3.2.24 \(\int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx\) [124]

3.2.24.1 Optimal result
3.2.24.2 Mathematica [A] (verified)
3.2.24.3 Rubi [A] (verified)
3.2.24.4 Maple [A] (verified)
3.2.24.5 Fricas [A] (verification not implemented)
3.2.24.6 Sympy [F]
3.2.24.7 Maxima [B] (verification not implemented)
3.2.24.8 Giac [A] (verification not implemented)
3.2.24.9 Mupad [B] (verification not implemented)

3.2.24.1 Optimal result

Integrand size = 18, antiderivative size = 139 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=-\frac {3 b^3}{4 c (1+c x)}+\frac {3 b^3 \text {arctanh}(c x)}{4 c}-\frac {3 b^2 (a+b \text {arctanh}(c x))}{2 c (1+c x)}+\frac {3 b (a+b \text {arctanh}(c x))^2}{4 c}-\frac {3 b (a+b \text {arctanh}(c x))^2}{2 c (1+c x)}+\frac {(a+b \text {arctanh}(c x))^3}{2 c}-\frac {(a+b \text {arctanh}(c x))^3}{c (1+c x)} \]

output
-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)
 
3.2.24.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=\frac {-8 a^3-12 a^2 b-12 a b^2-6 b^3-12 b \left (2 a^2+2 a b+b^2\right ) \text {arctanh}(c x)+6 b^2 (2 a+b) (-1+c x) \text {arctanh}(c x)^2+4 b^3 (-1+c x) \text {arctanh}(c x)^3-3 b \left (2 a^2+2 a b+b^2\right ) (1+c x) \log (1-c x)+6 a^2 b \log (1+c x)+6 a b^2 \log (1+c x)+3 b^3 \log (1+c x)+6 a^2 b c x \log (1+c x)+6 a b^2 c x \log (1+c x)+3 b^3 c x \log (1+c x)}{8 c (1+c x)} \]

input
Integrate[(a + b*ArcTanh[c*x])^3/(1 + c*x)^2,x]
 
output
(-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)*ArcTanh[c*x]^2 + 4*b^3*(-1 + c*x)*ArcTa 
nh[c*x]^3 - 3*b*(2*a^2 + 2*a*b + b^2)*(1 + c*x)*Log[1 - c*x] + 6*a^2*b*Log 
[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) 
)
 
3.2.24.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6480, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^3}{(c x+1)^2} \, dx\)

\(\Big \downarrow \) 6480

\(\displaystyle 3 b \int \left (\frac {(a+b \text {arctanh}(c x))^2}{2 \left (1-c^2 x^2\right )}+\frac {(a+b \text {arctanh}(c x))^2}{2 (c x+1)^2}\right )dx-\frac {(a+b \text {arctanh}(c x))^3}{c (c x+1)}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 b \left (\frac {(a+b \text {arctanh}(c x))^3}{6 b c}+\frac {(a+b \text {arctanh}(c x))^2}{4 c}-\frac {(a+b \text {arctanh}(c x))^2}{2 c (c x+1)}-\frac {b (a+b \text {arctanh}(c x))}{2 c (c x+1)}+\frac {b^2 \text {arctanh}(c x)}{4 c}-\frac {b^2}{4 c (c x+1)}\right )-\frac {(a+b \text {arctanh}(c x))^3}{c (c x+1)}\)

input
Int[(a + b*ArcTanh[c*x])^3/(1 + c*x)^2,x]
 
output
-((a + b*ArcTanh[c*x])^3/(c*(1 + c*x))) + 3*b*(-1/4*b^2/(c*(1 + c*x)) + (b 
^2*ArcTanh[c*x])/(4*c) - (b*(a + b*ArcTanh[c*x]))/(2*c*(1 + c*x)) + (a + b 
*ArcTanh[c*x])^2/(4*c) - (a + b*ArcTanh[c*x])^2/(2*c*(1 + c*x)) + (a + b*A 
rcTanh[c*x])^3/(6*b*c))
 

3.2.24.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6480
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_S 
ymbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])^p/(e*(q + 1))), x] - 
 Simp[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]
 
