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

Optimal. Leaf size=208 \[ -\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (c x+1)}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (c x+1)^2}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (c x+1)^2}-\frac{21 b^3}{64 c (c x+1)}-\frac{3 b^3}{64 c (c x+1)^2}+\frac{21 b^3 \tanh ^{-1}(c x)}{64 c} \]

[Out]

(-3*b^3)/(64*c*(1 + c*x)^2) - (21*b^3)/(64*c*(1 + c*x)) + (21*b^3*ArcTanh[c*x])/(64*c) - (3*b^2*(a + b*ArcTanh
[c*x]))/(16*c*(1 + c*x)^2) - (9*b^2*(a + b*ArcTanh[c*x]))/(16*c*(1 + c*x)) + (9*b*(a + b*ArcTanh[c*x])^2)/(32*
c) - (3*b*(a + b*ArcTanh[c*x])^2)/(8*c*(1 + c*x)^2) - (3*b*(a + b*ArcTanh[c*x])^2)/(8*c*(1 + c*x)) + (a + b*Ar
cTanh[c*x])^3/(8*c) - (a + b*ArcTanh[c*x])^3/(2*c*(1 + c*x)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.371169, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 24, 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} \[ -\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (c x+1)}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (c x+1)^2}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (c x+1)^2}-\frac{21 b^3}{64 c (c x+1)}-\frac{3 b^3}{64 c (c x+1)^2}+\frac{21 b^3 \tanh ^{-1}(c x)}{64 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^3)/(64*c*(1 + c*x)^2) - (21*b^3)/(64*c*(1 + c*x)) + (21*b^3*ArcTanh[c*x])/(64*c) - (3*b^2*(a + b*ArcTanh
[c*x]))/(16*c*(1 + c*x)^2) - (9*b^2*(a + b*ArcTanh[c*x]))/(16*c*(1 + c*x)) + (9*b*(a + b*ArcTanh[c*x])^2)/(32*
c) - (3*b*(a + b*ArcTanh[c*x])^2)/(8*c*(1 + c*x)^2) - (3*b*(a + b*ArcTanh[c*x])^2)/(8*c*(1 + c*x)) + (a + b*Ar
cTanh[c*x])^3/(8*c) - (a + b*ArcTanh[c*x])^3/(2*c*(1 + c*x)^2)

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)^3} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{2} (3 b) \int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 (1+c x)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 (1+c x)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{8} (3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx-\frac{1}{8} (3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{-1+c^2 x^2} \, dx+\frac{1}{4} (3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx\\ &=-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{4} \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{1}{4} \left (3 b^2\right ) \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{16} \left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{16} \left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac{1}{8} \left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx+\frac{1}{8} \left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{8} \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 )}{16 c (1+c x)^2}-\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{16} \left (3 b^3\right ) \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{16} \left (3 b^3\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx+\frac{1}{8} \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 )}{16 c (1+c x)^2}-\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{16} \left (3 b^3\right ) \int \frac{1}{(1-c x) (1+c x)^3} \, dx+\frac{1}{16} \left (3 b^3\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx+\frac{1}{8} \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 )}{16 c (1+c x)^2}-\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}+\frac{1}{16} \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{1}{16} \left (3 b^3\right ) \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{8} \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}{64 c (1+c x)^2}-\frac{21 b^3}{64 c (1+c x)}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)^2}-\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}-\frac{1}{64} \left (3 b^3\right ) \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{32} \left (3 b^3\right ) \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{16} \left (3 b^3\right ) \int \frac{1}{-1+c^2 x^2} \, dx\\ &=-\frac{3 b^3}{64 c (1+c x)^2}-\frac{21 b^3}{64 c (1+c x)}+\frac{21 b^3 \tanh ^{-1}(c x)}{64 c}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)^2}-\frac{9 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{16 c (1+c x)}+\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{32 c}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{8 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 c (1+c x)^2}\\ \end{align*}

