Integrand size = 18, antiderivative size = 135 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {3 b d^3 x}{5 c}+\frac {3 b d^3 (1+c x)^2}{20 c^2}+\frac {b d^3 (1+c x)^3}{20 c^2}+\frac {b d^3 (1+c x)^4}{20 c^2}-\frac {d^3 (1+c x)^4 (a+b \text {arctanh}(c x))}{4 c^2}+\frac {d^3 (1+c x)^5 (a+b \text {arctanh}(c x))}{5 c^2}+\frac {6 b d^3 \log (1-c x)}{5 c^2} \]
3/5*b*d^3*x/c+3/20*b*d^3*(c*x+1)^2/c^2+1/20*b*d^3*(c*x+1)^3/c^2+1/20*b*d^3 *(c*x+1)^4/c^2-1/4*d^3*(c*x+1)^4*(a+b*arctanh(c*x))/c^2+1/5*d^3*(c*x+1)^5* (a+b*arctanh(c*x))/c^2+6/5*b*d^3*ln(-c*x+1)/c^2
Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3 \left (50 b c x+20 a c^2 x^2+24 b c^2 x^2+40 a c^3 x^3+10 b c^3 x^3+30 a c^4 x^4+2 b c^4 x^4+8 a c^5 x^5+2 b c^2 x^2 \left (10+20 c x+15 c^2 x^2+4 c^3 x^3\right ) \text {arctanh}(c x)+49 b \log (1-c x)-b \log (1+c x)\right )}{40 c^2} \]
(d^3*(50*b*c*x + 20*a*c^2*x^2 + 24*b*c^2*x^2 + 40*a*c^3*x^3 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 2*b*c^4*x^4 + 8*a*c^5*x^5 + 2*b*c^2*x^2*(10 + 20*c*x + 1 5*c^2*x^2 + 4*c^3*x^3)*ArcTanh[c*x] + 49*b*Log[1 - c*x] - b*Log[1 + c*x])) /(40*c^2)
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 86, 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.
| \(\displaystyle \int x (c d x+d)^3 (a+b \text {arctanh}(c x)) \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int -\frac {d^3 (1-4 c x) (c x+1)^3}{20 c^2 (1-c x)}dx+\frac {d^3 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}-\frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b d^3 \int \frac {(1-4 c x) (c x+1)^3}{1-c x}dx}{20 c}+\frac {d^3 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}-\frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c^2}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {b d^3 \int \left (4 (c x+1)^3+3 (c x+1)^2+6 (c x+1)+\frac {24}{c x-1}+12\right )dx}{20 c}+\frac {d^3 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}-\frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}-\frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c^2}+\frac {b d^3 \left (\frac {(c x+1)^4}{c}+\frac {(c x+1)^3}{c}+\frac {3 (c x+1)^2}{c}+\frac {24 \log (1-c x)}{c}+12 x\right )}{20 c}\) |
-1/4*(d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/c^2 + (d^3*(1 + c*x)^5*(a + b* ArcTanh[c*x]))/(5*c^2) + (b*d^3*(12*x + (3*(1 + c*x)^2)/c + (1 + c*x)^3/c + (1 + c*x)^4/c + (24*Log[1 - c*x])/c))/(20*c)
3.1.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 ]))
Time = 1.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99
| method | result | size |
| parts | \(d^{3} a \left (\frac {1}{5} c^{3} x^{5}+\frac {3}{4} c^{2} x^{4}+c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {d^{3} b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {3 c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}+\frac {3 c^{2} x^{2}}{5}+\frac {5 c x}{4}+\frac {49 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{2}}\) | \(134\) |
| derivativedivides | \(\frac {d^{3} a \left (\frac {1}{5} c^{5} x^{5}+\frac {3}{4} c^{4} x^{4}+c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{3} b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {3 c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}+\frac {3 c^{2} x^{2}}{5}+\frac {5 c x}{4}+\frac {49 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{2}}\) | \(140\) |
| default | \(\frac {d^{3} a \left (\frac {1}{5} c^{5} x^{5}+\frac {3}{4} c^{4} x^{4}+c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{3} b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {3 c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}+\frac {3 c^{2} x^{2}}{5}+\frac {5 c x}{4}+\frac {49 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{2}}\) | \(140\) |
