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )-b^2 c d^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{3}{2} b^2 c d^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{3}{2} b^2 c d^3 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c^2 d^3 x+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+6 c d^3 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-6 b c d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+b^2 c^2 d^3 x \tanh ^{-1}(c x) \]

[Out]

a*b*c^2*d^3*x + b^2*c^2*d^3*x*ArcTanh[c*x] + (7*c*d^3*(a + b*ArcTanh[c*x])^2)/2 - (d^3*(a + b*ArcTanh[c*x])^2)
/x + 3*c^2*d^3*x*(a + b*ArcTanh[c*x])^2 + (c^3*d^3*x^2*(a + b*ArcTanh[c*x])^2)/2 + 6*c*d^3*(a + b*ArcTanh[c*x]
)^2*ArcTanh[1 - 2/(1 - c*x)] - 6*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)] + (b^2*c*d^3*Log[1 - c^2*x^2])/
2 + 2*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - 3*b^2*c*d^3*PolyLog[2, 1 - 2/(1 - c*x)] - 3*b*c*d^3*
(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)] + 3*b*c*d^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)]
- b^2*c*d^3*PolyLog[2, -1 + 2/(1 + c*x)] + (3*b^2*c*d^3*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (3*b^2*c*d^3*PolyLog
[3, -1 + 2/(1 - c*x)])/2

________________________________________________________________________________________

Rubi [A]  time = 0.776101, antiderivative size = 361, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 17, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.773, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5916, 5988, 5932, 2447, 5914, 6052, 5948, 6058, 6610, 5980, 260} \[ -3 b c d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+3 b c d^3 \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-3 b^2 c d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )-b^2 c d^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{3}{2} b^2 c d^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{3}{2} b^2 c d^3 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c^2 d^3 x+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+6 c d^3 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-6 b c d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+b^2 c^2 d^3 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*c^2*d^3*x + b^2*c^2*d^3*x*ArcTanh[c*x] + (7*c*d^3*(a + b*ArcTanh[c*x])^2)/2 - (d^3*(a + b*ArcTanh[c*x])^2)
/x + 3*c^2*d^3*x*(a + b*ArcTanh[c*x])^2 + (c^3*d^3*x^2*(a + b*ArcTanh[c*x])^2)/2 + 6*c*d^3*(a + b*ArcTanh[c*x]
)^2*ArcTanh[1 - 2/(1 - c*x)] - 6*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)] + (b^2*c*d^3*Log[1 - c^2*x^2])/
2 + 2*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - 3*b^2*c*d^3*PolyLog[2, 1 - 2/(1 - c*x)] - 3*b*c*d^3*
(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)] + 3*b*c*d^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)]
- b^2*c*d^3*PolyLog[2, -1 + 2/(1 + c*x)] + (3*b^2*c*d^3*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (3*b^2*c*d^3*PolyLog
[3, -1 + 2/(1 - c*x)])/2

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5910

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

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[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 5918

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

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 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 5914

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

Rule 6052

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

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 6058

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

Rule 5980

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

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]

Rubi steps

\begin{align*} \int \frac{(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx &=\int \left (3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2}+\frac{3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^3 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\left (3 c d^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx+\left (3 c^2 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+\left (c^3 d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )+\left (2 b c d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx-\left (12 b c^2 d^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b c^3 d^3\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\left (b c^4 d^3\right ) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=4 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )+\left (2 b c d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (b c^2 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\left (b c^2 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (6 b c^2 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (6 b c^2 d^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b c^2 d^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\\ &=a b c^2 d^3 x+\frac{7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )+\left (b^2 c^2 d^3\right ) \int \tanh ^{-1}(c x) \, dx-\left (2 b^2 c^2 d^3\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx+\left (3 b^2 c^2 d^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (3 b^2 c^2 d^3\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (6 b^2 c^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b c^2 d^3 x+b^2 c^2 d^3 x \tanh ^{-1}(c x)+\frac{7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )-b^2 c d^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )+\frac{3}{2} b^2 c d^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )-\frac{3}{2} b^2 c d^3 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )-\left (6 b^2 c d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )-\left (b^2 c^3 d^3\right ) \int \frac{x}{1-c^2 x^2} \, dx\\ &=a b c^2 d^3 x+b^2 c^2 d^3 x \tanh ^{-1}(c x)+\frac{7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )+\frac{1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-3 b^2 c d^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )-b^2 c d^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )+\frac{3}{2} b^2 c d^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )-\frac{3}{2} b^2 c d^3 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )\\ \end{align*}

