\(\int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx\) [23]

Optimal result

Integrand size = 17, antiderivative size = 84 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=b d^3 x+\frac {b d^3 (1+c x)^2}{4 c}+\frac {b d^3 (1+c x)^3}{12 c}+\frac {d^3 (1+c x)^4 (a+b \text {arctanh}(c x))}{4 c}+\frac {2 b d^3 \log (1-c x)}{c} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.37 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3 \left (24 a c x+42 b c x+36 a c^2 x^2+12 b c^2 x^2+24 a c^3 x^3+2 b c^3 x^3+6 a c^4 x^4+6 b c x \left (4+6 c x+4 c^2 x^2+c^3 x^3\right ) \text {arctanh}(c x)+45 b \log (1-c x)+3 b \log (1+c x)\right )}{24 c} \] Input:

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

Output:

(d^3*(24*a*c*x + 42*b*c*x + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 24*a*c^3*x^3 + 2 
*b*c^3*x^3 + 6*a*c^4*x^4 + 6*b*c*x*(4 + 6*c*x + 4*c^2*x^2 + c^3*x^3)*ArcTa 
nh[c*x] + 45*b*Log[1 - c*x] + 3*b*Log[1 + c*x]))/(24*c)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6478, 27, 456, 49, 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.

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

\(\Big \downarrow \) 6478

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 456

\(\displaystyle \frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c}-\frac {1}{4} b d^3 \int \frac {(c x+1)^3}{1-c x}dx\)

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c) - (b*d^3*(-4*x - (1 + c*x)^2/ 
c - (1 + c*x)^3/(3*c) - (8*Log[1 - c*x])/c))/4
 

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 49

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] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 456

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && 
EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] &&  !Integ 
erQ[n]))
 

rule 2009

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

rule 6478

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {d^{3} a \left (c x +1\right )^{4}}{4}+d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\)	\(101\)
default	\(\frac {\frac {d^{3} a \left (c x +1\right )^{4}}{4}+d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\)	\(101\)
parts	\(\frac {d^{3} a \left (c x +1\right )^{4}}{4 c}+\frac {d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\)	\(103\)
parallelrisch	\(\frac {3 d^{3} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+3 a \,c^{4} d^{3} x^{4}+12 d^{3} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+12 a \,c^{3} d^{3} x^{3}+b \,c^{3} d^{3} x^{3}+18 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{3}+18 a \,c^{2} d^{3} x^{2}+6 b \,c^{2} d^{3} x^{2}+12 b c \,d^{3} x \,\operatorname {arctanh}\left (c x \right )+12 d^{3} a c x +21 b c \,d^{3} x +24 \ln \left (c x -1\right ) b \,d^{3}+3 b \,d^{3} \operatorname {arctanh}\left (c x \right )}{12 c}\)	\(164\)
risch	\(\frac {d^{3} \left (c x +1\right )^{4} b \ln \left (c x +1\right )}{8 c}-\frac {d^{3} c^{3} b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {d^{3} c^{3} a \,x^{4}}{4}-\frac {d^{3} c^{2} b \,x^{3} \ln \left (-c x +1\right )}{2}+d^{3} c^{2} a \,x^{3}+\frac {d^{3} c^{2} b \,x^{3}}{12}-\frac {3 d^{3} c b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {3 d^{3} c a \,x^{2}}{2}+\frac {b c \,d^{3} x^{2}}{2}-\frac {b \,d^{3} x \ln \left (-c x +1\right )}{2}+a \,d^{3} x +\frac {7 b \,d^{3} x}{4}-\frac {\ln \left (-c x +1\right ) b \,d^{3}}{8 c}+\frac {2 d^{3} b \ln \left (c x -1\right )}{c}\)	\(192\)

Input:

int((c*d*x+d)^3*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 

Output:

1/c*(1/4*d^3*a*(c*x+1)^4+d^3*b*(1/4*arctanh(c*x)*c^4*x^4+arctanh(c*x)*c^3* 
x^3+3/2*arctanh(c*x)*c^2*x^2+arctanh(c*x)*c*x+1/4*arctanh(c*x)+1/12*x^3*c^ 
3+1/2*c^2*x^2+7/4*c*x+2*ln(c*x-1)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.77 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, a c^{4} d^{3} x^{4} + 2 \, {\left (12 \, a + b\right )} c^{3} d^{3} x^{3} + 12 \, {\left (3 \, a + b\right )} c^{2} d^{3} x^{2} + 6 \, {\left (4 \, a + 7 \, b\right )} c d^{3} x + 3 \, b d^{3} \log \left (c x + 1\right ) + 45 \, b d^{3} \log \left (c x - 1\right ) + 3 \, {\left (b c^{4} d^{3} x^{4} + 4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c} \] Input:

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

Output:

1/24*(6*a*c^4*d^3*x^4 + 2*(12*a + b)*c^3*d^3*x^3 + 12*(3*a + b)*c^2*d^3*x^ 
2 + 6*(4*a + 7*b)*c*d^3*x + 3*b*d^3*log(c*x + 1) + 45*b*d^3*log(c*x - 1) + 
 3*(b*c^4*d^3*x^4 + 4*b*c^3*d^3*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c*d^3*x)*log(- 
(c*x + 1)/(c*x - 1)))/c
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (73) = 146\).

