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

Optimal. Leaf size=139 \[ -\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c (c x+1)}+\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c (c x+1)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{c (c x+1)}-\frac{3 b^3}{4 c (c x+1)}+\frac{3 b^3 \tanh ^{-1}(c x)}{4 c} \]

[Out]

(-3*b^3)/(4*c*(1 + c*x)) + (3*b^3*ArcTanh[c*x])/(4*c) - (3*b^2*(a + b*ArcTanh[c*x]))/(2*c*(1 + c*x)) + (3*b*(a
+ b*ArcTanh[c*x])^2)/(4*c) - (3*b*(a + b*ArcTanh[c*x])^2)/(2*c*(1 + c*x)) + (a + b*ArcTanh[c*x])^3/(2*c) - (a
+ b*ArcTanh[c*x])^3/(c*(1 + c*x))

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Rubi [A]  time = 0.193227, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c (c x+1)}+\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c (c x+1)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{c (c x+1)}-\frac{3 b^3}{4 c (c x+1)}+\frac{3 b^3 \tanh ^{-1}(c x)}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/(1 + c*x)^2,x]

[Out]

(-3*b^3)/(4*c*(1 + c*x)) + (3*b^3*ArcTanh[c*x])/(4*c) - (3*b^2*(a + b*ArcTanh[c*x]))/(2*c*(1 + c*x)) + (3*b*(a
+ b*ArcTanh[c*x])^2)/(4*c) - (3*b*(a + b*ArcTanh[c*x])^2)/(2*c*(1 + c*x)) + (a + b*ArcTanh[c*x])^3/(2*c) - (a
+ b*ArcTanh[c*x])^3/(c*(1 + c*x))

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 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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]

Rubi steps

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

Mathematica [A]  time = 0.148799, size = 198, normalized size = 1.42 \[ \frac{-3 b \left (2 a^2+2 a b+b^2\right ) (c x+1) \log (1-c x)-12 b \left (2 a^2+2 a b+b^2\right ) \tanh ^{-1}(c x)+6 a^2 b \log (c x+1)+6 a^2 b c x \log (c x+1)-12 a^2 b-8 a^3+6 a b^2 \log (c x+1)+6 a b^2 c x \log (c x+1)+6 b^2 (2 a+b) (c x-1) \tanh ^{-1}(c x)^2-12 a b^2+3 b^3 \log (c x+1)+3 b^3 c x \log (c x+1)+4 b^3 (c x-1) \tanh ^{-1}(c x)^3-6 b^3}{8 c (c x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^3/(1 + c*x)^2,x]

[Out]

(-8*a^3 - 12*a^2*b - 12*a*b^2 - 6*b^3 - 12*b*(2*a^2 + 2*a*b + b^2)*ArcTanh[c*x] + 6*b^2*(2*a + b)*(-1 + c*x)*A
rcTanh[c*x]^2 + 4*b^3*(-1 + c*x)*ArcTanh[c*x]^3 - 3*b*(2*a^2 + 2*a*b + b^2)*(1 + c*x)*Log[1 - c*x] + 6*a^2*b*L
og[1 + c*x] + 6*a*b^2*Log[1 + c*x] + 3*b^3*Log[1 + c*x] + 6*a^2*b*c*x*Log[1 + c*x] + 6*a*b^2*c*x*Log[1 + c*x]
+ 3*b^3*c*x*Log[1 + c*x])/(8*c*(1 + c*x))

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Maple [C]  time = 0.377, size = 1895, normalized size = 13.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3/(c*x+1)^2,x)

[Out]

