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

Optimal. Leaf size=286 \[ -\frac{6 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^2}+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}-\frac{12 b d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{5 a b d^3 x}{2 c}+\frac{3 b^2 d^3 \log \left (1-c^2 x^2\right )}{2 c^2}-\frac{13 b^2 d^3 \tanh ^{-1}(c x)}{10 c^2}+\frac{1}{30} b^2 c d^3 x^3+\frac{13 b^2 d^3 x}{10 c}+\frac{5 b^2 d^3 x \tanh ^{-1}(c x)}{2 c}+\frac{1}{4} b^2 d^3 x^2 \]

[Out]

(5*a*b*d^3*x)/(2*c) + (13*b^2*d^3*x)/(10*c) + (b^2*d^3*x^2)/4 + (b^2*c*d^3*x^3)/30 - (13*b^2*d^3*ArcTanh[c*x])
/(10*c^2) + (5*b^2*d^3*x*ArcTanh[c*x])/(2*c) + (6*b*d^3*x^2*(a + b*ArcTanh[c*x]))/5 + (b*c*d^3*x^3*(a + b*ArcT
anh[c*x]))/2 + (b*c^2*d^3*x^4*(a + b*ArcTanh[c*x]))/10 - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x])^2)/(4*c^2) + (d
^3*(1 + c*x)^5*(a + b*ArcTanh[c*x])^2)/(5*c^2) - (12*b*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^2) + (3
*b^2*d^3*Log[1 - c^2*x^2])/(2*c^2) - (6*b^2*d^3*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.597233, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 38, number of rules used = 14, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5940, 5928, 5910, 260, 5916, 321, 206, 266, 43, 1586, 5918, 2402, 2315, 302} \[ -\frac{6 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^2}+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}-\frac{12 b d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{5 a b d^3 x}{2 c}+\frac{3 b^2 d^3 \log \left (1-c^2 x^2\right )}{2 c^2}-\frac{13 b^2 d^3 \tanh ^{-1}(c x)}{10 c^2}+\frac{1}{30} b^2 c d^3 x^3+\frac{13 b^2 d^3 x}{10 c}+\frac{5 b^2 d^3 x \tanh ^{-1}(c x)}{2 c}+\frac{1}{4} b^2 d^3 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*b*d^3*x)/(2*c) + (13*b^2*d^3*x)/(10*c) + (b^2*d^3*x^2)/4 + (b^2*c*d^3*x^3)/30 - (13*b^2*d^3*ArcTanh[c*x])
/(10*c^2) + (5*b^2*d^3*x*ArcTanh[c*x])/(2*c) + (6*b*d^3*x^2*(a + b*ArcTanh[c*x]))/5 + (b*c*d^3*x^3*(a + b*ArcT
anh[c*x]))/2 + (b*c^2*d^3*x^4*(a + b*ArcTanh[c*x]))/10 - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x])^2)/(4*c^2) + (d
^3*(1 + c*x)^5*(a + b*ArcTanh[c*x])^2)/(5*c^2) - (12*b*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^2) + (3
*b^2*d^3*Log[1 - c^2*x^2])/(2*c^2) - (6*b^2*d^3*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^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 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 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 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 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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 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]

