\(\int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx\) [12]

Optimal result

Integrand size = 18, antiderivative size = 129 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2+\frac {1}{12} b c d^2 x^3+\frac {1}{2} d^2 x^2 (a+b \text {arctanh}(c x))+\frac {2}{3} c d^2 x^3 (a+b \text {arctanh}(c x))+\frac {1}{4} c^2 d^2 x^4 (a+b \text {arctanh}(c x))+\frac {17 b d^2 \log (1-c x)}{24 c^2}-\frac {b d^2 \log (1+c x)}{24 c^2} \] Output:

3/4*b*d^2*x/c+1/3*b*d^2*x^2+1/12*b*c*d^2*x^3+1/2*d^2*x^2*(a+b*arctanh(c*x) 
)+2/3*c*d^2*x^3*(a+b*arctanh(c*x))+1/4*c^2*d^2*x^4*(a+b*arctanh(c*x))+17/2 
4*b*d^2*ln(-c*x+1)/c^2-1/24*b*d^2*ln(c*x+1)/c^2
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.83 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^2 \left (18 b c x+12 a c^2 x^2+8 b c^2 x^2+16 a c^3 x^3+2 b c^3 x^3+6 a c^4 x^4+2 b c^2 x^2 \left (6+8 c x+3 c^2 x^2\right ) \text {arctanh}(c x)+17 b \log (1-c x)-b \log (1+c x)\right )}{24 c^2} \] Input:

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

Output:

(d^2*(18*b*c*x + 12*a*c^2*x^2 + 8*b*c^2*x^2 + 16*a*c^3*x^3 + 2*b*c^3*x^3 + 
 6*a*c^4*x^4 + 2*b*c^2*x^2*(6 + 8*c*x + 3*c^2*x^2)*ArcTanh[c*x] + 17*b*Log 
[1 - c*x] - b*Log[1 + c*x]))/(24*c^2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 2333, 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.

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

\(\Big \downarrow \) 6498

\(\displaystyle -b c \int \frac {d^2 x^2 \left (3 c^2 x^2+8 c x+6\right )}{12 \left (1-c^2 x^2\right )}dx+\frac {1}{4} c^2 d^2 x^4 (a+b \text {arctanh}(c x))+\frac {2}{3} c d^2 x^3 (a+b \text {arctanh}(c x))+\frac {1}{2} d^2 x^2 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{12} b c d^2 \int \frac {x^2 \left (3 c^2 x^2+8 c x+6\right )}{1-c^2 x^2}dx+\frac {1}{4} c^2 d^2 x^4 (a+b \text {arctanh}(c x))+\frac {2}{3} c d^2 x^3 (a+b \text {arctanh}(c x))+\frac {1}{2} d^2 x^2 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 2333

\(\displaystyle -\frac {1}{12} b c d^2 \int \left (-3 x^2-\frac {8 x}{c}+\frac {8 c x+9}{c^2 \left (1-c^2 x^2\right )}-\frac {9}{c^2}\right )dx+\frac {1}{4} c^2 d^2 x^4 (a+b \text {arctanh}(c x))+\frac {2}{3} c d^2 x^3 (a+b \text {arctanh}(c x))+\frac {1}{2} d^2 x^2 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} c^2 d^2 x^4 (a+b \text {arctanh}(c x))+\frac {2}{3} c d^2 x^3 (a+b \text {arctanh}(c x))+\frac {1}{2} d^2 x^2 (a+b \text {arctanh}(c x))-\frac {1}{12} b c d^2 \left (\frac {9 \text {arctanh}(c x)}{c^3}-\frac {9 x}{c^2}-\frac {4 \log \left (1-c^2 x^2\right )}{c^3}-\frac {4 x^2}{c}-x^3\right )\)

Input:

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

Output:

(d^2*x^2*(a + b*ArcTanh[c*x]))/2 + (2*c*d^2*x^3*(a + b*ArcTanh[c*x]))/3 + 
(c^2*d^2*x^4*(a + b*ArcTanh[c*x]))/4 - (b*c*d^2*((-9*x)/c^2 - (4*x^2)/c - 
x^3 + (9*ArcTanh[c*x])/c^3 - (4*Log[1 - c^2*x^2])/c^3))/12
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 6498