Mathematica [A]  time = 0.212471, size = 215, normalized size = 1.03 \[ \frac{-6 b \left (8 a^2+12 a b+7 b^2\right ) (c x+1)-3 b \left (8 a^2+12 a b+7 b^2\right ) (c x+1)^2 \log (1-c x)+3 b \left (8 a^2+12 a b+7 b^2\right ) (c x+1)^2 \log (c x+1)-24 b \tanh ^{-1}(c x) \left (8 a^2+4 a b (c x+2)+b^2 (3 c x+4)\right )-2 \left (24 a^2 b+32 a^3+12 a b^2+3 b^3\right )+12 b^2 (c x-1) \tanh ^{-1}(c x)^2 (4 a (c x+3)+b (3 c x+5))+16 b^3 \left (c^2 x^2+2 c x-3\right ) \tanh ^{-1}(c x)^3}{128 c (c x+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(32*a^3 + 24*a^2*b + 12*a*b^2 + 3*b^3) - 6*b*(8*a^2 + 12*a*b + 7*b^2)*(1 + c*x) - 24*b*(8*a^2 + 4*a*b*(2 +
c*x) + b^2*(4 + 3*c*x))*ArcTanh[c*x] + 12*b^2*(-1 + c*x)*(4*a*(3 + c*x) + b*(5 + 3*c*x))*ArcTanh[c*x]^2 + 16*
b^3*(-3 + 2*c*x + c^2*x^2)*ArcTanh[c*x]^3 - 3*b*(8*a^2 + 12*a*b + 7*b^2)*(1 + c*x)^2*Log[1 - c*x] + 3*b*(8*a^2
+ 12*a*b + 7*b^2)*(1 + c*x)^2*Log[1 + c*x])/(128*c*(1 + c*x)^2)

________________________________________________________________________________________

Maple [C]  time = 0.432, size = 2752, normalized size = 13.2 \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)^3,x)

[Out]

-1/2/c*a^3/(c*x+1)^2+1/8*c*b^3/(c*x+1)^2*arctanh(c*x)^3*x^2-3/4/c*a*b^2/(c*x+1)^2*arctanh(c*x)-3/2/c*a*b^2/(c*
x+1)^2*arctanh(c*x)^2-3/2/c*a^2*b/(c*x+1)^2*arctanh(c*x)+9/32*c*b^3/(c*x+1)^2*arctanh(c*x)^2*x^2+21/64*c*b^3/(
c*x+1)^2*arctanh(c*x)*x^2+3/128*b^3/(c*x+1)^2*x+3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2
*x^2-1))*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)+1))*x^2-9/16/
c*a*b^2/(c*x+1)-3/8/c*a^2*b/(c*x+1)-3/8/c*b^3*arctanh(c*x)^2/(c*x+1)-3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*
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))*x^2+3/16*I*b^3/(
c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*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)+1))*x-3/16*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)
)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*x^2+3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-
1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*x^2-3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*cs
gn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*x^2+3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*
csgn(I/((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)/((c*x+1)^2/(-c^2
*x^2+1)+1))-3/8/c*a*b^2*arctanh(c*x)*ln(c*x-1)+3/16/c*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-51/256*b^3/c/(c*x+1)^2+3
/8/c*a*b^2*arctanh(c*x)*ln(c*x+1)+3/16/c*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/16/c*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2
+1/2*c*x)-3/32/c*a*b^2*ln(c*x-1)^2-3/16/c*b^3*arctanh(c*x)^2*ln(c*x-1)-3/16/c*a^2*b*ln(c*x-1)-9/32/c*a*b^2*ln(
c*x-1)+9/32/c*a*b^2*ln(c*x+1)+3/16*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*x^2-3/16*I/c*b^3/(c*x+1)^2*Pi*arctanh(c
*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2-3/16*I*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)
)^3*x-3/16*I*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*x+3/8*
I*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*x-3/8*I*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi
*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*x-3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))
^3+3/16*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3-3/8*I*b^3/(c*x+1)^2*arctanh(c
*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*x+3/16*I*b^3/(c*x+1)^2*arctanh(c*x
)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*x-3/16*I*b^3/(
c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*x-3/16*I*b^3/(c*
x+1)^2*arctanh(c*x)^2*Pi*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))*x-3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3
*x^2+3/16*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*x^2-3/16*I*c*b^3/(c*x+1)^2*
arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*x^2-3/32*I*c*b^3/(c*x+1)^2*arctanh(c*x)^2*Pi*csgn(I*(c*
x+1)^2/(c^2*x^2-1))^3*x^2+3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1
)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2-3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^
2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))-3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x
^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2-3/16*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*
csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2-3/16/c*a*b^2/(c*x+1)^2-3/8/c*a^2*b/(c*x+1)^
2+45/256*c*b^3/(c*x+1)^2*x^2+1/4*b^3/(c*x+1)^2*arctanh(c*x)^3*x+9/16*b^3/(c*x+1)^2*arctanh(c*x)^2*x+3/32*b^3/(
c*x+1)^2*arctanh(c*x)*x-27/64/c*b^3/(c*x+1)^2*arctanh(c*x)-3/32/c*b^3*arctanh(c*x)^2/(c*x+1)^2-3/8/c*b^3/(c*x+
1)^2*arctanh(c*x)^3+3/16/c*b^3*arctanh(c*x)^2*ln(c*x+1)-3/8/c*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2)
)-3/32/c*a*b^2*ln(c*x+1)^2+3/16/c*a^2*b*ln(c*x+1)-3/4/c*a*b^2/(c*x+1)*arctanh(c*x)+3/8*I*b^3/(c*x+1)^2*arctanh
(c*x)^2*Pi*x+3/16*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2-3/32*I/c*b^3/(c*x+1)^2*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^
2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3