Rubi steps

\begin{align*} \int x (d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\int \left (-\frac{(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac{(d+c d x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}\right ) \, dx\\ &=-\frac{\int (d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c}+\frac{\int (d+c d x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{(2 b) \int \left (-15 d^5 \left (a+b \tanh ^{-1}(c x)\right )-11 c d^5 x \left (a+b \tanh ^{-1}(c x)\right )-5 c^2 d^5 x^2 \left (a+b \tanh ^{-1}(c x)\right )-c^3 d^5 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{16 \left (d^5+c d^5 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}\right ) \, dx}{5 c d^2}+\frac{b \int \left (-7 d^4 \left (a+b \tanh ^{-1}(c x)\right )-4 c d^4 x \left (a+b \tanh ^{-1}(c x)\right )-c^2 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{8 \left (d^4+c d^4 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}\right ) \, dx}{2 c d}\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{(32 b) \int \frac{\left (d^5+c d^5 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c d^2}+\frac{(4 b) \int \frac{\left (d^4+c d^4 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d}-\left (2 b d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\frac{1}{5} \left (22 b d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\frac{\left (7 b d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}+\frac{\left (6 b d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c}-\frac{1}{2} \left (b c d^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (2 b c d^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\frac{1}{5} \left (2 b c^2 d^3\right ) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=\frac{5 a b d^3 x}{2 c}+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{(32 b) \int \frac{a+b \tanh ^{-1}(c x)}{\frac{1}{d^5}-\frac{c x}{d^5}} \, dx}{5 c d^2}+\frac{(4 b) \int \frac{a+b \tanh ^{-1}(c x)}{\frac{1}{d^4}-\frac{c x}{d^4}} \, dx}{c d}-\frac{\left (7 b^2 d^3\right ) \int \tanh ^{-1}(c x) \, dx}{2 c}+\frac{\left (6 b^2 d^3\right ) \int \tanh ^{-1}(c x) \, dx}{c}+\left (b^2 c d^3\right ) \int \frac{x^2}{1-c^2 x^2} \, dx-\frac{1}{5} \left (11 b^2 c d^3\right ) \int \frac{x^2}{1-c^2 x^2} \, dx+\frac{1}{6} \left (b^2 c^2 d^3\right ) \int \frac{x^3}{1-c^2 x^2} \, dx-\frac{1}{3} \left (2 b^2 c^2 d^3\right ) \int \frac{x^3}{1-c^2 x^2} \, dx-\frac{1}{10} \left (b^2 c^3 d^3\right ) \int \frac{x^4}{1-c^2 x^2} \, dx\\ &=\frac{5 a b d^3 x}{2 c}+\frac{6 b^2 d^3 x}{5 c}+\frac{5 b^2 d^3 x \tanh ^{-1}(c x)}{2 c}+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^2}+\frac{1}{2} \left (7 b^2 d^3\right ) \int \frac{x}{1-c^2 x^2} \, dx-\left (6 b^2 d^3\right ) \int \frac{x}{1-c^2 x^2} \, dx+\frac{\left (b^2 d^3\right ) \int \frac{1}{1-c^2 x^2} \, dx}{c}-\frac{\left (11 b^2 d^3\right ) \int \frac{1}{1-c^2 x^2} \, dx}{5 c}-\frac{\left (4 b^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c}+\frac{\left (32 b^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c}+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )-\frac{1}{3} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )-\frac{1}{10} \left (b^2 c^3 d^3\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{5 a b d^3 x}{2 c}+\frac{13 b^2 d^3 x}{10 c}+\frac{1}{30} b^2 c d^3 x^3-\frac{6 b^2 d^3 \tanh ^{-1}(c x)}{5 c^2}+\frac{5 b^2 d^3 x \tanh ^{-1}(c x)}{2 c}+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^2}+\frac{5 b^2 d^3 \log \left (1-c^2 x^2\right )}{4 c^2}+\frac{\left (4 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c^2}-\frac{\left (32 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{5 c^2}-\frac{\left (b^2 d^3\right ) \int \frac{1}{1-c^2 x^2} \, dx}{10 c}+\frac{1}{12} \left (b^2 c^2 d^3\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{1}{3} \left (b^2 c^2 d^3\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{5 a b d^3 x}{2 c}+\frac{13 b^2 d^3 x}{10 c}+\frac{1}{4} b^2 d^3 x^2+\frac{1}{30} b^2 c d^3 x^3-\frac{13 b^2 d^3 \tanh ^{-1}(c x)}{10 c^2}+\frac{5 b^2 d^3 x \tanh ^{-1}(c x)}{2 c}+\frac{6}{5} b d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{10} b c^2 d^3 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^2}+\frac{d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^2}+\frac{3 b^2 d^3 \log \left (1-c^2 x^2\right )}{2 c^2}-\frac{6 b^2 d^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^2}\\ \end{align*}

