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

Optimal. Leaf size=278 \[ -b d^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b d^2 \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+\frac{1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c d^2 x+2 c d^2 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{3}{2} d^2 \left (a+b \tanh ^{-1}(c x)\right )^2+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-4 b d^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 d^2 \log \left (1-c^2 x^2\right )+b^2 c d^2 x \tanh ^{-1}(c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.586404, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 14, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5914, 6052, 5948, 6058, 6610, 5916, 5980, 260} \[ -b d^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b d^2 \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+\frac{1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c d^2 x+2 c d^2 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{3}{2} d^2 \left (a+b \tanh ^{-1}(c x)\right )^2+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-4 b d^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 d^2 \log \left (1-c^2 x^2\right )+b^2 c d^2 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.704943, size = 324, normalized size = 1.17 \[ \frac{1}{2} d^2 \left (2 a b (\text{PolyLog}(2,c x)-\text{PolyLog}(2,-c x))+4 b^2 \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \left ((c x-1) \tanh ^{-1}(c x)-2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )\right )+2 b^2 \left (\tanh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-\frac{2}{3} \tanh ^{-1}(c x)^3-\tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\frac{i \pi ^3}{24}\right )+a^2 c^2 x^2+4 a^2 c x+2 a^2 \log (c x)+a b \left (2 c^2 x^2 \tanh ^{-1}(c x)+2 c x+\log (1-c x)-\log (c x+1)\right )+4 a b \left (\log \left (1-c^2 x^2\right )+2 c x \tanh ^{-1}(c x)\right )+b^2 \left (\log \left (1-c^2 x^2\right )+\left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^2+2 c x \tanh ^{-1}(c x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(4*a^2*c*x + a^2*c^2*x^2 + 2*a^2*Log[c*x] + a*b*(2*c*x + 2*c^2*x^2*ArcTanh[c*x] + Log[1 - c*x] - Log[1 +
c*x]) + 4*a*b*(2*c*x*ArcTanh[c*x] + Log[1 - c^2*x^2]) + b^2*(2*c*x*ArcTanh[c*x] + (-1 + c^2*x^2)*ArcTanh[c*x]^
2 + Log[1 - c^2*x^2]) + 4*b^2*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c*x] - 2*Log[1 + E^(-2*ArcTanh[c*x])]) + PolyL
og[2, -E^(-2*ArcTanh[c*x])]) + 2*a*b*(-PolyLog[2, -(c*x)] + PolyLog[2, c*x]) + 2*b^2*((I/24)*Pi^3 - (2*ArcTanh
[c*x]^3)/3 - ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + ArcTan
h[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + PolyLog[3, -E^(-2*ArcT
anh[c*x])]/2 - PolyLog[3, E^(2*ArcTanh[c*x])]/2)))/2

________________________________________________________________________________________

Maple [C]  time = 0.853, size = 1082, normalized size = 3.9 \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)^2*(a+b*arctanh(c*x))^2/x,x)

[Out]

a*b*c*d^2*x+b^2*c*d^2*x*arctanh(c*x)+d^2*a^2*ln(c*x)+3/2*d^2*b^2*arctanh(c*x)^2-2*d^2*b^2*polylog(3,(c*x+1)/(-
c^2*x^2+1)^(1/2))-2*d^2*b^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*d^2*b^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+
1))-d^2*b^2*ln((c*x+1)^2/(-c^2*x^2+1)+1)-4*d^2*b^2*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-4*d^2*b^2*dilog(1-I*(
c*x+1)/(-c^2*x^2+1)^(1/2))+d^2*b^2*arctanh(c*x)+2*d^2*a*b*arctanh(c*x)*ln(c*x)-d^2*a*b*ln(c*x)*ln(c*x+1)+d^2*a
*b*arctanh(c*x)*c^2*x^2+4*d^2*a*b*arctanh(c*x)*c*x+1/2*I*d^2*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+1/2*d^2*a^2*c^2*x^2-4*d^2*b^2*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1
/2))+2*c*x*a^2*d^2+d^2*b^2*arctanh(c*x)^2*ln(c*x)-d^2*b^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+d^2*b^2*
arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*d^2*b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+d
^2*b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*d^2*b^2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^
(1/2))-d^2*b^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-4*d^2*b^2*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2
+1)^(1/2))-d^2*a*b*dilog(c*x)-d^2*a*b*dilog(c*x+1)+5/2*d^2*a*b*ln(c*x-1)+3/2*d^2*a*b*ln(c*x+1)+1/2*I*d^2*b^2*P
i*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+1/2*d^2*b^2*arctanh(c*x)^2*c^2*x^2+2*d^2*b^2*arctanh(c*x)^2*c*x-1/2*I*d
^2*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*a
rctanh(c*x)^2-1/2*I*d^2*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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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