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

Optimal. Leaf size=359 \[ -\frac{b^2 d \left (c^2 d^2+e^2\right ) \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^3}+\frac{a b e x \left (6 c^2 d^2+e^2\right )}{2 c^3}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (6 c^2 d^2 e^2+c^4 d^4+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}-\frac{2 b d \left (c^2 d^2+e^2\right ) \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac{b^2 e x \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}(c x)}{2 c^3}+\frac{b^2 d e^2 x}{c^2}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e^3 x^2}{12 c^2}+\frac{b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4} \]

[Out]

(b^2*d*e^2*x)/c^2 + (a*b*e*(6*c^2*d^2 + e^2)*x)/(2*c^3) + (b^2*e^3*x^2)/(12*c^2) - (b^2*d*e^2*ArcTanh[c*x])/c^
3 + (b^2*e*(6*c^2*d^2 + e^2)*x*ArcTanh[c*x])/(2*c^3) + (b*d*e^2*x^2*(a + b*ArcTanh[c*x]))/c + (b*e^3*x^3*(a +
b*ArcTanh[c*x]))/(6*c) + (d*(c^2*d^2 + e^2)*(a + b*ArcTanh[c*x])^2)/c^3 - ((c^4*d^4 + 6*c^2*d^2*e^2 + e^4)*(a
+ b*ArcTanh[c*x])^2)/(4*c^4*e) + ((d + e*x)^4*(a + b*ArcTanh[c*x])^2)/(4*e) - (2*b*d*(c^2*d^2 + e^2)*(a + b*Ar
cTanh[c*x])*Log[2/(1 - c*x)])/c^3 + (b^2*e^3*Log[1 - c^2*x^2])/(12*c^4) + (b^2*e*(6*c^2*d^2 + e^2)*Log[1 - c^2
*x^2])/(4*c^4) - (b^2*d*(c^2*d^2 + e^2)*PolyLog[2, 1 - 2/(1 - c*x)])/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.533985, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {5928, 5910, 260, 5916, 321, 206, 266, 43, 6048, 5948, 5984, 5918, 2402, 2315} \[ -\frac{b^2 d \left (c^2 d^2+e^2\right ) \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^3}+\frac{a b e x \left (6 c^2 d^2+e^2\right )}{2 c^3}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (6 c^2 d^2 e^2+c^4 d^4+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}-\frac{2 b d \left (c^2 d^2+e^2\right ) \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac{b^2 e x \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}(c x)}{2 c^3}+\frac{b^2 d e^2 x}{c^2}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e^3 x^2}{12 c^2}+\frac{b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*d*e^2*x)/c^2 + (a*b*e*(6*c^2*d^2 + e^2)*x)/(2*c^3) + (b^2*e^3*x^2)/(12*c^2) - (b^2*d*e^2*ArcTanh[c*x])/c^
3 + (b^2*e*(6*c^2*d^2 + e^2)*x*ArcTanh[c*x])/(2*c^3) + (b*d*e^2*x^2*(a + b*ArcTanh[c*x]))/c + (b*e^3*x^3*(a +
b*ArcTanh[c*x]))/(6*c) + (d*(c^2*d^2 + e^2)*(a + b*ArcTanh[c*x])^2)/c^3 - ((c^4*d^4 + 6*c^2*d^2*e^2 + e^4)*(a
+ b*ArcTanh[c*x])^2)/(4*c^4*e) + ((d + e*x)^4*(a + b*ArcTanh[c*x])^2)/(4*e) - (2*b*d*(c^2*d^2 + e^2)*(a + b*Ar
cTanh[c*x])*Log[2/(1 - c*x)])/c^3 + (b^2*e^3*Log[1 - c^2*x^2])/(12*c^4) + (b^2*e*(6*c^2*d^2 + e^2)*Log[1 - c^2
*x^2])/(4*c^4) - (b^2*d*(c^2*d^2 + e^2)*PolyLog[2, 1 - 2/(1 - c*x)])/c^3

