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

Optimal. Leaf size=329 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d}-\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{3 c^4 d}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}-\frac{a b x}{c^3 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}-\frac{8 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^4 d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}+\frac{b^2 x}{3 c^3 d}-\frac{b^2 x \tanh ^{-1}(c x)}{c^3 d}-\frac{b^2 \tanh ^{-1}(c x)}{3 c^4 d} \]

[Out]

-((a*b*x)/(c^3*d)) + (b^2*x)/(3*c^3*d) - (b^2*ArcTanh[c*x])/(3*c^4*d) - (b^2*x*ArcTanh[c*x])/(c^3*d) + (b*x^2*
(a + b*ArcTanh[c*x]))/(3*c^2*d) + (11*(a + b*ArcTanh[c*x])^2)/(6*c^4*d) + (x*(a + b*ArcTanh[c*x])^2)/(c^3*d) -
(x^2*(a + b*ArcTanh[c*x])^2)/(2*c^2*d) + (x^3*(a + b*ArcTanh[c*x])^2)/(3*c*d) - (8*b*(a + b*ArcTanh[c*x])*Log
[2/(1 - c*x)])/(3*c^4*d) + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^4*d) - (b^2*Log[1 - c^2*x^2])/(2*c^4*d
) - (4*b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(3*c^4*d) - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^4*
d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.84035, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 14, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {5930, 5916, 5980, 321, 206, 5984, 5918, 2402, 2315, 5910, 260, 5948, 6056, 6610} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d}-\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{3 c^4 d}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}-\frac{a b x}{c^3 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}-\frac{8 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^4 d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}+\frac{b^2 x}{3 c^3 d}-\frac{b^2 x \tanh ^{-1}(c x)}{c^3 d}-\frac{b^2 \tanh ^{-1}(c x)}{3 c^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*x)/(c^3*d)) + (b^2*x)/(3*c^3*d) - (b^2*ArcTanh[c*x])/(3*c^4*d) - (b^2*x*ArcTanh[c*x])/(c^3*d) + (b*x^2*
(a + b*ArcTanh[c*x]))/(3*c^2*d) + (11*(a + b*ArcTanh[c*x])^2)/(6*c^4*d) + (x*(a + b*ArcTanh[c*x])^2)/(c^3*d) -
(x^2*(a + b*ArcTanh[c*x])^2)/(2*c^2*d) + (x^3*(a + b*ArcTanh[c*x])^2)/(3*c*d) - (8*b*(a + b*ArcTanh[c*x])*Log
[2/(1 - c*x)])/(3*c^4*d) + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^4*d) - (b^2*Log[1 - c^2*x^2])/(2*c^4*d
) - (4*b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(3*c^4*d) - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^4*
d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^4*d)

Rule 5930

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f)/e, 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] && GtQ[m, 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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 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 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 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^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx &=-\frac{\int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c}+\frac{\int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}+\frac{\int \frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c^2}-\frac{(2 b) \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 d}-\frac{\int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^2 d}\\ &=-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c^3}+\frac{\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^3 d}+\frac{(2 b) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^2 d}-\frac{(2 b) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^2 d}+\frac{b \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d}\\ &=\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^4 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^4 d}-\frac{(2 b) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{3 c^3 d}-\frac{b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d}+\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^3 d}-\frac{(2 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^3 d}-\frac{(2 b) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^2 d}-\frac{b^2 \int \frac{x^2}{1-c^2 x^2} \, dx}{3 c d}\\ &=-\frac{a b x}{c^3 d}+\frac{b^2 x}{3 c^3 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{3 c^4 d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^4 d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^4 d}-\frac{(2 b) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^3 d}-\frac{b^2 \int \frac{1}{1-c^2 x^2} \, dx}{3 c^3 d}+\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{3 c^3 d}-\frac{b^2 \int \tanh ^{-1}(c x) \, dx}{c^3 d}+\frac{b^2 \int \frac{\text{Li}_2\left (1-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}\\ &=-\frac{a b x}{c^3 d}+\frac{b^2 x}{3 c^3 d}-\frac{b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{3 c^4 d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^4 d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^4 d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 c^4 d}-\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 )}{3 c^4 d}+\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}+\frac{b^2 \int \frac{x}{1-c^2 x^2} \, dx}{c^2 d}\\ &=-\frac{a b x}{c^3 d}+\frac{b^2 x}{3 c^3 d}-\frac{b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{3 c^4 d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^4 d}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{3 c^4 d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^4 d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 c^4 d}-\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^4 d}\\ &=-\frac{a b x}{c^3 d}+\frac{b^2 x}{3 c^3 d}-\frac{b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac{11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac{8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{3 c^4 d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{c^4 d}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}-\frac{4 b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{3 c^4 d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^4 d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [A]  time = 0.863035, size = 347, normalized size = 1.05 \[ \frac{a b \left (-3 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-8 \log \left (\frac{1}{\sqrt{1-c^2 x^2}}\right )+\left (1-c^2 x^2\right ) \left (-2 c x \tanh ^{-1}(c x)+3 \tanh ^{-1}(c x)-1\right )-3 c x+8 c x \tanh ^{-1}(c x)+6 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )}{3 c^4 d}+\frac{b^2 \left (\left (8-6 \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+6 \log \left (\frac{1}{\sqrt{1-c^2 x^2}}\right )-2 c x \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2+3 \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2-2 \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)+2 c x+8 c x \tanh ^{-1}(c x)^2-8 \tanh ^{-1}(c x)^2-6 c x \tanh ^{-1}(c x)+6 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-16 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )}{6 c^4 d}-\frac{a^2 x^2}{2 c^2 d}+\frac{a^2 x}{c^3 d}-\frac{a^2 \log (c x+1)}{c^4 d}+\frac{a^2 x^3}{3 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*x)/(c^3*d) - (a^2*x^2)/(2*c^2*d) + (a^2*x^3)/(3*c*d) - (a^2*Log[1 + c*x])/(c^4*d) + (a*b*(-3*c*x + 8*c*x*
ArcTanh[c*x] + (1 - c^2*x^2)*(-1 + 3*ArcTanh[c*x] - 2*c*x*ArcTanh[c*x]) + 6*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh
[c*x])] - 8*Log[1/Sqrt[1 - c^2*x^2]] - 3*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(3*c^4*d) + (b^2*(2*c*x - 6*c*x*Ar
cTanh[c*x] - 2*(1 - c^2*x^2)*ArcTanh[c*x] - 8*ArcTanh[c*x]^2 + 8*c*x*ArcTanh[c*x]^2 + 3*(1 - c^2*x^2)*ArcTanh[
c*x]^2 - 2*c*x*(1 - c^2*x^2)*ArcTanh[c*x]^2 - 16*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + 6*ArcTanh[c*x]^2*
Log[1 + E^(-2*ArcTanh[c*x])] + 6*Log[1/Sqrt[1 - c^2*x^2]] + (8 - 6*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x
])] - 3*PolyLog[3, -E^(-2*ArcTanh[c*x])]))/(6*c^4*d)

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Maple [C]  time = 0.964, size = 1298, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} x^{3}}{c x + 1}\, dx + \int \frac{b^{2} x^{3} \operatorname{atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac{2 a b x^{3} \operatorname{atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**3/(c*x + 1), x) + Integral(b**2*x**3*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*x**3*atanh
(c*x)/(c*x + 1), 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} x^{3}}{c d x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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