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]

-3/4*b^3/c/(c*x+1)+3/4*b^3*arctanh(c*x)/c-3/2*b^2*(a+b*arctanh(c*x))/c/(c*x+1)+3/4*b*(a+b*arctanh(c*x))^2/c-3/
2*b*(a+b*arctanh(c*x))^2/c/(c*x+1)+1/2*(a+b*arctanh(c*x))^3/c-(a+b*arctanh(c*x))^3/c/(c*x+1)

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Rubi [A]  time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.15, size = 198, normalized size = 1.42 \[ \frac {-8 a^3-3 b \left (2 a^2+2 a b+b^2\right ) (c x+1) \log (1-c x)-12 b \left (2 a^2+2 a b+b^2\right ) \tanh ^{-1}(c x)+6 a^2 b \log (c x+1)+6 a^2 b c x \log (c x+1)-12 a^2 b+6 a b^2 \log (c x+1)+6 a b^2 c x \log (c x+1)+6 b^2 (2 a+b) (c x-1) \tanh ^{-1}(c x)^2-12 a b^2+3 b^3 \log (c x+1)+3 b^3 c x \log (c x+1)+4 b^3 (c x-1) \tanh ^{-1}(c x)^3-6 b^3}{8 c (c x+1)} \]

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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fricas [A]  time = 0.87, size = 160, normalized size = 1.15 \[ \frac {{\left (b^{3} c x - b^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{3} - 16 \, a^{3} - 24 \, a^{2} b - 24 \, a b^{2} - 12 \, b^{3} - 3 \, {\left (2 \, a b^{2} + b^{3} - {\left (2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 6 \, {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{16 \, {\left (c^{2} x + c\right )}} \]

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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giac [A]  time = 0.21, size = 172, normalized size = 1.24 \[ \frac {1}{16} \, {\left (\frac {{\left (c x - 1\right )} b^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )^{3}}{{\left (c x + 1\right )} c^{2}} + \frac {3 \, {\left (2 \, a b^{2} + b^{3}\right )} {\left (c x - 1\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )} c^{2}} + \frac {6 \, {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (c x - 1\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )} c^{2}} + \frac {2 \, {\left (4 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 3 \, b^{3}\right )} {\left (c x - 1\right )}}{{\left (c x + 1\right )} c^{2}}\right )} c \]

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

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maple [C]  time = 0.70, size = 1895, normalized size = 13.63 \[ \text {result too large to display} \]

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

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maxima [B]  time = 0.35, size = 529, normalized size = 3.81 \[ -\frac {b^{3} \operatorname {artanh}\left (c x\right )^{3}}{c^{2} x + c} - \frac {3}{4} \, {\left (c {\left (\frac {2}{c^{3} x + c^{2}} - \frac {\log \left (c x + 1\right )}{c^{2}} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} + \frac {4 \, \operatorname {artanh}\left (c x\right )}{c^{2} x + c}\right )} a^{2} b - \frac {3}{8} \, {\left (4 \, c {\left (\frac {2}{c^{3} x + c^{2}} - \frac {\log \left (c x + 1\right )}{c^{2}} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {{\left ({\left (c x + 1\right )} \log \left (c x + 1\right )^{2} + {\left (c x + 1\right )} \log \left (c x - 1\right )^{2} - 2 \, {\left (c x + {\left (c x + 1\right )} \log \left (c x - 1\right ) + 1\right )} \log \left (c x + 1\right ) + 2 \, {\left (c x + 1\right )} \log \left (c x - 1\right ) + 4\right )} c^{2}}{c^{4} x + c^{3}}\right )} a b^{2} - \frac {1}{16} \, {\left (12 \, c {\left (\frac {2}{c^{3} x + c^{2}} - \frac {\log \left (c x + 1\right )}{c^{2}} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} \operatorname {artanh}\left (c x\right )^{2} - {\left (\frac {{\left ({\left (c x + 1\right )} \log \left (c x + 1\right )^{3} - {\left (c x + 1\right )} \log \left (c x - 1\right )^{3} - 3 \, {\left (c x + {\left (c x + 1\right )} \log \left (c x - 1\right ) + 1\right )} \log \left (c x + 1\right )^{2} - 3 \, {\left (c x + 1\right )} \log \left (c x - 1\right )^{2} + 3 \, {\left ({\left (c x + 1\right )} \log \left (c x - 1\right )^{2} + 2 \, c x + 2 \, {\left (c x + 1\right )} \log \left (c x - 1\right ) + 2\right )} \log \left (c x + 1\right ) - 6 \, {\left (c x + 1\right )} \log \left (c x - 1\right ) - 12\right )} c^{2}}{c^{5} x + c^{4}} - \frac {6 \, {\left ({\left (c x + 1\right )} \log \left (c x + 1\right )^{2} + {\left (c x + 1\right )} \log \left (c x - 1\right )^{2} - 2 \, {\left (c x + {\left (c x + 1\right )} \log \left (c x - 1\right ) + 1\right )} \log \left (c x + 1\right ) + 2 \, {\left (c x + 1\right )} \log \left (c x - 1\right ) + 4\right )} c \operatorname {artanh}\left (c x\right )}{c^{4} x + c^{3}}\right )} c\right )} b^{3} - \frac {3 \, a b^{2} \operatorname {artanh}\left (c x\right )^{2}}{c^{2} x + c} - \frac {a^{3}}{c^{2} x + c} \]