Mathematica [C]  time = 0.635039, size = 479, normalized size = 1.33 \[ \frac{d^3 \left (-24 a b c x \text{PolyLog}(2,-c x)+24 a b c x \text{PolyLog}(2,c x)+24 b^2 c x \left (\tanh ^{-1}(c x)+1\right ) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-8 b^2 c x \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+24 b^2 c x \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+12 b^2 c x \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )-12 b^2 c x \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )+4 a^2 c^3 x^3+24 a^2 c^2 x^2+24 a^2 c x \log (x)-8 a^2+8 a b c^2 x^2+16 a b c x \log \left (1-c^2 x^2\right )+8 a b c^3 x^3 \tanh ^{-1}(c x)+48 a b c^2 x^2 \tanh ^{-1}(c x)+16 a b c x \log (c x)+4 a b c x \log (1-c x)-4 a b c x \log (c x+1)-16 a b \tanh ^{-1}(c x)+4 b^2 c x \log \left (1-c^2 x^2\right )+4 b^2 c^3 x^3 \tanh ^{-1}(c x)^2+24 b^2 c^2 x^2 \tanh ^{-1}(c x)^2+8 b^2 c^2 x^2 \tanh ^{-1}(c x)+i \pi ^3 b^2 c x-16 b^2 c x \tanh ^{-1}(c x)^3-20 b^2 c x \tanh ^{-1}(c x)^2-8 b^2 \tanh ^{-1}(c x)^2+16 b^2 c x \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-24 b^2 c x \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-48 b^2 c x \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+24 b^2 c x \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )\right )}{8 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-8*a^2 + I*b^2*c*Pi^3*x + 24*a^2*c^2*x^2 + 8*a*b*c^2*x^2 + 4*a^2*c^3*x^3 - 16*a*b*ArcTanh[c*x] + 48*a*b*
c^2*x^2*ArcTanh[c*x] + 8*b^2*c^2*x^2*ArcTanh[c*x] + 8*a*b*c^3*x^3*ArcTanh[c*x] - 8*b^2*ArcTanh[c*x]^2 - 20*b^2
*c*x*ArcTanh[c*x]^2 + 24*b^2*c^2*x^2*ArcTanh[c*x]^2 + 4*b^2*c^3*x^3*ArcTanh[c*x]^2 - 16*b^2*c*x*ArcTanh[c*x]^3
+ 16*b^2*c*x*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - 48*b^2*c*x*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])]
- 24*b^2*c*x*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 24*b^2*c*x*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x
])] + 24*a^2*c*x*Log[x] + 16*a*b*c*x*Log[c*x] + 4*a*b*c*x*Log[1 - c*x] - 4*a*b*c*x*Log[1 + c*x] + 16*a*b*c*x*L
og[1 - c^2*x^2] + 4*b^2*c*x*Log[1 - c^2*x^2] + 24*b^2*c*x*(1 + ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])]
- 8*b^2*c*x*PolyLog[2, E^(-2*ArcTanh[c*x])] + 24*b^2*c*x*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 24*a*b*
c*x*PolyLog[2, -(c*x)] + 24*a*b*c*x*PolyLog[2, c*x] + 12*b^2*c*x*PolyLog[3, -E^(-2*ArcTanh[c*x])] - 12*b^2*c*x
*PolyLog[3, E^(2*ArcTanh[c*x])]))/(8*x)