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 6048

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 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 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 (d+e x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{(b c) \int \left (-\frac{e^2 \left (6 c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4}-\frac{4 d e^3 x \left (a+b \tanh ^{-1}(c x)\right )}{c^2}-\frac{e^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c^2}+\frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{b \int \frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c^3 e}+\frac{\left (2 b d e^2\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c}+\frac{\left (b e^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}+\frac{\left (b e \left (6 c^2 d^2+e^2\right )\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^3}\\ &=\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{b \int \left (\frac{c^4 d^4 \left (1+\frac{6 c^2 d^2 e^2+e^4}{c^4 d^4}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}+\frac{4 c^2 d e \left (c^2 d^2+e^2\right ) x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}\right ) \, dx}{2 c^3 e}-\left (b^2 d e^2\right ) \int \frac{x^2}{1-c^2 x^2} \, dx-\frac{1}{6} \left (b^2 e^3\right ) \int \frac{x^3}{1-c^2 x^2} \, dx+\frac{\left (b^2 e \left (6 c^2 d^2+e^2\right )\right ) \int \tanh ^{-1}(c x) \, dx}{2 c^3}\\ &=\frac{b^2 d e^2 x}{c^2}+\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{\left (b^2 d e^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{c^2}-\frac{1}{12} \left (b^2 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )-\frac{\left (2 b d \left (c^2 d^2+e^2\right )\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c}-\frac{\left (b^2 e \left (6 c^2 d^2+e^2\right )\right ) \int \frac{x}{1-c^2 x^2} \, dx}{2 c^2}-\frac{\left (b \left (c^4 d^4+6 c^2 d^2 e^2+e^4\right )\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{2 c^3 e}\\ &=\frac{b^2 d e^2 x}{c^2}+\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac{1}{12} \left (b^2 e^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{\left (2 b d \left (c^2 d^2+e^2\right )\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^2}\\ &=\frac{b^2 d e^2 x}{c^2}+\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3}+\frac{b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac{\left (2 b^2 d \left (c^2 d^2+e^2\right )\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^2}\\ &=\frac{b^2 d e^2 x}{c^2}+\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3}+\frac{b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac{\left (2 b^2 d \left (c^2 d^2+e^2\right )\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c^3}\\ &=\frac{b^2 d e^2 x}{c^2}+\frac{a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac{b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac{d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac{2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^3}+\frac{b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac{b^2 d \left (c^2 d^2+e^2\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.886709, size = 506, normalized size = 1.41 \[ \frac{12 b^2 c d \left (c^2 d^2+e^2\right ) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+18 a^2 c^4 d^2 e x^2+12 a^2 c^4 d^3 x+12 a^2 c^4 d e^2 x^3+3 a^2 c^4 e^3 x^4+2 b c \tanh ^{-1}(c x) \left (3 a c^3 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+b e \left (18 c^2 d^2 x+6 d e \left (c^2 x^2-1\right )+e^2 x \left (c^2 x^2+3\right )\right )-12 b d \left (c^2 d^2+e^2\right ) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+36 a b c^3 d^2 e x+18 a b c^2 d^2 e \log (1-c x)-18 a b c^2 d^2 e \log (c x+1)+12 a b c^3 d^3 \log \left (1-c^2 x^2\right )+12 a b c^3 d e^2 x^2+12 a b c d e^2 \log \left (c^2 x^2-1\right )+2 a b c^3 e^3 x^3+6 a b c e^3 x+3 a b e^3 \log (1-c x)-3 a b e^3 \log (c x+1)+3 b^2 \tanh ^{-1}(c x)^2 \left (c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )-6 c^2 d^2 e-4 c^3 d^3-4 c d e^2-e^3\right )+18 b^2 c^2 d^2 e \log \left (1-c^2 x^2\right )+12 b^2 c^2 d e^2 x+b^2 c^2 e^3 x^2+4 b^2 e^3 \log \left (1-c^2 x^2\right )-b^2 e^3}{12 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b^2*e^3) + 12*a^2*c^4*d^3*x + 36*a*b*c^3*d^2*e*x + 12*b^2*c^2*d*e^2*x + 6*a*b*c*e^3*x + 18*a^2*c^4*d^2*e*x^
2 + 12*a*b*c^3*d*e^2*x^2 + b^2*c^2*e^3*x^2 + 12*a^2*c^4*d*e^2*x^3 + 2*a*b*c^3*e^3*x^3 + 3*a^2*c^4*e^3*x^4 + 3*
b^2*(-4*c^3*d^3 - 6*c^2*d^2*e - 4*c*d*e^2 - e^3 + c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3))*ArcTanh[c
*x]^2 + 2*b*c*ArcTanh[c*x]*(3*a*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b*e*(18*c^2*d^2*x + 6*d*e*
(-1 + c^2*x^2) + e^2*x*(3 + c^2*x^2)) - 12*b*d*(c^2*d^2 + e^2)*Log[1 + E^(-2*ArcTanh[c*x])]) + 18*a*b*c^2*d^2*
e*Log[1 - c*x] + 3*a*b*e^3*Log[1 - c*x] - 18*a*b*c^2*d^2*e*Log[1 + c*x] - 3*a*b*e^3*Log[1 + c*x] + 12*a*b*c^3*
d^3*Log[1 - c^2*x^2] + 18*b^2*c^2*d^2*e*Log[1 - c^2*x^2] + 4*b^2*e^3*Log[1 - c^2*x^2] + 12*a*b*c*d*e^2*Log[-1
+ c^2*x^2] + 12*b^2*c*d*(c^2*d^2 + e^2)*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(12*c^4)