-3/8*I/c*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*arctanh(c*x)^2*Pi-3/4*
I/c*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*arctanh(c*x)^2*Pi+3/8*I/c*b
^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*
x)^2*Pi-3/4*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*arctanh(c*x)^2*Pi
*x-3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)
+1))*arctanh(c*x)^2*Pi*x+3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I*(
c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2*Pi*x-3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2
*x^2+1)^(1/2))^2*arctanh(c*x)^2*Pi*x-3/8*I/c*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+
1))^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2*Pi-1/c*a^3/(c*x+1)+3/8*b^3/(c*x+1)*x+3/4*I*b^3/(c*x+1)
*arctanh(c*x)^2*Pi*x+3/4*I/c*b^3/(c*x+1)*arctanh(c*x)^2*Pi-3/2/c*a*b^2/(c*x+1)-3/2/c*a^2*b/(c*x+1)-3/4/c*b^3/(
c*x+1)*arctanh(c*x)-3/4/c*b^3*arctanh(c*x)^2/(c*x+1)-1/2/c*b^3/(c*x+1)*arctanh(c*x)^3+1/2*b^3/(c*x+1)*arctanh(
c*x)^3*x+3/4*b^3/(c*x+1)*arctanh(c*x)*x+3/4*b^3/(c*x+1)*arctanh(c*x)^2*x-3/2/c*a*b^2*arctanh(c*x)*ln(c*x-1)+3/
4/c*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-3/8*b^3/c/(c*x+1)+3/2/c*a*b^2*arctanh(c*x)*ln(c*x+1)+3/4/c*a*b^2*ln(-1/2*c
*x+1/2)*ln(c*x+1)-3/4/c*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-3/8/c*a*b^2*ln(c*x-1)^2-3/4/c*b^3*arctanh(c*x)^
2*ln(c*x-1)-3/4/c*a^2*b*ln(c*x-1)-3/4/c*a*b^2*ln(c*x-1)+3/4/c*a*b^2*ln(c*x+1)+3/8*I/c*b^3/(c*x+1)*csgn(I*(c*x+
1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*
arctanh(c*x)^2*Pi+3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/
(c^2*x^2-1))*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2*Pi*x-3/8*I/c*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*
x^2-1))^3*arctanh(c*x)^2*Pi-3/8*I/c*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arc
tanh(c*x)^2*Pi-3/4*I/c*b^3/(c*x+1)*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2*Pi+3/4/c*b^3*arctanh(c*
x)^2*ln(c*x+1)-3/2/c*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-3/8/c*a*b^2*ln(c*x+1)^2+3/4/c*a^2*b*ln(
c*x+1)-3/c*a*b^2/(c*x+1)*arctanh(c*x)^2-3/c*a^2*b/(c*x+1)*arctanh(c*x)-3/c*a*b^2/(c*x+1)*arctanh(c*x)+3/4*I*b^
3/(c*x+1)*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2*Pi*x-3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2
-1))^3*arctanh(c*x)^2*Pi*x-3/8*I*b^3/(c*x+1)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctan
h(c*x)^2*Pi*x-3/4*I*b^3/(c*x+1)*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2*Pi*x+3/4*I/c*b^3/(c*x+1)*c
sgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2*Pi

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Maxima [B]  time = 1.06134, size = 714, normalized size = 5.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="maxima")

[Out]

-b^3*arctanh(c*x)^3/(c^2*x + c) - 3/4*(c*(2/(c^3*x + c^2) - log(c*x + 1)/c^2 + log(c*x - 1)/c^2) + 4*arctanh(c
*x)/(c^2*x + c))*a^2*b - 3/8*(4*c*(2/(c^3*x + c^2) - log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*arctanh(c*x) + ((c*x
+ 1)*log(c*x + 1)^2 + (c*x + 1)*log(c*x - 1)^2 - 2*(c*x + (c*x + 1)*log(c*x - 1) + 1)*log(c*x + 1) + 2*(c*x +
1)*log(c*x - 1) + 4)*c^2/(c^4*x + c^3))*a*b^2 - 1/16*(12*c*(2/(c^3*x + c^2) - log(c*x + 1)/c^2 + log(c*x - 1)
/c^2)*arctanh(c*x)^2 - (((c*x + 1)*log(c*x + 1)^3 - (c*x + 1)*log(c*x - 1)^3 - 3*(c*x + (c*x + 1)*log(c*x - 1)
+ 1)*log(c*x + 1)^2 - 3*(c*x + 1)*log(c*x - 1)^2 + 3*((c*x + 1)*log(c*x - 1)^2 + 2*c*x + 2*(c*x + 1)*log(c*x
- 1) + 2)*log(c*x + 1) - 6*(c*x + 1)*log(c*x - 1) - 12)*c^2/(c^5*x + c^4) - 6*((c*x + 1)*log(c*x + 1)^2 + (c*x
+ 1)*log(c*x - 1)^2 - 2*(c*x + (c*x + 1)*log(c*x - 1) + 1)*log(c*x + 1) + 2*(c*x + 1)*log(c*x - 1) + 4)*c*arc
tanh(c*x)/(c^4*x + c^3))*c)*b^3 - 3*a*b^2*arctanh(c*x)^2/(c^2*x + c) - a^3/(c^2*x + c)

