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

Optimal. Leaf size=270 \[ -\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^4}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{a b d x}{2 c^3}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}-\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^4}+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{b^2 d x^2}{12 c^2}+\frac{b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}+\frac{3 b^2 d x}{10 c^3}+\frac{b^2 d x \tanh ^{-1}(c x)}{2 c^3}-\frac{3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 d x^3}{30 c} \]

[Out]

(a*b*d*x)/(2*c^3) + (3*b^2*d*x)/(10*c^3) + (b^2*d*x^2)/(12*c^2) + (b^2*d*x^3)/(30*c) - (3*b^2*d*ArcTanh[c*x])/
(10*c^4) + (b^2*d*x*ArcTanh[c*x])/(2*c^3) + (b*d*x^2*(a + b*ArcTanh[c*x]))/(5*c^2) + (b*d*x^3*(a + b*ArcTanh[c
*x]))/(6*c) + (b*d*x^4*(a + b*ArcTanh[c*x]))/10 - (d*(a + b*ArcTanh[c*x])^2)/(20*c^4) + (d*x^4*(a + b*ArcTanh[
c*x])^2)/4 + (c*d*x^5*(a + b*ArcTanh[c*x])^2)/5 - (2*b*d*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^4) + (b^2
*d*Log[1 - c^2*x^2])/(3*c^4) - (b^2*d*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^4)

________________________________________________________________________________________

Rubi [A]  time = 0.649668, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 15, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5940, 5916, 5980, 266, 43, 5910, 260, 5948, 302, 206, 321, 5984, 5918, 2402, 2315} \[ -\frac{b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^4}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{a b d x}{2 c^3}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}-\frac{2 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^4}+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{b^2 d x^2}{12 c^2}+\frac{b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}+\frac{3 b^2 d x}{10 c^3}+\frac{b^2 d x \tanh ^{-1}(c x)}{2 c^3}-\frac{3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 d x^3}{30 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*d*x)/(2*c^3) + (3*b^2*d*x)/(10*c^3) + (b^2*d*x^2)/(12*c^2) + (b^2*d*x^3)/(30*c) - (3*b^2*d*ArcTanh[c*x])/
(10*c^4) + (b^2*d*x*ArcTanh[c*x])/(2*c^3) + (b*d*x^2*(a + b*ArcTanh[c*x]))/(5*c^2) + (b*d*x^3*(a + b*ArcTanh[c
*x]))/(6*c) + (b*d*x^4*(a + b*ArcTanh[c*x]))/10 - (d*(a + b*ArcTanh[c*x])^2)/(20*c^4) + (d*x^4*(a + b*ArcTanh[
c*x])^2)/4 + (c*d*x^5*(a + b*ArcTanh[c*x])^2)/5 - (2*b*d*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^4) + (b^2
*d*Log[1 - c^2*x^2])/(3*c^4) - (b^2*d*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^4)

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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 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 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]

Rubi steps

\begin{align*} \int x^3 (d+c d x) \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\int \left (d x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+c d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+(c d) \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{2} (b c d) \int \frac{x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\frac{1}{5} \left (2 b c^2 d\right ) \int \frac{x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} (2 b d) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\frac{1}{5} (2 b d) \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac{(b d) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}-\frac{(b d) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{6} \left (b^2 d\right ) \int \frac{x^3}{1-c^2 x^2} \, dx+\frac{(b d) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac{(b d) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{2 c^3}+\frac{(2 b d) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^2}-\frac{(2 b d) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}-\frac{1}{10} \left (b^2 c d\right ) \int \frac{x^4}{1-c^2 x^2} \, dx\\ &=\frac{a b d x}{2 c^3}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{12} \left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )-\frac{(2 b d) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^3}+\frac{\left (b^2 d\right ) \int \tanh ^{-1}(c x) \, dx}{2 c^3}-\frac{\left (b^2 d\right ) \int \frac{x^2}{1-c^2 x^2} \, dx}{5 c}-\frac{1}{10} \left (b^2 c d\right ) \int \left (-\frac{1}{c^4}-\frac{x^2}{c^2}+\frac{1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac{a b d x}{2 c^3}+\frac{3 b^2 d x}{10 c^3}+\frac{b^2 d x^3}{30 c}+\frac{b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^4}-\frac{1}{12} \left (b^2 d\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac{\left (b^2 d\right ) \int \frac{1}{1-c^2 x^2} \, dx}{10 c^3}-\frac{\left (b^2 d\right ) \int \frac{1}{1-c^2 x^2} \, dx}{5 c^3}+\frac{\left (2 b^2 d\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^3}-\frac{\left (b^2 d\right ) \int \frac{x}{1-c^2 x^2} \, dx}{2 c^2}\\ &=\frac{a b d x}{2 c^3}+\frac{3 b^2 d x}{10 c^3}+\frac{b^2 d x^2}{12 c^2}+\frac{b^2 d x^3}{30 c}-\frac{3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^4}+\frac{b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}-\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 )}{5 c^4}\\ &=\frac{a b d x}{2 c^3}+\frac{3 b^2 d x}{10 c^3}+\frac{b^2 d x^2}{12 c^2}+\frac{b^2 d x^3}{30 c}-\frac{3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac{1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^4}+\frac{b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}-\frac{b^2 d \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^4}\\ \end{align*}

