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

Optimal. Leaf size=260 \[ -\frac{2 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{c^3 d^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (c x+1)}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (c x+1)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}+\frac{2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2}-\frac{b^2}{2 c^3 d^2 (c x+1)}+\frac{b^2 \tanh ^{-1}(c x)}{2 c^3 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.475892, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 18, 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, 5928, 5926, 627, 44, 207, 5948, 6056, 6610} \[ -\frac{2 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{c^3 d^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (c x+1)}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (c x+1)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}+\frac{2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2}-\frac{b^2}{2 c^3 d^2 (c x+1)}+\frac{b^2 \tanh ^{-1}(c x)}{2 c^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b^2/(2*c^3*d^2*(1 + c*x)) + (b^2*ArcTanh[c*x])/(2*c^3*d^2) - (b*(a + b*ArcTanh[c*x]))/(c^3*d^2*(1 + c*x)) + (
3*(a + b*ArcTanh[c*x])^2)/(2*c^3*d^2) + (x*(a + b*ArcTanh[c*x])^2)/(c^2*d^2) - (a + b*ArcTanh[c*x])^2/(c^3*d^2
*(1 + c*x)) - (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(c^3*d^2) + (2*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x
)])/(c^3*d^2) - (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(c^3*d^2) - (2*b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 +
c*x)])/(c^3*d^2) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(c^3*d^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 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[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 5926

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 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{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^2} \, dx &=\int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2 (1+c x)^2}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2 (1+c x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^2 d^2}+\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{c^2 d^2}-\frac{2 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{c^2 d^2}\\ &=\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}+\frac{(2 b) \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^2}-\frac{(4 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^2}-\frac{(2 b) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d^2}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}+\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^2 d^2}-\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{c^2 d^2}-\frac{(2 b) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^2 d^2}+\frac{\left (2 b^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^2}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3 d^2}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}+\frac{b^2 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^2 d^2}+\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^2}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3 d^2}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c^3 d^2}+\frac{b^2 \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{c^2 d^2}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3 d^2}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}+\frac{b^2 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^2}\\ &=-\frac{b^2}{2 c^3 d^2 (1+c x)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3 d^2}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \int \frac{1}{-1+c^2 x^2} \, dx}{2 c^2 d^2}\\ &=-\frac{b^2}{2 c^3 d^2 (1+c x)}+\frac{b^2 \tanh ^{-1}(c x)}{2 c^3 d^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2 (1+c x)}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3 d^2}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c^3 d^2}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{c^3 d^2}\\ \end{align*}

Mathematica [A]  time = 0.918927, size = 295, normalized size = 1.13 \[ \frac{2 a b \left (-4 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+2 \log \left (1-c^2 x^2\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \left (2 c x+4 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )\right )\right )+b^2 \left (\left (4-8 \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-4 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+4 c x \tanh ^{-1}(c x)^2-4 \tanh ^{-1}(c x)^2+8 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-8 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+2 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )-2 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )-2 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )\right )+4 a^2 c x-\frac{4 a^2}{c x+1}-8 a^2 \log (c x+1)}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a^2*c*x - (4*a^2)/(1 + c*x) - 8*a^2*Log[1 + c*x] + b^2*(-4*ArcTanh[c*x]^2 + 4*c*x*ArcTanh[c*x]^2 - Cosh[2*A
rcTanh[c*x]] - 2*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] - 2*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] - 8*ArcTanh[c*x]*Lo
g[1 + E^(-2*ArcTanh[c*x])] + 8*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + (4 - 8*ArcTanh[c*x])*PolyLog[2, -
E^(-2*ArcTanh[c*x])] - 4*PolyLog[3, -E^(-2*ArcTanh[c*x])] + Sinh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]*Sinh[2*ArcTa
nh[c*x]] + 2*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]]) + 2*a*b*(-Cosh[2*ArcTanh[c*x]] + 2*Log[1 - c^2*x^2] - 4*Poly
Log[2, -E^(-2*ArcTanh[c*x])] + Sinh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]*(2*c*x - Cosh[2*ArcTanh[c*x]] + 4*Log[1 +
E^(-2*ArcTanh[c*x])] + Sinh[2*ArcTanh[c*x]])))/(4*c^3*d^2)

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Maple [C]  time = 0.507, size = 5542, normalized size = 21.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} x^{2}}{c^{2} x^{2} + 2 c x + 1}\, dx + \int \frac{b^{2} x^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 2 c x + 1}\, dx + \int \frac{2 a b x^{2} \operatorname{atanh}{\left (c x \right )}}{c^{2} x^{2} + 2 c x + 1}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**2/(c**2*x**2 + 2*c*x + 1), x) + Integral(b**2*x**2*atanh(c*x)**2/(c**2*x**2 + 2*c*x + 1), x)
+ Integral(2*a*b*x**2*atanh(c*x)/(c**2*x**2 + 2*c*x + 1), x))/d**2

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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} x^{2}}{{\left (c d x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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