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( 
x_))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[( 
a + b*ArcTanh[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/(1 - c^2*x 
^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && Intege 
rQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0 
]))
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84

method	result	size
parts	\(d^{2} a \left (\frac {1}{4} c^{2} x^{4}+\frac {2}{3} c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{3}+\frac {3 c x}{4}+\frac {17 \ln \left (c x -1\right )}{24}-\frac {\ln \left (c x +1\right )}{24}\right )}{c^{2}}\)	\(108\)
derivativedivides	\(\frac {d^{2} a \left (\frac {1}{4} c^{4} x^{4}+\frac {2}{3} x^{3} c^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{3}+\frac {3 c x}{4}+\frac {17 \ln \left (c x -1\right )}{24}-\frac {\ln \left (c x +1\right )}{24}\right )}{c^{2}}\)	\(114\)
default	\(\frac {d^{2} a \left (\frac {1}{4} c^{4} x^{4}+\frac {2}{3} x^{3} c^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{3}+\frac {3 c x}{4}+\frac {17 \ln \left (c x -1\right )}{24}-\frac {\ln \left (c x +1\right )}{24}\right )}{c^{2}}\)	\(114\)
parallelrisch	\(\frac {3 d^{2} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+3 a \,c^{4} d^{2} x^{4}+8 d^{2} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+8 a \,c^{3} d^{2} x^{3}+b \,c^{3} d^{2} x^{3}+6 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{2}+6 a \,c^{2} d^{2} x^{2}+4 b \,c^{2} d^{2} x^{2}+9 b c \,d^{2} x +8 \ln \left (c x -1\right ) b \,d^{2}-b \,d^{2} \operatorname {arctanh}\left (c x \right )}{12 c^{2}}\)	\(144\)
risch	\(\frac {d^{2} b \,x^{2} \left (3 c^{2} x^{2}+8 c x +6\right ) \ln \left (c x +1\right )}{24}-\frac {d^{2} c^{2} b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {d^{2} c^{2} a \,x^{4}}{4}-\frac {d^{2} c b \,x^{3} \ln \left (-c x +1\right )}{3}+\frac {2 d^{2} c a \,x^{3}}{3}+\frac {b c \,d^{2} x^{3}}{12}-\frac {d^{2} b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {d^{2} a \,x^{2}}{2}+\frac {b \,d^{2} x^{2}}{3}+\frac {3 b \,d^{2} x}{4 c}+\frac {17 b \,d^{2} \ln \left (-c x +1\right )}{24 c^{2}}-\frac {b \,d^{2} \ln \left (c x +1\right )}{24 c^{2}}\)	\(174\)

Input:

int(x*(c*d*x+d)^2*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 

Output:

d^2*a*(1/4*c^2*x^4+2/3*c*x^3+1/2*x^2)+d^2*b/c^2*(1/4*arctanh(c*x)*c^4*x^4+ 
2/3*arctanh(c*x)*c^3*x^3+1/2*arctanh(c*x)*c^2*x^2+1/12*x^3*c^3+1/3*c^2*x^2 
+3/4*c*x+17/24*ln(c*x-1)-1/24*ln(c*x+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.06 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, a c^{4} d^{2} x^{4} + 2 \, {\left (8 \, a + b\right )} c^{3} d^{2} x^{3} + 4 \, {\left (3 \, a + 2 \, b\right )} c^{2} d^{2} x^{2} + 18 \, b c d^{2} x - b d^{2} \log \left (c x + 1\right ) + 17 \, b d^{2} \log \left (c x - 1\right ) + {\left (3 \, b c^{4} d^{2} x^{4} + 8 \, b c^{3} d^{2} x^{3} + 6 \, b c^{2} d^{2} x^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{2}} \] Input:

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

Output:

1/24*(6*a*c^4*d^2*x^4 + 2*(8*a + b)*c^3*d^2*x^3 + 4*(3*a + 2*b)*c^2*d^2*x^ 
2 + 18*b*c*d^2*x - b*d^2*log(c*x + 1) + 17*b*d^2*log(c*x - 1) + (3*b*c^4*d 
^2*x^4 + 8*b*c^3*d^2*x^3 + 6*b*c^2*d^2*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2
 

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.29 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{2} d^{2} x^{4}}{4} + \frac {2 a c d^{2} x^{3}}{3} + \frac {a d^{2} x^{2}}{2} + \frac {b c^{2} d^{2} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {2 b c d^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b c d^{2} x^{3}}{12} + \frac {b d^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b d^{2} x^{2}}{3} + \frac {3 b d^{2} x}{4 c} + \frac {2 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{3 c^{2}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{12 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*c**2*d**2*x**4/4 + 2*a*c*d**2*x**3/3 + a*d**2*x**2/2 + b*c**2 
*d**2*x**4*atanh(c*x)/4 + 2*b*c*d**2*x**3*atanh(c*x)/3 + b*c*d**2*x**3/12 
+ b*d**2*x**2*atanh(c*x)/2 + b*d**2*x**2/3 + 3*b*d**2*x/(4*c) + 2*b*d**2*l 
og(x - 1/c)/(3*c**2) - b*d**2*atanh(c*x)/(12*c**2), Ne(c, 0)), (a*d**2*x** 
2/2, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.39 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{4} \, a c^{2} d^{2} x^{4} + \frac {2}{3} \, a c d^{2} x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{2} + \frac {1}{2} \, a d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{2} \] Input:

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

Output:

1/4*a*c^2*d^2*x^4 + 2/3*a*c*d^2*x^3 + 1/24*(6*x^4*arctanh(c*x) + c*(2*(c^2 
*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*c^2*d^2 + 1/ 
3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c*d^2 + 1/2* 
a*d^2*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log( 
c*x - 1)/c^3))*b*d^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (113) = 226\).

Time = 0.12 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.29 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=-\frac {1}{3} \, c {\left (\frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{3}} - \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}} - \frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{3}} - \frac {\frac {24 \, {\left (c x + 1\right )}^{3} a d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {24 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {16 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} - 4 \, a d^{2} + \frac {10 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {23 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {18 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - 5 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}}\right )} \] Input:

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

Output:

-1/3*c*(2*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^3 - 2*(6*(c*x + 1)^3*b*d^2 
/(c*x - 1)^3 - 6*(c*x + 1)^2*b*d^2/(c*x - 1)^2 + 4*(c*x + 1)*b*d^2/(c*x - 
1) - b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4*c^3/(c*x - 1)^4 - 4*(c* 
x + 1)^3*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*c^3/(c*x - 1)^2 - 4*(c*x + 1)*c^3 
/(c*x - 1) + c^3) - 2*b*d^2*log(-(c*x + 1)/(c*x - 1))/c^3 - (24*(c*x + 1)^ 
3*a*d^2/(c*x - 1)^3 - 24*(c*x + 1)^2*a*d^2/(c*x - 1)^2 + 16*(c*x + 1)*a*d^ 
2/(c*x - 1) - 4*a*d^2 + 10*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 23*(c*x + 1)^2* 
b*d^2/(c*x - 1)^2 + 18*(c*x + 1)*b*d^2/(c*x - 1) - 5*b*d^2)/((c*x + 1)^4*c 
^3/(c*x - 1)^4 - 4*(c*x + 1)^3*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*c^3/(c*x - 
1)^2 - 4*(c*x + 1)*c^3/(c*x - 1) + c^3))
 

Mupad [B] (verification not implemented)

Time = 3.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^2\,\left (6\,a\,x^2+4\,b\,x^2+6\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{12}-\frac {\frac {d^2\,\left (9\,b\,\mathrm {atanh}\left (c\,x\right )-4\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12}-\frac {3\,b\,c\,d^2\,x}{4}}{c^2}+\frac {c^2\,d^2\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c\,d^2\,\left (8\,a\,x^3+b\,x^3+8\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \] Input:

int(x*(a + b*atanh(c*x))*(d + c*d*x)^2,x)
 

Output:

(d^2*(6*a*x^2 + 4*b*x^2 + 6*b*x^2*atanh(c*x)))/12 - ((d^2*(9*b*atanh(c*x) 
- 4*b*log(c^2*x^2 - 1)))/12 - (3*b*c*d^2*x)/4)/c^2 + (c^2*d^2*(3*a*x^4 + 3 
*b*x^4*atanh(c*x)))/12 + (c*d^2*(8*a*x^3 + b*x^3 + 8*b*x^3*atanh(c*x)))/12
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91 \[ \int x (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{2} \left (3 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+8 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+6 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-\mathit {atanh} \left (c x \right ) b +8 \,\mathrm {log}\left (c^{2} x -c \right ) b +3 a \,c^{4} x^{4}+8 a \,c^{3} x^{3}+6 a \,c^{2} x^{2}+b \,c^{3} x^{3}+4 b \,c^{2} x^{2}+9 b c x \right )}{12 c^{2}} \] Input:

int(x*(c*d*x+d)^2*(a+b*atanh(c*x)),x)
 

Output:

(d**2*(3*atanh(c*x)*b*c**4*x**4 + 8*atanh(c*x)*b*c**3*x**3 + 6*atanh(c*x)* 
b*c**2*x**2 - atanh(c*x)*b + 8*log(c**2*x - c)*b + 3*a*c**4*x**4 + 8*a*c** 
3*x**3 + 6*a*c**2*x**2 + b*c**3*x**3 + 4*b*c**2*x**2 + 9*b*c*x))/(12*c**2)