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Fricas [A]  time = 2.02394, size = 354, normalized size = 2.55 \begin{align*} \frac{{\left (b^{3} c x - b^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{3} - 16 \, a^{3} - 24 \, a^{2} b - 24 \, a b^{2} - 12 \, b^{3} - 3 \,{\left (2 \, a b^{2} + b^{3} -{\left (2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} - 6 \,{\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3} -{\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{16 \,{\left (c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="fricas")

[Out]

1/16*((b^3*c*x - b^3)*log(-(c*x + 1)/(c*x - 1))^3 - 16*a^3 - 24*a^2*b - 24*a*b^2 - 12*b^3 - 3*(2*a*b^2 + b^3 -
(2*a*b^2 + b^3)*c*x)*log(-(c*x + 1)/(c*x - 1))^2 - 6*(2*a^2*b + 2*a*b^2 + b^3 - (2*a^2*b + 2*a*b^2 + b^3)*c*x
)*log(-(c*x + 1)/(c*x - 1)))/(c^2*x + c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}}{\left (c x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**3/(c*x+1)**2,x)

[Out]

Integral((a + b*atanh(c*x))**3/(c*x + 1)**2, x)

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Giac [A]  time = 1.19108, size = 279, normalized size = 2.01 \begin{align*} \frac{1}{16} \,{\left (\frac{b^{3}}{c} - \frac{2 \, b^{3}}{{\left (c x + 1\right )} c}\right )} \log \left (\frac{1}{\frac{2}{c x + 1} - 1}\right )^{3} + \frac{3}{16} \,{\left (\frac{2 \, a b^{2} + b^{3}}{c} - \frac{2 \,{\left (2 \, a b^{2} + b^{3}\right )}}{{\left (c x + 1\right )} c}\right )} \log \left (\frac{1}{\frac{2}{c x + 1} - 1}\right )^{2} - \frac{3 \,{\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \log \left (-\frac{2}{c x + 1} + 1\right )}{8 \, c} - \frac{3 \,{\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \log \left (\frac{1}{\frac{2}{c x + 1} - 1}\right )}{4 \,{\left (c x + 1\right )} c} - \frac{4 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 3 \, b^{3}}{4 \,{\left (c x + 1\right )} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="giac")

[Out]

1/16*(b^3/c - 2*b^3/((c*x + 1)*c))*log(1/(2/(c*x + 1) - 1))^3 + 3/16*((2*a*b^2 + b^3)/c - 2*(2*a*b^2 + b^3)/((
c*x + 1)*c))*log(1/(2/(c*x + 1) - 1))^2 - 3/8*(2*a^2*b + 2*a*b^2 + b^3)*log(-2/(c*x + 1) + 1)/c - 3/4*(2*a^2*b
+ 2*a*b^2 + b^3)*log(1/(2/(c*x + 1) - 1))/((c*x + 1)*c) - 1/4*(4*a^3 + 6*a^2*b + 6*a*b^2 + 3*b^3)/((c*x + 1)*
c)