Mathematica [A]  time = 0.752875, size = 271, normalized size = 1. \[ \frac{d \left (12 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+12 a^2 c^5 x^5+15 a^2 c^4 x^4+6 a b c^4 x^4+10 a b c^3 x^3+12 a b c^2 x^2+12 a b \log \left (c^2 x^2-1\right )+2 b \tanh ^{-1}(c x) \left (3 a c^4 x^4 (4 c x+5)+b \left (3 c^4 x^4+5 c^3 x^3+6 c^2 x^2+15 c x-9\right )-12 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+30 a b c x+15 a b \log (1-c x)-15 a b \log (c x+1)-18 a b+2 b^2 c^3 x^3+5 b^2 c^2 x^2+20 b^2 \log \left (1-c^2 x^2\right )+3 b^2 \left (4 c^5 x^5+5 c^4 x^4-9\right ) \tanh ^{-1}(c x)^2+18 b^2 c x-5 b^2\right )}{60 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*(-18*a*b - 5*b^2 + 30*a*b*c*x + 18*b^2*c*x + 12*a*b*c^2*x^2 + 5*b^2*c^2*x^2 + 10*a*b*c^3*x^3 + 2*b^2*c^3*x^
3 + 15*a^2*c^4*x^4 + 6*a*b*c^4*x^4 + 12*a^2*c^5*x^5 + 3*b^2*(-9 + 5*c^4*x^4 + 4*c^5*x^5)*ArcTanh[c*x]^2 + 2*b*
ArcTanh[c*x]*(3*a*c^4*x^4*(5 + 4*c*x) + b*(-9 + 15*c*x + 6*c^2*x^2 + 5*c^3*x^3 + 3*c^4*x^4) - 12*b*Log[1 + E^(
-2*ArcTanh[c*x])]) + 15*a*b*Log[1 - c*x] - 15*a*b*Log[1 + c*x] + 20*b^2*Log[1 - c^2*x^2] + 12*a*b*Log[-1 + c^2
*x^2] + 12*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(60*c^4)