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Maple [B]  time = 0.058, size = 1430, 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((e*x+d)^3*(a+b*arctanh(c*x))^2,x)

[Out]

a^2*e^2*x^3*d+3/2*a^2*e*x^2*d^2+1/2*a*b/c^3*x*e^3+1/4*a^2/e*d^4+3/c*b^2*e*arctanh(c*x)*x*d^2+1/c^3*b^2*e^2*arc
tanh(c*x)*ln(c*x+1)*d-1/2/c^3*b^2*e^2*ln(1/2+1/2*c*x)*ln(c*x-1)*d+1/c^3*b^2*e^2*arctanh(c*x)*ln(c*x-1)*d+1/c*b
^2*e^2*arctanh(c*x)*x^2*d+3/2/c^2*b^2*e*arctanh(c*x)*ln(c*x-1)*d^2+3/4/c^2*b^2*e*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1
/2)*d^2-3/2/c^2*b^2*e*arctanh(c*x)*ln(c*x+1)*d^2+1/c^3*a*b*e^2*ln(c*x+1)*d-3/2/c^2*a*b*e*ln(c*x+1)*d^2+3/2/c^2
*a*b*e*ln(c*x-1)*d^2+1/c^3*a*b*e^2*ln(c*x-1)*d+1/4*a^2*e^3*x^4+a^2*x*d^3+1/16/c^4*b^2*e^3*ln(c*x+1)^2+1/3/c^4*
b^2*e^3*ln(c*x-1)+1/3/c^4*b^2*e^3*ln(c*x+1)-1/4/c*b^2*ln(c*x+1)^2*d^3-1/c*b^2*dilog(1/2+1/2*c*x)*d^3+1/4/c*b^2
*ln(c*x-1)^2*d^3+1/16/c^4*b^2*e^3*ln(c*x-1)^2+1/4*b^2*e^3*arctanh(c*x)^2*x^4+1/16*b^2/e*ln(c*x+1)^2*d^4+1/16*b
^2/e*ln(c*x-1)^2*d^4+1/4*b^2/e*arctanh(c*x)^2*d^4+3*a*b/c*x*d^2*e-3/4/c^2*b^2*e*ln(1/2+1/2*c*x)*ln(c*x-1)*d^2+
2*a*b*e^2*arctanh(c*x)*x^3*d+3*a*b*e*arctanh(c*x)*x^2*d^2+1/c*a*b*x^2*d*e^2-1/2/c^3*b^2*e^2*ln(1/2+1/2*c*x)*ln
(-1/2*c*x+1/2)*d-3/4/c^2*b^2*e*ln(-1/2*c*x+1/2)*ln(c*x+1)*d^2+1/2/c^3*b^2*e^2*ln(-1/2*c*x+1/2)*ln(c*x+1)*d-1/2
/c*b^2*ln(1/2+1/2*c*x)*ln(c*x-1)*d^3+1/6/c*b^2*e^3*arctanh(c*x)*x^3-1/2/c*b^2*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)
*d^3+1/2/c*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)*d^3+1/c*b^2*arctanh(c*x)*ln(c*x-1)*d^3-1/2/c^3*b^2*e^2*ln(c*x+1)*d-1
/c^3*b^2*e^2*dilog(1/2+1/2*c*x)*d+3/8/c^2*b^2*e*ln(c*x+1)^2*d^2+1/6/c*a*b*x^3*e^3+1/4*b^2/e*arctanh(c*x)*ln(c*
x-1)*d^4-1/4*b^2/e*arctanh(c*x)*ln(c*x+1)*d^4+1/8*b^2/e*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)*d^4+b^2*e^2*arctanh(c
*x)^2*x^3*d-1/8*b^2/e*ln(-1/2*c*x+1/2)*ln(c*x+1)*d^4-1/8*b^2/e*ln(1/2+1/2*c*x)*ln(c*x-1)*d^4+2*a*b*arctanh(c*x
)*x*d^3+3/2*b^2*e*arctanh(c*x)^2*x^2*d^2+1/2*a*b*e^3*arctanh(c*x)*x^4+1/2*a*b/e*arctanh(c*x)*d^4+1/2/c^3*b^2*e
^2*ln(c*x-1)*d+1/2/c^3*b^2*e^3*arctanh(c*x)*x-1/8/c^4*b^2*e^3*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/8/c^4*b^2*e^3*ln(1/
2+1/2*c*x)*ln(c*x-1)+1/4*a*b/e*ln(c*x-1)*d^4-1/4*a*b/e*ln(c*x+1)*d^4+1/c*a*b*ln(c*x+1)*d^3+1/c*a*b*ln(c*x-1)*d
^3+1/c*b^2*arctanh(c*x)*ln(c*x+1)*d^3+1/4/c^4*b^2*e^3*arctanh(c*x)*ln(c*x-1)-1/4/c^4*b^2*e^3*arctanh(c*x)*ln(c
*x+1)+1/8/c^4*b^2*e^3*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)-1/4/c^3*b^2*e^2*ln(c*x+1)^2*d+1/4/c^4*a*b*e^3*ln(c*x-1)
-1/4/c^4*a*b*e^3*ln(c*x+1)+3/8/c^2*b^2*e*ln(c*x-1)^2*d^2+1/4/c^3*b^2*e^2*ln(c*x-1)^2*d+3/2/c^2*b^2*e*ln(c*x-1)
*d^2+3/2/c^2*b^2*e*ln(c*x+1)*d^2+b^2*arctanh(c*x)^2*x*d^3+1/12*b^2*e^3*x^2/c^2+b^2*d*e^2*x/c^2