Mathematica [A]  time = 1.18882, size = 325, normalized size = 1.14 \[ \frac{d^3 \left (72 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+12 a^2 c^5 x^5+45 a^2 c^4 x^4+60 a^2 c^3 x^3+30 a^2 c^2 x^2+6 a b c^4 x^4+30 a b c^3 x^3+72 a b c^2 x^2+72 a b \log \left (c^2 x^2-1\right )+6 b \tanh ^{-1}(c x) \left (a c^2 x^2 \left (4 c^3 x^3+15 c^2 x^2+20 c x+10\right )+b \left (c^4 x^4+5 c^3 x^3+12 c^2 x^2+25 c x-13\right )-24 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+150 a b c x+75 a b \log (1-c x)-75 a b \log (c x+1)-18 a b+2 b^2 c^3 x^3+15 b^2 c^2 x^2+90 b^2 \log \left (1-c^2 x^2\right )+3 b^2 \left (4 c^5 x^5+15 c^4 x^4+20 c^3 x^3+10 c^2 x^2-49\right ) \tanh ^{-1}(c x)^2+78 b^2 c x-15 b^2\right )}{60 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-18*a*b - 15*b^2 + 150*a*b*c*x + 78*b^2*c*x + 30*a^2*c^2*x^2 + 72*a*b*c^2*x^2 + 15*b^2*c^2*x^2 + 60*a^2*
c^3*x^3 + 30*a*b*c^3*x^3 + 2*b^2*c^3*x^3 + 45*a^2*c^4*x^4 + 6*a*b*c^4*x^4 + 12*a^2*c^5*x^5 + 3*b^2*(-49 + 10*c
^2*x^2 + 20*c^3*x^3 + 15*c^4*x^4 + 4*c^5*x^5)*ArcTanh[c*x]^2 + 6*b*ArcTanh[c*x]*(a*c^2*x^2*(10 + 20*c*x + 15*c
^2*x^2 + 4*c^3*x^3) + b*(-13 + 25*c*x + 12*c^2*x^2 + 5*c^3*x^3 + c^4*x^4) - 24*b*Log[1 + E^(-2*ArcTanh[c*x])])
+ 75*a*b*Log[1 - c*x] - 75*a*b*Log[1 + c*x] + 90*b^2*Log[1 - c^2*x^2] + 72*a*b*Log[-1 + c^2*x^2] + 72*b^2*Pol
yLog[2, -E^(-2*ArcTanh[c*x])]))/(60*c^2)

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Maple [B]  time = 0.054, size = 570, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/10*b^2*d^3*x/c+1/30*b^2*c*d^3*x^3+3/4*c^2*d^3*a^2*x^4+c*d^3*a^2*x^3+1/5*c^3*d^3*a^2*x^5+49/80/c^2*d^3*b^2*l
n(c*x-1)^2+1/80/c^2*d^3*b^2*ln(c*x+1)^2-6/5/c^2*d^3*b^2*dilog(1/2+1/2*c*x)+17/20/c^2*d^3*b^2*ln(c*x+1)+43/20/c
^2*d^3*b^2*ln(c*x-1)+6/5*d^3*b^2*arctanh(c*x)*x^2+1/2*d^3*b^2*arctanh(c*x)^2*x^2+6/5*d^3*a*b*x^2+1/2*c*d^3*a*b
*x^3-1/20/c^2*d^3*a*b*ln(c*x+1)+3/4*c^2*d^3*b^2*arctanh(c*x)^2*x^4+1/10*c^2*d^3*b^2*arctanh(c*x)*x^4+1/2*c*d^3
*b^2*arctanh(c*x)*x^3+49/20/c^2*d^3*b^2*arctanh(c*x)*ln(c*x-1)-1/20/c^2*d^3*b^2*arctanh(c*x)*ln(c*x+1)-1/40/c^
2*d^3*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/10*c^2*d^3*a*b*x^4-49/40/c^2*d^3*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+c*d^3*b^
2*arctanh(c*x)^2*x^3+49/20/c^2*d^3*a*b*ln(c*x-1)+d^3*a*b*arctanh(c*x)*x^2+1/40/c^2*d^3*b^2*ln(-1/2*c*x+1/2)*ln
(1/2+1/2*c*x)+1/5*c^3*d^3*b^2*arctanh(c*x)^2*x^5+1/2*d^3*a^2*x^2+1/4*b^2*d^3*x^2+5/2*a*b*d^3*x/c+5/2*b^2*d^3*x
*arctanh(c*x)/c+2/5*c^3*d^3*a*b*arctanh(c*x)*x^5+2*c*d^3*a*b*arctanh(c*x)*x^3+3/2*c^2*d^3*a*b*arctanh(c*x)*x^4