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Maple [A]  time = 0.054, size = 422, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}dx{\it Artanh} \left ( cx \right ) }{2\,{c}^{3}}}+{\frac{3\,{b}^{2}dx}{10\,{c}^{3}}}+{\frac{d{b}^{2}{x}^{2}}{12\,{c}^{2}}}+{\frac{d{b}^{2}{x}^{3}}{30\,c}}+{\frac{2\,cdab{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+{\frac{{a}^{2}d{x}^{4}}{4}}+{\frac{d{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{80\,{c}^{4}}}+{\frac{29\,d{b}^{2}\ln \left ( cx-1 \right ) }{60\,{c}^{4}}}+{\frac{9\,d{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{80\,{c}^{4}}}-{\frac{d{b}^{2}}{5\,{c}^{4}}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{11\,d{b}^{2}\ln \left ( cx+1 \right ) }{60\,{c}^{4}}}+{\frac{c{a}^{2}d{x}^{5}}{5}}+{\frac{abdx}{2\,{c}^{3}}}+{\frac{dab{x}^{3}}{6\,c}}+{\frac{dab{x}^{2}}{5\,{c}^{2}}}-{\frac{d{b}^{2}\ln \left ( cx+1 \right ) }{40\,{c}^{4}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{dab\ln \left ( cx+1 \right ) }{20\,{c}^{4}}}+{\frac{9\,dab\ln \left ( cx-1 \right ) }{20\,{c}^{4}}}-{\frac{9\,d{b}^{2}\ln \left ( cx-1 \right ) }{40\,{c}^{4}}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{d{b}^{2}}{40\,{c}^{4}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{dab{\it Artanh} \left ( cx \right ){x}^{4}}{2}}+{\frac{d{b}^{2}{\it Artanh} \left ( cx \right ){x}^{3}}{6\,c}}+{\frac{d{b}^{2}{\it Artanh} \left ( cx \right ){x}^{2}}{5\,{c}^{2}}}+{\frac{cd{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}{x}^{5}}{5}}+{\frac{9\,d{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{20\,{c}^{4}}}-{\frac{d{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{20\,{c}^{4}}}+{\frac{dab{x}^{4}}{10}}+{\frac{d{b}^{2}{\it Artanh} \left ( cx \right ){x}^{4}}{10}}+{\frac{d{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*b^2*d*x/c^3+1/12*b^2*d*x^2/c^2+1/30*b^2*d*x^3/c+2/5*c*d*a*b*arctanh(c*x)*x^5+1/4*a^2*d*x^4+1/10*d*b^2*arc
tanh(c*x)*x^4+1/4*d*b^2*arctanh(c*x)^2*x^4+1/80/c^4*d*b^2*ln(c*x+1)^2+29/60/c^4*d*b^2*ln(c*x-1)+9/80/c^4*d*b^2
*ln(c*x-1)^2-1/5/c^4*d*b^2*dilog(1/2+1/2*c*x)+11/60/c^4*d*b^2*ln(c*x+1)+1/5*c*a^2*d*x^5+1/2*a*b*d*x/c^3+1/2*b^
2*d*x*arctanh(c*x)/c^3+1/6/c*d*a*b*x^3+1/5/c^2*d*a*b*x^2+1/2*d*a*b*arctanh(c*x)*x^4-1/40/c^4*d*b^2*ln(-1/2*c*x
+1/2)*ln(c*x+1)-1/20/c^4*d*a*b*ln(c*x+1)+9/20/c^4*d*a*b*ln(c*x-1)+1/6/c*d*b^2*arctanh(c*x)*x^3+1/5/c^2*d*b^2*a
rctanh(c*x)*x^2+1/5*c*d*b^2*arctanh(c*x)^2*x^5+9/20/c^4*d*b^2*arctanh(c*x)*ln(c*x-1)-1/20/c^4*d*b^2*arctanh(c*
x)*ln(c*x+1)-9/40/c^4*d*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/40/c^4*d*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/10*d*a
*b*x^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*c*d*x^5 + 1/4*b^2*d*x^4*arctanh(c*x)^2 + 1/4*a^2*d*x^4 + 1/10*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^
2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*a*b*c*d - 1/36000*(24*c^6*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^10 - 15*log(c*
x + 1)/c^11 + 15*log(c*x - 1)/c^11) - 45*c^5*((c^2*x^4 + 2*x^2)/c^8 + 2*log(c^2*x^2 - 1)/c^10) - 1080000*c^5*i
ntegrate(1/150*x^5*log(c*x + 1)/(c^6*x^2 - c^4), x) + 50*c^4*(2*(c^2*x^3 + 3*x)/c^8 - 3*log(c*x + 1)/c^9 + 3*l
og(c*x - 1)/c^9) - 300*c^3*(x^2/c^6 + log(c^2*x^2 - 1)/c^8) + 900*c^2*(2*x/c^6 - log(c*x + 1)/c^7 + log(c*x -
1)/c^7) - 540000*c*integrate(1/150*x*log(c*x + 1)/(c^6*x^2 - c^4), x) - 60*(30*c^5*x^5*log(c*x + 1)^2 + (12*c^
5*x^5 - 15*c^4*x^4 + 20*c^3*x^3 - 30*c^2*x^2 + 60*c*x - 60*(c^5*x^5 + 1)*log(c*x + 1))*log(-c*x + 1))/c^5 - (7
2*(c*x - 1)^5*(25*log(-c*x + 1)^2 - 10*log(-c*x + 1) + 2) + 1125*(c*x - 1)^4*(8*log(-c*x + 1)^2 - 4*log(-c*x +
1) + 1) + 2000*(c*x - 1)^3*(9*log(-c*x + 1)^2 - 6*log(-c*x + 1) + 2) + 9000*(c*x - 1)^2*(2*log(-c*x + 1)^2 -
2*log(-c*x + 1) + 1) + 9000*(c*x - 1)*(log(-c*x + 1)^2 - 2*log(-c*x + 1) + 2))/c^5 + 1800*log(150*c^6*x^2 - 15
0*c^4)/c^5 - 540000*integrate(1/150*log(c*x + 1)/(c^6*x^2 - c^4), x))*b^2*c*d + 1/12*(6*x^4*arctanh(c*x) + c*(
2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*a*b*d + 1/48*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3
*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)*arctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*lo
g(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16*log(c*x - 1))/c^4)*b^2*d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{2} c d x^{4} + a^{2} d x^{3} +{\left (b^{2} c d x^{4} + b^{2} d x^{3}\right )} \operatorname{artanh}\left (c x\right )^{2} + 2 \,{\left (a b c d x^{4} + a b d x^{3}\right )} \operatorname{artanh}\left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*c*d*x^4 + a^2*d*x^3 + (b^2*c*d*x^4 + b^2*d*x^3)*arctanh(c*x)^2 + 2*(a*b*c*d*x^4 + a*b*d*x^3)*arct
anh(c*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d \left (\int a^{2} x^{3}\, dx + \int a^{2} c x^{4}\, dx + \int b^{2} x^{3} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b x^{3} \operatorname{atanh}{\left (c x \right )}\, dx + \int b^{2} c x^{4} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b c x^{4} \operatorname{atanh}{\left (c x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(Integral(a**2*x**3, x) + Integral(a**2*c*x**4, x) + Integral(b**2*x**3*atanh(c*x)**2, x) + Integral(2*a*b*x
**3*atanh(c*x), x) + Integral(b**2*c*x**4*atanh(c*x)**2, x) + Integral(2*a*b*c*x**4*atanh(c*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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