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Maxima [B]  time = 1.79891, size = 1056, normalized size = 2.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*e^3*x^4 + a^2*d*e^2*x^3 + 3/2*a^2*d^2*e*x^2 + 3/2*(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^2*e + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*d*e^2 + 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*e^3 + a^2*d^3*x +
(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*d^3/c + (c^2*d^3 + d*e^2)*(log(c*x + 1)*log(-1/2*c*x + 1/2) + di
log(1/2*c*x + 1/2))*b^2/c^3 + 1/6*(9*c^2*d^2*e - 3*c*d*e^2 + 2*e^3)*b^2*log(c*x + 1)/c^4 + 1/6*(9*c^2*d^2*e +
3*c*d*e^2 + 2*e^3)*b^2*log(c*x - 1)/c^4 + 1/48*(4*b^2*c^2*e^3*x^2 + 48*b^2*c^2*d*e^2*x + 3*(b^2*c^4*e^3*x^4 +
4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x + (4*c^3*d^3 - 6*c^2*d^2*e + 4*c*d*e^2 - e^3)*b^2)
*log(c*x + 1)^2 + 3*(b^2*c^4*e^3*x^4 + 4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x - (4*c^3*d^
3 + 6*c^2*d^2*e + 4*c*d*e^2 + e^3)*b^2)*log(-c*x + 1)^2 + 4*(b^2*c^3*e^3*x^3 + 6*b^2*c^3*d*e^2*x^2 + 3*(6*c^3*
d^2*e + c*e^3)*b^2*x)*log(c*x + 1) - 2*(2*b^2*c^3*e^3*x^3 + 12*b^2*c^3*d*e^2*x^2 + 6*(6*c^3*d^2*e + c*e^3)*b^2
*x + 3*(b^2*c^4*e^3*x^4 + 4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x + (4*c^3*d^3 - 6*c^2*d^2
*e + 4*c*d*e^2 - e^3)*b^2)*log(c*x + 1))*log(-c*x + 1))/c^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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