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Maxima [B]  time = 2.15455, size = 1053, normalized size = 3.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*c^3*d^3*x^5 + 3/4*a^2*c^2*d^3*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^3*d^3 + a^2*c*d^3*x^3 + 1/2*b^2*d^3*x^2*arctanh(c*x)^2 + 1/4*(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*c^2*d^3 + (2*x^3*arctanh(c*x) + c*(x^2/c^2 +
log(c^2*x^2 - 1)/c^4))*a*b*c*d^3 + 1/2*a^2*d^3*x^2 + 1/2*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 +
log(c*x - 1)/c^3))*a*b*d^3 + 1/8*(4*c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3)*arctanh(c*x) - (2*(log(
c*x - 1) - 2)*log(c*x + 1) - log(c*x + 1)^2 - log(c*x - 1)^2 - 4*log(c*x - 1))/c^2)*b^2*d^3 + 6/5*(log(c*x + 1
)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*d^3/c^2 + 7/20*b^2*d^3*log(c*x + 1)/c^2 + 33/20*b^2*d^3*log(
c*x - 1)/c^2 + 1/240*(8*b^2*c^3*d^3*x^3 + 60*b^2*c^2*d^3*x^2 + 312*b^2*c*d^3*x + 3*(4*b^2*c^5*d^3*x^5 + 15*b^2
*c^4*d^3*x^4 + 20*b^2*c^3*d^3*x^3 + 9*b^2*d^3)*log(c*x + 1)^2 + 3*(4*b^2*c^5*d^3*x^5 + 15*b^2*c^4*d^3*x^4 + 20
*b^2*c^3*d^3*x^3 - 39*b^2*d^3)*log(-c*x + 1)^2 + 12*(b^2*c^4*d^3*x^4 + 5*b^2*c^3*d^3*x^3 + 12*b^2*c^2*d^3*x^2
+ 15*b^2*c*d^3*x)*log(c*x + 1) - 6*(2*b^2*c^4*d^3*x^4 + 10*b^2*c^3*d^3*x^3 + 24*b^2*c^2*d^3*x^2 + 30*b^2*c*d^3
*x + (4*b^2*c^5*d^3*x^5 + 15*b^2*c^4*d^3*x^4 + 20*b^2*c^3*d^3*x^3 + 9*b^2*d^3)*log(c*x + 1))*log(-c*x + 1))/c^
2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int a^{2} x\, dx + \int 3 a^{2} c x^{2}\, dx + \int 3 a^{2} c^{2} x^{3}\, dx + \int a^{2} c^{3} x^{4}\, dx + \int b^{2} x \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b x \operatorname{atanh}{\left (c x \right )}\, dx + \int 3 b^{2} c x^{2} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 3 b^{2} c^{2} x^{3} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int b^{2} c^{3} x^{4} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 6 a b c x^{2} \operatorname{atanh}{\left (c x \right )}\, dx + \int 6 a b c^{2} x^{3} \operatorname{atanh}{\left (c x \right )}\, dx + \int 2 a b c^{3} 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*(c*d*x+d)**3*(a+b*atanh(c*x))**2,x)

[Out]

d**3*(Integral(a**2*x, x) + Integral(3*a**2*c*x**2, x) + Integral(3*a**2*c**2*x**3, x) + Integral(a**2*c**3*x*
*4, x) + Integral(b**2*x*atanh(c*x)**2, x) + Integral(2*a*b*x*atanh(c*x), x) + Integral(3*b**2*c*x**2*atanh(c*
x)**2, x) + Integral(3*b**2*c**2*x**3*atanh(c*x)**2, x) + Integral(b**2*c**3*x**4*atanh(c*x)**2, x) + Integral
(6*a*b*c*x**2*atanh(c*x), x) + Integral(6*a*b*c**2*x**3*atanh(c*x), x) + Integral(2*a*b*c**3*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 )}^{3}{\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(x*(c*d*x+d)^3*(a+b*arctanh(c*x))^2,x, algorithm="giac")

[Out]

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