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

Optimal. Leaf size=159 \[ \frac{e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac{b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac{e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac{1}{2} a b e^3 x+\frac{b^2 e^3 (c+d x)^2}{12 d}+\frac{b^2 e^3 \log \left (1-(c+d x)^2\right )}{3 d}+\frac{b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.247547, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {6107, 12, 5916, 5980, 266, 43, 5910, 260, 5948} \[ \frac{e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac{b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac{e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac{1}{2} a b e^3 x+\frac{b^2 e^3 (c+d x)^2}{12 d}+\frac{b^2 e^3 \log \left (1-(c+d x)^2\right )}{3 d}+\frac{b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

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

Mathematica [A]  time = 0.180823, size = 148, normalized size = 0.93 \[ \frac{e^3 \left (3 a^2 (c+d x)^4+2 a b (c+d x)^3+6 a b (c+d x)+b (3 a+4 b) \log (-c-d x+1)+b (4 b-3 a) \log (c+d x+1)+2 b (c+d x) \tanh ^{-1}(c+d x) \left (3 a (c+d x)^3+b (c+d x)^2+3 b\right )+b^2 (c+d x)^2+3 b^2 \left ((c+d x)^4-1\right ) \tanh ^{-1}(c+d x)^2\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.054, size = 732, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.9763, size = 1116, normalized size = 7.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*d^3*e^3*x^4 + a^2*c*d^2*e^3*x^3 + 3/2*a^2*c^2*d*e^3*x^2 + 3/2*(2*x^2*arctanh(d*x + c) + d*(2*x/d^2 - (
c^2 + 2*c + 1)*log(d*x + c + 1)/d^3 + (c^2 - 2*c + 1)*log(d*x + c - 1)/d^3))*a*b*c^2*d*e^3 + (2*x^3*arctanh(d*
x + c) + d*((d*x^2 - 4*c*x)/d^3 + (c^3 + 3*c^2 + 3*c + 1)*log(d*x + c + 1)/d^4 - (c^3 - 3*c^2 + 3*c - 1)*log(d
*x + c - 1)/d^4))*a*b*c*d^2*e^3 + 1/12*(6*x^4*arctanh(d*x + c) + d*(2*(d^2*x^3 - 3*c*d*x^2 + 3*(3*c^2 + 1)*x)/
d^4 - 3*(c^4 + 4*c^3 + 6*c^2 + 4*c + 1)*log(d*x + c + 1)/d^5 + 3*(c^4 - 4*c^3 + 6*c^2 - 4*c + 1)*log(d*x + c -
1)/d^5))*a*b*d^3*e^3 + a^2*c^3*e^3*x + (2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*a*b*c^3*e^3/d +
1/48*(4*b^2*d^2*e^3*x^2 + 8*b^2*c*d*e^3*x + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2
+ 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(d*x + c + 1)^2 + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b
^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(-d*x - c + 1)^2 + 4*(b^2*d^3*e^3*x^3 + 3*b^2
*c*d^2*e^3*x^2 + 3*(c^2*d*e^3 + d*e^3)*b^2*x + (c^3*e^3 + 3*c*e^3 + 4*e^3)*b^2)*log(d*x + c + 1) - 2*(2*b^2*d^
3*e^3*x^3 + 6*b^2*c*d^2*e^3*x^2 + 6*(c^2*d*e^3 + d*e^3)*b^2*x + 2*(c^3*e^3 + 3*c*e^3 - 4*e^3)*b^2 + 3*(b^2*d^4
*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(d*x + c
+ 1))*log(-d*x - c + 1))/d