| parallelrisch | \(\frac {4 b \,c^{5} d^{3} \operatorname {arctanh}\left (c x \right ) x^{5}+4 a \,c^{5} d^{3} x^{5}+15 x^{4} \operatorname {arctanh}\left (c x \right ) b \,c^{4} d^{3}+15 a \,c^{4} d^{3} x^{4}+b \,c^{4} d^{3} x^{4}+20 x^{3} \operatorname {arctanh}\left (c x \right ) b \,d^{3} c^{3}+20 a \,c^{3} d^{3} x^{3}+5 b \,c^{3} d^{3} x^{3}+10 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{3}+10 a \,c^{2} d^{3} x^{2}+12 b \,c^{2} d^{3} x^{2}+25 b c \,d^{3} x +24 \ln \left (c x -1\right ) b \,d^{3}-b \,d^{3} \operatorname {arctanh}\left (c x \right )}{20 c^{2}}\) | \(184\) |
| risch | \(\frac {d^{3} b \,x^{2} \left (4 c^{3} x^{3}+15 c^{2} x^{2}+20 c x +10\right ) \ln \left (c x +1\right )}{40}-\frac {d^{3} c^{3} b \,x^{5} \ln \left (-c x +1\right )}{10}+\frac {a \,c^{3} d^{3} x^{5}}{5}-\frac {3 d^{3} c^{2} b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {3 a \,c^{2} d^{3} x^{4}}{4}+\frac {b \,c^{2} d^{3} x^{4}}{20}-\frac {d^{3} c b \,x^{3} \ln \left (-c x +1\right )}{2}+a c \,d^{3} x^{3}+\frac {b c \,d^{3} x^{3}}{4}-\frac {d^{3} b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {a \,d^{3} x^{2}}{2}+\frac {3 b \,d^{3} x^{2}}{5}+\frac {5 b \,d^{3} x}{4 c}-\frac {d^{3} b \ln \left (c x +1\right )}{40 c^{2}}+\frac {49 b \,d^{3} \ln \left (-c x +1\right )}{40 c^{2}}\) | \(224\) |
d^3*a*(1/5*c^3*x^5+3/4*c^2*x^4+c*x^3+1/2*x^2)+d^3*b/c^2*(1/5*c^5*x^5*arcta nh(c*x)+3/4*c^4*x^4*arctanh(c*x)+c^3*x^3*arctanh(c*x)+1/2*c^2*x^2*arctanh( c*x)+1/20*c^4*x^4+1/4*c^3*x^3+3/5*c^2*x^2+5/4*c*x+49/40*ln(c*x-1)-1/40*ln( c*x+1))
Time = 0.27 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.22 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {8 \, a c^{5} d^{3} x^{5} + 2 \, {\left (15 \, a + b\right )} c^{4} d^{3} x^{4} + 10 \, {\left (4 \, a + b\right )} c^{3} d^{3} x^{3} + 4 \, {\left (5 \, a + 6 \, b\right )} c^{2} d^{3} x^{2} + 50 \, b c d^{3} x - b d^{3} \log \left (c x + 1\right ) + 49 \, b d^{3} \log \left (c x - 1\right ) + {\left (4 \, b c^{5} d^{3} x^{5} + 15 \, b c^{4} d^{3} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 10 \, b c^{2} d^{3} x^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{40 \, c^{2}} \]
1/40*(8*a*c^5*d^3*x^5 + 2*(15*a + b)*c^4*d^3*x^4 + 10*(4*a + b)*c^3*d^3*x^ 3 + 4*(5*a + 6*b)*c^2*d^3*x^2 + 50*b*c*d^3*x - b*d^3*log(c*x + 1) + 49*b*d ^3*log(c*x - 1) + (4*b*c^5*d^3*x^5 + 15*b*c^4*d^3*x^4 + 20*b*c^3*d^3*x^3 + 10*b*c^2*d^3*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2
Time = 0.40 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.56 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{3} d^{3} x^{5}}{5} + \frac {3 a c^{2} d^{3} x^{4}}{4} + a c d^{3} x^{3} + \frac {a d^{3} x^{2}}{2} + \frac {b c^{3} d^{3} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {3 b c^{2} d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b c^{2} d^{3} x^{4}}{20} + b c d^{3} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c d^{3} x^{3}}{4} + \frac {b d^{3} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {3 b d^{3} x^{2}}{5} + \frac {5 b d^{3} x}{4 c} + \frac {6 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{5 c^{2}} - \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{20 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{3} x^{2}}{2} & \text {otherwise} \end {cases} \]
Piecewise((a*c**3*d**3*x**5/5 + 3*a*c**2*d**3*x**4/4 + a*c*d**3*x**3 + a*d **3*x**2/2 + b*c**3*d**3*x**5*atanh(c*x)/5 + 3*b*c**2*d**3*x**4*atanh(c*x) /4 + b*c**2*d**3*x**4/20 + b*c*d**3*x**3*atanh(c*x) + b*c*d**3*x**3/4 + b* d**3*x**2*atanh(c*x)/2 + 3*b*d**3*x**2/5 + 5*b*d**3*x/(4*c) + 6*b*d**3*log (x - 1/c)/(5*c**2) - b*d**3*atanh(c*x)/(20*c**2), Ne(c, 0)), (a*d**3*x**2/ 2, True))
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (121) = 242\).
