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

Optimal. Leaf size=244 \[ -\frac{3 b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 e \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{2 c^2}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}-\frac{3 b^2 e \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2}-\frac{\left (\frac{e^2}{c^2}+d^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{3 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.601792, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5928, 5910, 5984, 5918, 2402, 2315, 6048, 5948, 6058, 6610} \[ -\frac{3 b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 e \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{2 c^2}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}-\frac{3 b^2 e \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2}-\frac{\left (\frac{e^2}{c^2}+d^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{3 b d \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 (d+e x) \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{(3 b c) \int \left (-\frac{e^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2}+\frac{\left (c^2 d^2+e^2+2 c^2 d e x\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{(3 b) \int \frac{\left (c^2 d^2+e^2+2 c^2 d e x\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c e}+\frac{(3 b e) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c}\\ &=\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{(3 b) \int \left (\frac{c^2 d^2 \left (1+\frac{e^2}{c^2 d^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}+\frac{2 c^2 d e x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx}{2 c e}-\left (3 b^2 e\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-(3 b c d) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx-\frac{\left (3 b^2 e\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c}-\frac{\left (3 b \left (c^2 d^2+e^2\right )\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c e}\\ &=\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{\left (d^2+\frac{e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^2}-(3 b d) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx+\frac{\left (3 b^3 e\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c}\\ &=\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{\left (d^2+\frac{e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^2}-\frac{3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}+\left (6 b^2 d\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx-\frac{\left (3 b^3 e\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c^2}\\ &=\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{\left (d^2+\frac{e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^2}-\frac{3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b^3 e \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{2 c^2}-\frac{3 b^2 d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\left (3 b^3 d\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac{3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{\left (d^2+\frac{e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac{3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^2}-\frac{3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{3 b^3 e \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{2 c^2}-\frac{3 b^2 d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.762874, size = 331, normalized size = 1.36 \[ \frac{12 a b^2 c d \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \left ((c x-1) \tanh ^{-1}(c x)-2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )\right )-2 b^3 e \left (\tanh ^{-1}(c x) \left (\left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2+(3-3 c x) \tanh ^{-1}(c x)+6 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )-3 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+4 b^3 c d \left (3 \tanh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x)^2 \left ((c x-1) \tanh ^{-1}(c x)-3 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )\right )+6 a^2 b c^2 x \tanh ^{-1}(c x) (2 d+e x)+2 a^2 c x (2 a c d+3 b e)+3 a^2 b (2 c d+e) \log (1-c x)+3 a^2 b (2 c d-e) \log (c x+1)+2 a^3 c^2 e x^2+6 a b^2 e \left (\log \left (1-c^2 x^2\right )+\left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^2+2 c x \tanh ^{-1}(c x)\right )}{4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.954, size = 12404, normalized size = 50.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{3} e x^{2} + \frac{3}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a^{2} b e + a^{3} d x + \frac{3 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a^{2} b d}{2 \, c} - \frac{{\left (b^{3} c^{2} e x^{2} + 2 \, b^{3} c^{2} d x -{\left (2 \, c d + e\right )} b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \,{\left (2 \, a b^{2} c^{2} e x^{2} + 2 \,{\left (2 \, a b^{2} c^{2} d + b^{3} c e\right )} x +{\left (b^{3} c^{2} e x^{2} + 2 \, b^{3} c^{2} d x +{\left (2 \, c d - e\right )} b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{16 \, c^{2}} - \int -\frac{{\left (b^{3} c^{2} e x^{2} - b^{3} c d +{\left (c^{2} d - c e\right )} b^{3} x\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c^{2} e x^{2} - a b^{2} c d +{\left (c^{2} d - c e\right )} a b^{2} x\right )} \log \left (c x + 1\right )^{2} - 3 \,{\left (2 \, a b^{2} c^{2} e x^{2} +{\left (b^{3} c^{2} e x^{2} - b^{3} c d +{\left (c^{2} d - c e\right )} b^{3} x\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (2 \, a b^{2} c^{2} d + b^{3} c e\right )} x -{\left (4 \, a b^{2} c d -{\left (2 \, c d - e\right )} b^{3} -{\left (4 \, a b^{2} c^{2} e + b^{3} c^{2} e\right )} x^{2} - 2 \,{\left (b^{3} c^{2} d + 2 \,{\left (c^{2} d - c e\right )} a b^{2}\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c^{2} x - c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*e*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*e + a^3*d*x
+ 3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b*d/c - 1/16*((b^3*c^2*e*x^2 + 2*b^3*c^2*d*x - (2*c*d + e)
*b^3)*log(-c*x + 1)^3 - 3*(2*a*b^2*c^2*e*x^2 + 2*(2*a*b^2*c^2*d + b^3*c*e)*x + (b^3*c^2*e*x^2 + 2*b^3*c^2*d*x
+ (2*c*d - e)*b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c^2 - integrate(-1/8*((b^3*c^2*e*x^2 - b^3*c*d + (c^2*d - c*
e)*b^3*x)*log(c*x + 1)^3 + 6*(a*b^2*c^2*e*x^2 - a*b^2*c*d + (c^2*d - c*e)*a*b^2*x)*log(c*x + 1)^2 - 3*(2*a*b^2
*c^2*e*x^2 + (b^3*c^2*e*x^2 - b^3*c*d + (c^2*d - c*e)*b^3*x)*log(c*x + 1)^2 + 2*(2*a*b^2*c^2*d + b^3*c*e)*x -
(4*a*b^2*c*d - (2*c*d - e)*b^3 - (4*a*b^2*c^2*e + b^3*c^2*e)*x^2 - 2*(b^3*c^2*d + 2*(c^2*d - c*e)*a*b^2)*x)*lo
g(c*x + 1))*log(-c*x + 1))/(c^2*x - c), x)

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

[Out]

integral(a^3*e*x + a^3*d + (b^3*e*x + b^3*d)*arctanh(c*x)^3 + 3*(a*b^2*e*x + a*b^2*d)*arctanh(c*x)^2 + 3*(a^2*
b*e*x + a^2*b*d)*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 )^{3} \left (d + e x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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