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Fricas [B]  time = 2.12868, size = 830, normalized size = 5.22 \begin{align*} \frac{12 \, a^{2} d^{4} e^{3} x^{4} + 8 \,{\left (6 \, a^{2} c + a b\right )} d^{3} e^{3} x^{3} + 4 \,{\left (18 \, a^{2} c^{2} + 6 \, a b c + b^{2}\right )} d^{2} e^{3} x^{2} + 8 \,{\left (6 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c + 3 \, a b\right )} d e^{3} x + 4 \,{\left (3 \, a b c^{4} + b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b + 4 \, b^{2}\right )} e^{3} \log \left (d x + c + 1\right ) - 4 \,{\left (3 \, a b c^{4} + b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b - 4 \, b^{2}\right )} e^{3} \log \left (d x + c - 1\right ) + 3 \,{\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x +{\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 4 \,{\left (3 \, a b d^{4} e^{3} x^{4} +{\left (12 \, a b c + b^{2}\right )} d^{3} e^{3} x^{3} + 3 \,{\left (6 \, a b c^{2} + b^{2} c\right )} d^{2} e^{3} x^{2} + 3 \,{\left (4 \, a b c^{3} + b^{2} c^{2} + b^{2}\right )} d e^{3} x\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*a^2*d^4*e^3*x^4 + 8*(6*a^2*c + a*b)*d^3*e^3*x^3 + 4*(18*a^2*c^2 + 6*a*b*c + b^2)*d^2*e^3*x^2 + 8*(6*a
^2*c^3 + 3*a*b*c^2 + b^2*c + 3*a*b)*d*e^3*x + 4*(3*a*b*c^4 + b^2*c^3 + 3*b^2*c - 3*a*b + 4*b^2)*e^3*log(d*x +
c + 1) - 4*(3*a*b*c^4 + b^2*c^3 + 3*b^2*c - 3*a*b - 4*b^2)*e^3*log(d*x + c - 1) + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c
*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (b^2*c^4 - b^2)*e^3)*log(-(d*x + c + 1)/(d*x + c -
1))^2 + 4*(3*a*b*d^4*e^3*x^4 + (12*a*b*c + b^2)*d^3*e^3*x^3 + 3*(6*a*b*c^2 + b^2*c)*d^2*e^3*x^2 + 3*(4*a*b*c^3
+ b^2*c^2 + b^2)*d*e^3*x)*log(-(d*x + c + 1)/(d*x + c - 1)))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.5108, size = 944, normalized size = 5.94 \begin{align*} \frac{3 \, b^{2} d^{4} x^{4} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 12 \, a b d^{4} x^{4} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 12 \, b^{2} c d^{3} x^{3} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 12 \, a^{2} d^{4} x^{4} e^{3} + 48 \, a b c d^{3} x^{3} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 18 \, b^{2} c^{2} d^{2} x^{2} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 48 \, a^{2} c d^{3} x^{3} e^{3} + 72 \, a b c^{2} d^{2} x^{2} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 4 \, b^{2} d^{3} x^{3} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 12 \, b^{2} c^{3} d x e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 72 \, a^{2} c^{2} d^{2} x^{2} e^{3} + 8 \, a b d^{3} x^{3} e^{3} + 48 \, a b c^{3} d x e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 12 \, b^{2} c d^{2} x^{2} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 3 \, b^{2} c^{4} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 48 \, a^{2} c^{3} d x e^{3} + 24 \, a b c d^{2} x^{2} e^{3} + 12 \, a b c^{4} e^{3} \log \left (d x + c + 1\right ) - 12 \, a b c^{4} e^{3} \log \left (d x + c - 1\right ) + 12 \, b^{2} c^{2} d x e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 24 \, a b c^{2} d x e^{3} + 4 \, b^{2} d^{2} x^{2} e^{3} + 4 \, b^{2} c^{3} e^{3} \log \left (d x + c + 1\right ) - 4 \, b^{2} c^{3} e^{3} \log \left (d x + c - 1\right ) + 8 \, b^{2} c d x e^{3} + 12 \, b^{2} d x e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 24 \, a b d x e^{3} + 12 \, b^{2} c e^{3} \log \left (d x + c + 1\right ) - 12 \, b^{2} c e^{3} \log \left (d x + c - 1\right ) - 3 \, b^{2} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} - 12 \, a b e^{3} \log \left (d x + c + 1\right ) + 16 \, b^{2} e^{3} \log \left (d x + c + 1\right ) + 12 \, a b e^{3} \log \left (d x + c - 1\right ) + 16 \, b^{2} e^{3} \log \left (d x + c - 1\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*b^2*d^4*x^4*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 12*a*b*d^4*x^4*e^3*log(-(d*x + c + 1)/(d*x + c -
1)) + 12*b^2*c*d^3*x^3*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 12*a^2*d^4*x^4*e^3 + 48*a*b*c*d^3*x^3*e^3*lo
g(-(d*x + c + 1)/(d*x + c - 1)) + 18*b^2*c^2*d^2*x^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 48*a^2*c*d^3*x^
3*e^3 + 72*a*b*c^2*d^2*x^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1)) + 4*b^2*d^3*x^3*e^3*log(-(d*x + c + 1)/(d*x +
c - 1)) + 12*b^2*c^3*d*x*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 72*a^2*c^2*d^2*x^2*e^3 + 8*a*b*d^3*x^3*e^3
+ 48*a*b*c^3*d*x*e^3*log(-(d*x + c + 1)/(d*x + c - 1)) + 12*b^2*c*d^2*x^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1
)) + 3*b^2*c^4*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 48*a^2*c^3*d*x*e^3 + 24*a*b*c*d^2*x^2*e^3 + 12*a*b*c^
4*e^3*log(d*x + c + 1) - 12*a*b*c^4*e^3*log(d*x + c - 1) + 12*b^2*c^2*d*x*e^3*log(-(d*x + c + 1)/(d*x + c - 1)
) + 24*a*b*c^2*d*x*e^3 + 4*b^2*d^2*x^2*e^3 + 4*b^2*c^3*e^3*log(d*x + c + 1) - 4*b^2*c^3*e^3*log(d*x + c - 1) +
8*b^2*c*d*x*e^3 + 12*b^2*d*x*e^3*log(-(d*x + c + 1)/(d*x + c - 1)) + 24*a*b*d*x*e^3 + 12*b^2*c*e^3*log(d*x +
c + 1) - 12*b^2*c*e^3*log(d*x + c - 1) - 3*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2 - 12*a*b*e^3*log(d*x +
c + 1) + 16*b^2*e^3*log(d*x + c + 1) + 12*a*b*e^3*log(d*x + c - 1) + 16*b^2*e^3*log(d*x + c - 1))/d