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

Optimal. Leaf size=385 \[ \frac{b d^2 \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^3}-\frac{b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}+\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e^3}-\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^3}+\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c e^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac{d^2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^3}+\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}-\frac{d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}+\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e^2}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{a b x}{c e}+\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^2 e}+\frac{b^2 x \tanh ^{-1}(c x)}{c e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.427147, antiderivative size = 385, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 5922} \[ \frac{b d^2 \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^3}-\frac{b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}+\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e^3}-\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^3}+\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c e^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac{d^2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^3}+\frac{d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}-\frac{d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}+\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e^2}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac{a b x}{c e}+\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^2 e}+\frac{b^2 x \tanh ^{-1}(c x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2,
0]

Rubi steps

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

Mathematica [C]  time = 16.0594, size = 1297, normalized size = 3.37 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a^2*d*x)/e^2) + (a^2*x^2)/(2*e) + (a^2*d^2*Log[d + e*x])/e^3 + (a*b*(c*e^2*x + I*c^2*d^2*Pi*ArcTanh[c*x] -
2*c^2*d*e*x*ArcTanh[c*x] - e^2*(1 - c^2*x^2)*ArcTanh[c*x] + 2*c^2*d^2*ArcTanh[(c*d)/e]*ArcTanh[c*x] - c^2*d^2*
ArcTanh[c*x]^2 + c*d*e*ArcTanh[c*x]^2 - (c*d*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] - 2*
c^2*d^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - I*c^2*d^2*Pi*Log[1 + E^(2*ArcTanh[c*x])] + 2*c^2*d^2*ArcTa
nh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 2*c^2*d^2*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[
(c*d)/e] + ArcTanh[c*x]))] + 2*c*d*e*Log[1/Sqrt[1 - c^2*x^2]] + I*c^2*d^2*Pi*Log[1/Sqrt[1 - c^2*x^2]] - 2*c^2*
d^2*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + c^2*d^2*PolyLog[2, -E^(-2*ArcTanh[c*x])] -
c^2*d^2*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c^2*e^3) + (b^2*((6*c*e^2*x*ArcTanh[c*x] + 6*
c*d*e*ArcTanh[c*x]^2 - 6*c^2*d*e*x*ArcTanh[c*x]^2 - 3*e^2*(1 - c^2*x^2)*ArcTanh[c*x]^2 - 2*c^2*d^2*ArcTanh[c*x
]^3 + 2*c*d*e*ArcTanh[c*x]^3 + 12*c*d*e*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 6*c^2*d^2*ArcTanh[c*x]^2*L
og[1 + E^(-2*ArcTanh[c*x])] - 6*e^2*Log[1/Sqrt[1 - c^2*x^2]] + 6*c*d*(-e + c*d*ArcTanh[c*x])*PolyLog[2, -E^(-2
*ArcTanh[c*x])] + 3*c^2*d^2*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(6*e^3) - (c*d*(-(c*d) + e)*(c*d + e)*(-6*c*d*Ar
cTanh[c*x]^3 + 2*e*ArcTanh[c*x]^3 - (4*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - (6*I)*c*
d*Pi*ArcTanh[c*x]*Log[(1 + E^(2*ArcTanh[c*x]))/(2*E^ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[1 + ((c*d + e)*E
^(2*ArcTanh[c*x]))/(c*d - e)] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcT
anh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*(ArcTanh[(c*d)/e]
+ ArcTanh[c*x]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 +
E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])))] + 6*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c*x])) + c*d*(1 + E
^(2*ArcTanh[c*x])))/(2*E^ArcTanh[c*x])] + (6*I)*c*d*Pi*ArcTanh[c*x]*Log[1/Sqrt[1 - c^2*x^2]] - 6*c*d*ArcTanh[c
*x]^2*Log[(c*d)/Sqrt[1 - c^2*x^2] + (c*e*x)/Sqrt[1 - c^2*x^2]] - 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Si
nh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - 6*c*d*ArcTanh[c*x]*PolyLog[2, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e
))] + 12*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 12*c*d*ArcTanh[c*x]*PolyLog[2, E^
(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] +
3*c*d*PolyLog[3, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e))] - 12*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTa
nh[c*x])] - 12*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 3*c*d*PolyLog[3, E^(2*(ArcTanh[(c*d)/e] +
ArcTanh[c*x]))]))/(e^3*(6*c^2*d^2 - 6*e^2))))/c^2

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Maple [C]  time = 1.298, size = 1656, normalized size = 4.3 \begin{align*} \text{result 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/(e*x+d),x)

[Out]

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

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

[Out]

1/2*a^2*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 1/8*(b^2*e*x^2 - 2*b^2*d*x)*log(-c*x + 1)^2/e^2 - int
egrate(-1/4*((b^2*c*e^2*x^3 - b^2*e^2*x^2)*log(c*x + 1)^2 + 4*(a*b*c*e^2*x^3 - a*b*e^2*x^2)*log(c*x + 1) + (2*
b^2*c*d^2*x - (4*a*b*c*e^2 + b^2*c*e^2)*x^3 + (b^2*c*d*e + 4*a*b*e^2)*x^2 - 2*(b^2*c*e^2*x^3 - b^2*e^2*x^2)*lo
g(c*x + 1))*log(-c*x + 1))/(c*e^3*x^2 - d*e^2 + (c*d*e^2 - e^3)*x), 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}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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