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Maxima [B]  time = 1.09875, size = 1075, normalized size = 5.17 \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)^3,x, algorithm="maxima")

[Out]

-1/2*b^3*arctanh(c*x)^3/(c^3*x^2 + 2*c^2*x + c) - 3/16*(c*(2*(c*x + 2)/(c^4*x^2 + 2*c^3*x + c^2) - log(c*x + 1
)/c^2 + log(c*x - 1)/c^2) + 8*arctanh(c*x)/(c^3*x^2 + 2*c^2*x + c))*a^2*b - 3/32*(4*c*(2*(c*x + 2)/(c^4*x^2 +
2*c^3*x + c^2) - log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*arctanh(c*x) + ((c^2*x^2 + 2*c*x + 1)*log(c*x + 1)^2 + (
c^2*x^2 + 2*c*x + 1)*log(c*x - 1)^2 + 6*c*x - (3*c^2*x^2 + 6*c*x + 2*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 3)*l
og(c*x + 1) + 3*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 8)*c^2/(c^5*x^2 + 2*c^4*x + c^3))*a*b^2 - 1/128*(24*c*(2*
(c*x + 2)/(c^4*x^2 + 2*c^3*x + c^2) - log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*arctanh(c*x)^2 - ((2*(c^2*x^2 + 2*c
*x + 1)*log(c*x + 1)^3 - 2*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1)^3 - 3*(3*c^2*x^2 + 6*c*x + 2*(c^2*x^2 + 2*c*x +
1)*log(c*x - 1) + 3)*log(c*x + 1)^2 - 9*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1)^2 - 42*c*x + 3*(7*c^2*x^2 + 2*(c^2*
x^2 + 2*c*x + 1)*log(c*x - 1)^2 + 14*c*x + 6*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 7)*log(c*x + 1) - 21*(c^2*x^
2 + 2*c*x + 1)*log(c*x - 1) - 48)*c^2/(c^6*x^2 + 2*c^5*x + c^4) - 12*((c^2*x^2 + 2*c*x + 1)*log(c*x + 1)^2 + (
c^2*x^2 + 2*c*x + 1)*log(c*x - 1)^2 + 6*c*x - (3*c^2*x^2 + 6*c*x + 2*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 3)*l
og(c*x + 1) + 3*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 8)*c*arctanh(c*x)/(c^5*x^2 + 2*c^4*x + c^3))*c)*b^3 - 3/2
*a*b^2*arctanh(c*x)^2/(c^3*x^2 + 2*c^2*x + c) - 1/2*a^3/(c^3*x^2 + 2*c^2*x + c)

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Fricas [A]  time = 1.90281, size = 554, normalized size = 2.66 \begin{align*} \frac{2 \,{\left (b^{3} c^{2} x^{2} + 2 \, b^{3} c x - 3 \, b^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{3} - 64 \, a^{3} - 96 \, a^{2} b - 96 \, a b^{2} - 48 \, b^{3} - 6 \,{\left (8 \, a^{2} b + 12 \, a b^{2} + 7 \, b^{3}\right )} c x + 3 \,{\left ({\left (4 \, a b^{2} + 3 \, b^{3}\right )} c^{2} x^{2} - 12 \, a b^{2} - 5 \, b^{3} + 2 \,{\left (4 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} + 3 \,{\left ({\left (8 \, a^{2} b + 12 \, a b^{2} + 7 \, b^{3}\right )} c^{2} x^{2} - 24 \, a^{2} b - 20 \, a b^{2} - 9 \, b^{3} + 2 \,{\left (8 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{128 \,{\left (c^{3} x^{2} + 2 \, 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)^3,x, algorithm="fricas")

[Out]

1/128*(2*(b^3*c^2*x^2 + 2*b^3*c*x - 3*b^3)*log(-(c*x + 1)/(c*x - 1))^3 - 64*a^3 - 96*a^2*b - 96*a*b^2 - 48*b^3
- 6*(8*a^2*b + 12*a*b^2 + 7*b^3)*c*x + 3*((4*a*b^2 + 3*b^3)*c^2*x^2 - 12*a*b^2 - 5*b^3 + 2*(4*a*b^2 + b^3)*c*
x)*log(-(c*x + 1)/(c*x - 1))^2 + 3*((8*a^2*b + 12*a*b^2 + 7*b^3)*c^2*x^2 - 24*a^2*b - 20*a*b^2 - 9*b^3 + 2*(8*
a^2*b + 4*a*b^2 + b^3)*c*x)*log(-(c*x + 1)/(c*x - 1)))/(c^3*x^2 + 2*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 )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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