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Maple [C]  time = 1.339, size = 1270, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*I*c*d^3*b^2*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+a*b*c^2*d^3*
x+b^2*c^2*d^3*x*arctanh(c*x)-3/2*I*c*d^3*b^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*arctanh(c*x)^2-3/2*I*c*d^3*b^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*arctanh(c*x)^2+3/2*I*c*d^3*b^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))*arctanh(c*x)^2-d^3*a^2/x-6*c*d^3*b^2*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*dilog(1-I*(c*x+1)
/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c*d^3*b^2*dilog(1+(c*x+1)/(-c^2*x^2+1
)^(1/2))-2*c*d^3*b^2*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))-d^3*b^2*arctanh(c*x)^2/x+1/2*d^3*a^2*c^3*x^2+3*c^2*x*a^
2*d^3+3*c*d^3*a^2*ln(c*x)+3/2*c*d^3*b^2*arctanh(c*x)^2+c*d^3*b^2*arctanh(c*x)+3/2*c*d^3*b^2*polylog(3,-(c*x+1)
^2/(-c^2*x^2+1))-c*d^3*b^2*ln((c*x+1)^2/(-c^2*x^2+1)+1)-6*c*d^3*b^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))-2*d
^3*a*b*arctanh(c*x)/x+5/2*c*d^3*a*b*ln(c*x-1)+2*c*d^3*a*b*ln(c*x)+3*d^3*b^2*arctanh(c*x)^2*c^2*x-3*c*d^3*b^2*a
rctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3*c*d^3*b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+6*c
*d^3*b^2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2
+1)^(1/2))-6*c*d^3*b^2*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c*d^3*b^2*arctanh(c*x)*ln(1+(c*x+1)/(
-c^2*x^2+1)^(1/2))-3*c*d^3*b^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+3*c*d^3*b^2*arctanh(c*x)^2*ln(1-(c*
x+1)/(-c^2*x^2+1)^(1/2))+6*c*d^3*b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*d^3*b^2*arctanh(c*
x)^2*ln(c*x)+1/2*d^3*b^2*arctanh(c*x)^2*c^3*x^2-3*c*d^3*a*b*dilog(c*x)-3*c*d^3*a*b*dilog(c*x+1)+3/2*c*d^3*a*b*
ln(c*x+1)+6*c*d^3*a*b*arctanh(c*x)*ln(c*x)-3*c*d^3*a*b*ln(c*x)*ln(c*x+1)+6*d^3*a*b*arctanh(c*x)*c^2*x+d^3*a*b*
arctanh(c*x)*c^3*x^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2} c^{3} d^{3} x^{2} + 3 \, a^{2} c^{2} d^{3} x + 3 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a b c d^{3} + 3 \, a^{2} c d^{3} \log \left (x\right ) -{\left (c{\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x}\right )} a b d^{3} - \frac{a^{2} d^{3}}{x} + \frac{{\left (b^{2} c^{3} d^{3} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} - 2 \, b^{2} d^{3}\right )} \log \left (-c x + 1\right )^{2}}{8 \, x} - \int -\frac{{\left (b^{2} c^{4} d^{3} x^{4} + 2 \, b^{2} c^{3} d^{3} x^{3} - 2 \, b^{2} c d^{3} x - b^{2} d^{3}\right )} \log \left (c x + 1\right )^{2} + 4 \,{\left (a b c^{4} d^{3} x^{4} - a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} - 3 \, a b c d^{3} x\right )} \log \left (c x + 1\right ) -{\left (12 \, a b c^{2} d^{3} x^{2} +{\left (4 \, a b c^{4} d^{3} + b^{2} c^{4} d^{3}\right )} x^{4} - 2 \,{\left (2 \, a b c^{3} d^{3} - 3 \, b^{2} c^{3} d^{3}\right )} x^{3} - 2 \,{\left (6 \, a b c d^{3} + b^{2} c d^{3}\right )} x + 2 \,{\left (b^{2} c^{4} d^{3} x^{4} + 2 \, b^{2} c^{3} d^{3} x^{3} - 2 \, b^{2} c d^{3} x - b^{2} d^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c x^{3} - x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*c^3*d^3*x^2 + 3*a^2*c^2*d^3*x + 3*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*c*d^3 + 3*a^2*c*d^3*log
(x) - (c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*d^3 - a^2*d^3/x + 1/8*(b^2*c^3*d^3*x^3 + 6*b^2*
c^2*d^3*x^2 - 2*b^2*d^3)*log(-c*x + 1)^2/x - integrate(-1/4*((b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^
3*x - b^2*d^3)*log(c*x + 1)^2 + 4*(a*b*c^4*d^3*x^4 - a*b*c^3*d^3*x^3 + 3*a*b*c^2*d^3*x^2 - 3*a*b*c*d^3*x)*log(
c*x + 1) - (12*a*b*c^2*d^3*x^2 + (4*a*b*c^4*d^3 + b^2*c^4*d^3)*x^4 - 2*(2*a*b*c^3*d^3 - 3*b^2*c^3*d^3)*x^3 - 2
*(6*a*b*c*d^3 + b^2*c*d^3)*x + 2*(b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^3*x - b^2*d^3)*log(c*x + 1))
*log(-c*x + 1))/(c*x^3 - x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int 3 a^{2} c^{2}\, dx + \int \frac{a^{2}}{x^{2}}\, dx + \int \frac{3 a^{2} c}{x}\, dx + \int a^{2} c^{3} x\, dx + \int 3 b^{2} c^{2} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int 6 a b c^{2} \operatorname{atanh}{\left (c x \right )}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{3 b^{2} c \operatorname{atanh}^{2}{\left (c x \right )}}{x}\, dx + \int b^{2} c^{3} x \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int \frac{6 a b c \operatorname{atanh}{\left (c x \right )}}{x}\, dx + \int 2 a b c^{3} x \operatorname{atanh}{\left (c x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(3*a**2*c**2, x) + Integral(a**2/x**2, x) + Integral(3*a**2*c/x, x) + Integral(a**2*c**3*x, x) +
Integral(3*b**2*c**2*atanh(c*x)**2, x) + Integral(b**2*atanh(c*x)**2/x**2, x) + Integral(6*a*b*c**2*atanh(c*x
), x) + Integral(2*a*b*atanh(c*x)/x**2, x) + Integral(3*b**2*c*atanh(c*x)**2/x, x) + Integral(b**2*c**3*x*atan
h(c*x)**2, x) + Integral(6*a*b*c*atanh(c*x)/x, x) + Integral(2*a*b*c**3*x*atanh(c*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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