3.2.24.4 Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.31

method result size
parallelrisch \(-\frac {-2 \operatorname {arctanh}\left (c x \right )^{3} b^{3} c x -6 \operatorname {arctanh}\left (c x \right )^{2} a \,b^{2} c x -3 b^{3} \operatorname {arctanh}\left (c x \right )^{2} x c -6 \,\operatorname {arctanh}\left (c x \right ) a^{2} b c x -6 a \,b^{2} \operatorname {arctanh}\left (c x \right ) x c -3 b^{3} c x \,\operatorname {arctanh}\left (c x \right )+2 b^{3} \operatorname {arctanh}\left (c x \right )^{3}-4 a^{3} c x -6 a^{2} b c x -6 a \,b^{2} c x -3 b^{3} c x +6 a \,b^{2} \operatorname {arctanh}\left (c x \right )^{2}+3 b^{3} \operatorname {arctanh}\left (c x \right )^{2}+6 a^{2} b \,\operatorname {arctanh}\left (c x \right )+6 a \,b^{2} \operatorname {arctanh}\left (c x \right )+3 b^{3} \operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right ) c}\) \(182\)
risch \(\frac {\left (c x -1\right ) b^{3} \ln \left (c x +1\right )^{3}}{16 c \left (c x +1\right )}+\frac {3 b^{2} \left (-b c x \ln \left (-c x +1\right )+2 c x a +b c x +b \ln \left (-c x +1\right )-2 a -b \right ) \ln \left (c x +1\right )^{2}}{16 c \left (c x +1\right )}-\frac {3 b \left (-\ln \left (-c x +1\right )^{2} b^{2} c x +4 \ln \left (-c x +1\right ) a b c x +2 b^{2} c x \ln \left (-c x +1\right )+b^{2} \ln \left (-c x +1\right )^{2}-4 b \ln \left (-c x +1\right ) a -2 b^{2} \ln \left (-c x +1\right )+8 a^{2}+8 a b +4 b^{2}\right ) \ln \left (c x +1\right )}{16 c \left (c x +1\right )}+\frac {-16 a^{3}-12 b^{3}-24 a^{2} b -12 a \,b^{2} \ln \left (c x -1\right )-12 a^{2} b \ln \left (c x -1\right )-24 a \,b^{2}-6 \ln \left (-c x +1\right )^{2} a \,b^{2}+24 \ln \left (-c x +1\right ) a^{2} b +\ln \left (-c x +1\right )^{3} b^{3}-\ln \left (-c x +1\right )^{3} b^{3} c x +6 \ln \left (-c x +1\right )^{2} a \,b^{2} c x +24 \ln \left (-c x +1\right ) a \,b^{2}+6 \ln \left (-c x -1\right ) b^{3} c x -6 \ln \left (c x -1\right ) b^{3} c x +3 b^{3} c x \ln \left (-c x +1\right )^{2}+12 b^{3} \ln \left (-c x +1\right )-3 b^{3} \ln \left (-c x +1\right )^{2}+6 b^{3} \ln \left (-c x -1\right )-6 b^{3} \ln \left (c x -1\right )+12 \ln \left (-c x -1\right ) a^{2} b +12 \ln \left (-c x -1\right ) a \,b^{2}+12 \ln \left (-c x -1\right ) a^{2} b c x +12 \ln \left (-c x -1\right ) a \,b^{2} c x -12 \ln \left (c x -1\right ) a^{2} b c x -12 \ln \left (c x -1\right ) a \,b^{2} c x}{16 c \left (c x +1\right )}\) \(525\)
derivativedivides \(\text {Expression too large to display}\) \(909\)
default \(\text {Expression too large to display}\) \(909\)
parts \(\text {Expression too large to display}\) \(917\)

input
int((a+b*arctanh(c*x))^3/(c*x+1)^2,x,method=_RETURNVERBOSE)
 
output
-1/4*(-2*arctanh(c*x)^3*b^3*c*x-6*arctanh(c*x)^2*a*b^2*c*x-3*b^3*arctanh(c 
*x)^2*x*c-6*arctanh(c*x)*a^2*b*c*x-6*a*b^2*arctanh(c*x)*x*c-3*b^3*c*x*arct 
anh(c*x)+2*b^3*arctanh(c*x)^3-4*a^3*c*x-6*a^2*b*c*x-6*a*b^2*c*x-3*b^3*c*x+ 
6*a*b^2*arctanh(c*x)^2+3*b^3*arctanh(c*x)^2+6*a^2*b*arctanh(c*x)+6*a*b^2*a 
rctanh(c*x)+3*b^3*arctanh(c*x))/(c*x+1)/c
 
3.2.24.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=\frac {{\left (b^{3} c x - b^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{3} - 16 \, a^{3} - 24 \, a^{2} b - 24 \, a b^{2} - 12 \, b^{3} - 3 \, {\left (2 \, a b^{2} + b^{3} - {\left (2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 6 \, {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{16 \, {\left (c^{2} x + c\right )}} \]

input
integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="fricas")
 
output
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)
 
3.2.24.6 Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{\left (c x + 1\right )^{2}}\, dx \]

input
integrate((a+b*atanh(c*x))**3/(c*x+1)**2,x)
 
output
Integral((a + b*atanh(c*x))**3/(c*x + 1)**2, x)
 
3.2.24.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (127) = 254\).