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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mupad [B]  time = 2.30, size = 582, normalized size = 4.19 \[ \ln \left (1-c\,x\right )\,\left (\ln \left (c\,x+1\right )\,\left (\frac {3\,b^3\,x+\frac {3\,b^2\,\left (4\,a+b\right )}{c}}{8\,c\,x+8}-\frac {3\,\left (b^3+a\,b^2\right )}{4\,c}+\frac {6\,b^3}{c\,\left (8\,c\,x+8\right )}\right )+\frac {3\,b^3\,x+\frac {3\,b^2\,\left (4\,a+b\right )}{c}}{8\,c\,x+8}-{\ln \left (c\,x+1\right )}^2\,\left (\frac {3\,b^3}{16\,c}-\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}\right )-\frac {3\,b^3\,x+\frac {3\,\left (-4\,a^2\,b+4\,a\,b^2+5\,b^3\right )}{c}}{8\,c\,x+8}+\frac {6\,b^3}{c\,\left (8\,c\,x+8\right )}+\frac {3\,\left (8\,c\,x+24\right )\,\left (b^3+a\,b^2\right )}{4\,c\,\left (8\,c\,x+8\right )}\right )-{\ln \left (1-c\,x\right )}^2\,\left (\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}-\frac {3\,\left (b^3+a\,b^2\right )}{8\,c}-\ln \left (c\,x+1\right )\,\left (\frac {3\,b^3}{16\,c}-\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}\right )+\frac {3\,b^3\,\left (8\,c\,x+24\right )}{16\,c\,\left (8\,c\,x+8\right )}+\frac {3\,b^2\,\left (2\,a-b\right )}{c\,\left (8\,c\,x+8\right )}\right )-{\ln \left (c\,x+1\right )}^2\,\left (\frac {\frac {3\,b^3\,x}{16\,c}+\frac {3\,b^2\,\left (4\,a+3\,b\right )}{16\,c^2}}{x+\frac {1}{c}}-\frac {3\,b^2\,\left (a+b\right )}{8\,c}\right )-{\ln \left (1-c\,x\right )}^3\,\left (\frac {b^3}{16\,c}-\frac {b^3}{c\,\left (8\,c\,x+8\right )}\right )+{\ln \left (c\,x+1\right )}^3\,\left (\frac {b^3}{16\,c}-\frac {b^3}{8\,c^2\,\left (x+\frac {1}{c}\right )}\right )-\frac {\ln \left (c\,x+1\right )\,\left (\frac {3\,b\,\left (2\,a^2+3\,a\,b+2\,b^2\right )}{4\,c^2}+\frac {3\,b^2\,x\,\left (a+b\right )}{4\,c}\right )}{x+\frac {1}{c}}-\frac {4\,a^3+6\,a^2\,b+6\,a\,b^2+3\,b^3}{2\,c\,\left (2\,c\,x+2\right )}-\frac {b\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,\left (2\,a^2+4\,a\,b+3\,b^2\right )\,3{}\mathrm {i}}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - c*x)*(log(c*x + 1)*((3*b^3*x + (3*b^2*(4*a + b))/c)/(8*c*x + 8) - (3*(a*b^2 + b^3))/(4*c) + (6*b^3)/(c
*(8*c*x + 8))) + (3*b^3*x + (3*b^2*(4*a + b))/c)/(8*c*x + 8) - log(c*x + 1)^2*((3*b^3)/(16*c) - (3*b^3)/(c*(8*
c*x + 8))) - (3*b^3*x + (3*(4*a*b^2 - 4*a^2*b + 5*b^3))/c)/(8*c*x + 8) + (6*b^3)/(c*(8*c*x + 8)) + (3*(8*c*x +
24)*(a*b^2 + b^3))/(4*c*(8*c*x + 8))) - log(1 - c*x)^2*((3*b^3)/(c*(8*c*x + 8)) - (3*(a*b^2 + b^3))/(8*c) - l
og(c*x + 1)*((3*b^3)/(16*c) - (3*b^3)/(c*(8*c*x + 8))) + (3*b^3*(8*c*x + 24))/(16*c*(8*c*x + 8)) + (3*b^2*(2*a
- b))/(c*(8*c*x + 8))) - log(c*x + 1)^2*(((3*b^3*x)/(16*c) + (3*b^2*(4*a + 3*b))/(16*c^2))/(x + 1/c) - (3*b^2
*(a + b))/(8*c)) - log(1 - c*x)^3*(b^3/(16*c) - b^3/(c*(8*c*x + 8))) + log(c*x + 1)^3*(b^3/(16*c) - b^3/(8*c^2
*(x + 1/c))) - (log(c*x + 1)*((3*b*(3*a*b + 2*a^2 + 2*b^2))/(4*c^2) + (3*b^2*x*(a + b))/(4*c)))/(x + 1/c) - (6
*a*b^2 + 6*a^2*b + 4*a^3 + 3*b^3)/(2*c*(2*c*x + 2)) - (b*atan(c*x*1i)*(4*a*b + 2*a^2 + 3*b^2)*3i)/(4*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{\left (c x + 1\right )^{2}}\, dx \]

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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