Time = 0.34 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.17 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{3} d^{3} x^{4}}{4} + a c^{2} d^{3} x^{3} + \frac {3 a c d^{3} x^{2}}{2} + a d^{3} x + \frac {b c^{3} d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + b c^{2} d^{3} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{2} d^{3} x^{3}}{12} + \frac {3 b c d^{3} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b c d^{3} x^{2}}{2} + b d^{3} x \operatorname {atanh}{\left (c x \right )} + \frac {7 b d^{3} x}{4} + \frac {2 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{4 c} & \text {for}\: c \neq 0 \\a d^{3} x & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*c**3*d**3*x**4/4 + a*c**2*d**3*x**3 + 3*a*c*d**3*x**2/2 + a*d 
**3*x + b*c**3*d**3*x**4*atanh(c*x)/4 + b*c**2*d**3*x**3*atanh(c*x) + b*c* 
*2*d**3*x**3/12 + 3*b*c*d**3*x**2*atanh(c*x)/2 + b*c*d**3*x**2/2 + b*d**3* 
x*atanh(c*x) + 7*b*d**3*x/4 + 2*b*d**3*log(x - 1/c)/c + b*d**3*atanh(c*x)/ 
(4*c), Ne(c, 0)), (a*d**3*x, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (78) = 156\).

Time = 0.03 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.61 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{4} \, a c^{3} d^{3} x^{4} + a c^{2} d^{3} x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{3} d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{3} + \frac {3}{2} \, a c d^{3} x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{3} + a d^{3} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (78) = 156\).

Time = 0.13 (sec) , antiderivative size = 425, normalized size of antiderivative = 5.06 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=-\frac {1}{3} \, {\left (\frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {6 \, {\left (\frac {4 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} - \frac {\frac {48 \, {\left (c x + 1\right )}^{3} a d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {72 \, {\left (c x + 1\right )}^{2} a d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {48 \, {\left (c x + 1\right )} a d^{3}}{c x - 1} - 12 \, a d^{3} + \frac {18 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {45 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {38 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - 11 \, b d^{3}}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}}\right )} c \] Input:

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

Output:

-1/3*(6*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 6*b*d^3*log(-(c*x + 1)/( 
c*x - 1))/c^2 - 6*(4*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 6*(c*x + 1)^2*b*d^3/( 
c*x - 1)^2 + 4*(c*x + 1)*b*d^3/(c*x - 1) - b*d^3)*log(-(c*x + 1)/(c*x - 1) 
)/((c*x + 1)^4*c^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 
1)^2*c^2/(c*x - 1)^2 - 4*(c*x + 1)*c^2/(c*x - 1) + c^2) - (48*(c*x + 1)^3* 
a*d^3/(c*x - 1)^3 - 72*(c*x + 1)^2*a*d^3/(c*x - 1)^2 + 48*(c*x + 1)*a*d^3/ 
(c*x - 1) - 12*a*d^3 + 18*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 45*(c*x + 1)^2*b 
*d^3/(c*x - 1)^2 + 38*(c*x + 1)*b*d^3/(c*x - 1) - 11*b*d^3)/((c*x + 1)^4*c 
^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 1)^2*c^2/(c*x - 
1)^2 - 4*(c*x + 1)*c^2/(c*x - 1) + c^2))*c
 

Mupad [B] (verification not implemented)

Time = 3.70 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3\,\left (12\,a\,x+21\,b\,x+12\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^3\,d^3\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12}-\frac {d^3\,\left (21\,b\,\mathrm {atanh}\left (c\,x\right )-12\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12\,c}+\frac {c\,d^3\,\left (18\,a\,x^2+6\,b\,x^2+18\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^2\,d^3\,\left (12\,a\,x^3+b\,x^3+12\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \] Input:

int((a + b*atanh(c*x))*(d + c*d*x)^3,x)
 

Output:

(d^3*(12*a*x + 21*b*x + 12*b*x*atanh(c*x)))/12 + (c^3*d^3*(3*a*x^4 + 3*b*x 
^4*atanh(c*x)))/12 - (d^3*(21*b*atanh(c*x) - 12*b*log(c^2*x^2 - 1)))/(12*c 
) + (c*d^3*(18*a*x^2 + 6*b*x^2 + 18*b*x^2*atanh(c*x)))/12 + (c^2*d^3*(12*a 
*x^3 + b*x^3 + 12*b*x^3*atanh(c*x)))/12
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.56 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{3} \left (3 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+12 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+18 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}+12 \mathit {atanh} \left (c x \right ) b c x +3 \mathit {atanh} \left (c x \right ) b +24 \,\mathrm {log}\left (c^{2} x -c \right ) b +3 a \,c^{4} x^{4}+12 a \,c^{3} x^{3}+18 a \,c^{2} x^{2}+12 a c x +b \,c^{3} x^{3}+6 b \,c^{2} x^{2}+21 b c x \right )}{12 c} \] Input:

int((c*d*x+d)^3*(a+b*atanh(c*x)),x)
 

Output:

(d**3*(3*atanh(c*x)*b*c**4*x**4 + 12*atanh(c*x)*b*c**3*x**3 + 18*atanh(c*x 
)*b*c**2*x**2 + 12*atanh(c*x)*b*c*x + 3*atanh(c*x)*b + 24*log(c**2*x - c)* 
b + 3*a*c**4*x**4 + 12*a*c**3*x**3 + 18*a*c**2*x**2 + 12*a*c*x + b*c**3*x* 
*3 + 6*b*c**2*x**2 + 21*b*c*x))/(12*c)