Time = 0.21 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.81 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=-\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} \]

input
integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="maxima")
 
output
-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)*lo 
g(c*x - 1) + 1)*log(c*x + 1) + 2*(c*x + 1)*log(c*x - 1) + 4)*c*arctanh(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)
 
3.2.24.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=\frac {1}{16} \, {\left (\frac {{\left (c x - 1\right )} b^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )^{3}}{{\left (c x + 1\right )} c^{2}} + \frac {3 \, {\left (2 \, a b^{2} + b^{3}\right )} {\left (c x - 1\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )} c^{2}} + \frac {6 \, {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (c x - 1\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )} c^{2}} + \frac {2 \, {\left (4 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 3 \, b^{3}\right )} {\left (c x - 1\right )}}{{\left (c x + 1\right )} c^{2}}\right )} c \]

input
integrate((a+b*arctanh(c*x))^3/(c*x+1)^2,x, algorithm="giac")
 
output
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
 
3.2.24.9 Mupad [B] (verification not implemented)

Time = 4.90 (sec) , antiderivative size = 582, normalized size of antiderivative = 4.19 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{(1+c x)^2} \, dx=\ln \left (1-c\,x\right )\,\left (\ln \left (c\,x+1\right )\,\left (\frac {3\,b^3\,x+\frac {3\,b^2\,\left (4\,a+b\right )}{c}}{8\,c\,x+8}-\frac {3\,\left (b^3+a\,b^2\right )}{4\,c}+\frac {6\,b^3}{c\,\left (8\,c\,x+8\right )}\right )+\frac {3\,b^3\,x+\frac {3\,b^2\,\left (4\,a+b\right )}{c}}{8\,c\,x+8}-{\ln \left (c\,x+1\right )}^2\,\left (\frac {3\,b^3}{16\,c}-\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}\right )-\frac {3\,b^3\,x+\frac {3\,\left (-4\,a^2\,b+4\,a\,b^2+5\,b^3\right )}{c}}{8\,c\,x+8}+\frac {6\,b^3}{c\,\left (8\,c\,x+8\right )}+\frac {3\,\left (8\,c\,x+24\right )\,\left (b^3+a\,b^2\right )}{4\,c\,\left (8\,c\,x+8\right )}\right )-{\ln \left (1-c\,x\right )}^2\,\left (\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}-\frac {3\,\left (b^3+a\,b^2\right )}{8\,c}-\ln \left (c\,x+1\right )\,\left (\frac {3\,b^3}{16\,c}-\frac {3\,b^3}{c\,\left (8\,c\,x+8\right )}\right )+\frac {3\,b^3\,\left (8\,c\,x+24\right )}{16\,c\,\left (8\,c\,x+8\right )}+\frac {3\,b^2\,\left (2\,a-b\right )}{c\,\left (8\,c\,x+8\right )}\right )-{\ln \left (c\,x+1\right )}^2\,\left (\frac {\frac {3\,b^3\,x}{16\,c}+\frac {3\,b^2\,\left (4\,a+3\,b\right )}{16\,c^2}}{x+\frac {1}{c}}-\frac {3\,b^2\,\left (a+b\right )}{8\,c}\right )-{\ln \left (1-c\,x\right )}^3\,\left (\frac {b^3}{16\,c}-\frac {b^3}{c\,\left (8\,c\,x+8\right )}\right )+{\ln \left (c\,x+1\right )}^3\,\left (\frac {b^3}{16\,c}-\frac {b^3}{8\,c^2\,\left (x+\frac {1}{c}\right )}\right )-\frac {\ln \left (c\,x+1\right )\,\left (\frac {3\,b\,\left (2\,a^2+3\,a\,b+2\,b^2\right )}{4\,c^2}+\frac {3\,b^2\,x\,\left (a+b\right )}{4\,c}\right )}{x+\frac {1}{c}}-\frac {4\,a^3+6\,a^2\,b+6\,a\,b^2+3\,b^3}{2\,c\,\left (2\,c\,x+2\right )}-\frac {b\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,\left (2\,a^2+4\,a\,b+3\,b^2\right )\,3{}\mathrm {i}}{4\,c} \]

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