Time = 0.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.81 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, a c^{3} d^{3} x^{5} + \frac {3}{4} \, a c^{2} d^{3} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{3} + a c d^{3} x^{3} + \frac {1}{8} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{3} + \frac {1}{2} \, a d^{3} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{3} \]
1/5*a*c^3*d^3*x^5 + 3/4*a*c^2*d^3*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2 *x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*b*c^3*d^3 + a*c*d^3*x^3 + 1/8 *(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*l og(c*x - 1)/c^5))*b*c^2*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*d^3 + 1/2*a*d^3*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^3
Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (121) = 242\).
Time = 0.30 (sec) , antiderivative size = 527, normalized size of antiderivative = 3.90 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=-\frac {1}{5} \, {\left (\frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{3}} - \frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{3}} - \frac {2 \, {\left (\frac {20 \, {\left (c x + 1\right )}^{4} b d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {30 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {30 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {15 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} + 3 \, b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{3}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{3}}{c x - 1} - c^{3}} - \frac {\frac {80 \, {\left (c x + 1\right )}^{4} a d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {120 \, {\left (c x + 1\right )}^{3} a d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {120 \, {\left (c x + 1\right )}^{2} a d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {60 \, {\left (c x + 1\right )} a d^{3}}{c x - 1} + 12 \, a d^{3} + \frac {34 \, {\left (c x + 1\right )}^{4} b d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {103 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {123 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {69 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} + 15 \, b d^{3}}{\frac {{\left (c x + 1\right )}^{5} c^{3}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{3}}{c x - 1} - c^{3}}\right )} c \]
-1/5*(6*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^3 - 6*b*d^3*log(-(c*x + 1)/( c*x - 1))/c^3 - 2*(20*(c*x + 1)^4*b*d^3/(c*x - 1)^4 - 30*(c*x + 1)^3*b*d^3 /(c*x - 1)^3 + 30*(c*x + 1)^2*b*d^3/(c*x - 1)^2 - 15*(c*x + 1)*b*d^3/(c*x - 1) + 3*b*d^3)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^3/(c*x - 1)^5 - 5 *(c*x + 1)^4*c^3/(c*x - 1)^4 + 10*(c*x + 1)^3*c^3/(c*x - 1)^3 - 10*(c*x + 1)^2*c^3/(c*x - 1)^2 + 5*(c*x + 1)*c^3/(c*x - 1) - c^3) - (80*(c*x + 1)^4* a*d^3/(c*x - 1)^4 - 120*(c*x + 1)^3*a*d^3/(c*x - 1)^3 + 120*(c*x + 1)^2*a* d^3/(c*x - 1)^2 - 60*(c*x + 1)*a*d^3/(c*x - 1) + 12*a*d^3 + 34*(c*x + 1)^4 *b*d^3/(c*x - 1)^4 - 103*(c*x + 1)^3*b*d^3/(c*x - 1)^3 + 123*(c*x + 1)^2*b *d^3/(c*x - 1)^2 - 69*(c*x + 1)*b*d^3/(c*x - 1) + 15*b*d^3)/((c*x + 1)^5*c ^3/(c*x - 1)^5 - 5*(c*x + 1)^4*c^3/(c*x - 1)^4 + 10*(c*x + 1)^3*c^3/(c*x - 1)^3 - 10*(c*x + 1)^2*c^3/(c*x - 1)^2 + 5*(c*x + 1)*c^3/(c*x - 1) - c^3)) *c
Time = 3.48 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.13 \[ \int x (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3\,\left (10\,a\,x^2+12\,b\,x^2+10\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{20}-\frac {\frac {d^3\,\left (25\,b\,\mathrm {atanh}\left (c\,x\right )-12\,b\,\ln \left (c^2\,x^2-1\right )\right )}{20}-\frac {5\,b\,c\,d^3\,x}{4}}{c^2}+\frac {c^3\,d^3\,\left (4\,a\,x^5+4\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{20}+\frac {c\,d^3\,\left (20\,a\,x^3+5\,b\,x^3+20\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{20}+\frac {c^2\,d^3\,\left (15\,a\,x^4+b\,x^4+15\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{20} \]