\(\int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx\) [10]

Optimal result

Integrand size = 20, antiderivative size = 157 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {5 b d^2 x}{12 c^3}+\frac {b d^2 x^2}{5 c^2}+\frac {5 b d^2 x^3}{36 c}+\frac {1}{10} b d^2 x^4+\frac {1}{30} b c d^2 x^5+\frac {1}{4} d^2 x^4 (a+b \text {arctanh}(c x))+\frac {2}{5} c d^2 x^5 (a+b \text {arctanh}(c x))+\frac {1}{6} c^2 d^2 x^6 (a+b \text {arctanh}(c x))+\frac {49 b d^2 \log (1-c x)}{120 c^4}-\frac {b d^2 \log (1+c x)}{120 c^4} \] Output:

5/12*b*d^2*x/c^3+1/5*b*d^2*x^2/c^2+5/36*b*d^2*x^3/c+1/10*b*d^2*x^4+1/30*b* 
c*d^2*x^5+1/4*d^2*x^4*(a+b*arctanh(c*x))+2/5*c*d^2*x^5*(a+b*arctanh(c*x))+ 
1/6*c^2*d^2*x^6*(a+b*arctanh(c*x))+49/120*b*d^2*ln(-c*x+1)/c^4-1/120*b*d^2 
*ln(c*x+1)/c^4
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.80 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^2 \left (150 b c x+72 b c^2 x^2+50 b c^3 x^3+90 a c^4 x^4+36 b c^4 x^4+144 a c^5 x^5+12 b c^5 x^5+60 a c^6 x^6+6 b c^4 x^4 \left (15+24 c x+10 c^2 x^2\right ) \text {arctanh}(c x)+147 b \log (1-c x)-3 b \log (1+c x)\right )}{360 c^4} \] Input:

Integrate[x^3*(d + c*d*x)^2*(a + b*ArcTanh[c*x]),x]
 

Output:

(d^2*(150*b*c*x + 72*b*c^2*x^2 + 50*b*c^3*x^3 + 90*a*c^4*x^4 + 36*b*c^4*x^ 
4 + 144*a*c^5*x^5 + 12*b*c^5*x^5 + 60*a*c^6*x^6 + 6*b*c^4*x^4*(15 + 24*c*x 
 + 10*c^2*x^2)*ArcTanh[c*x] + 147*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(360 
*c^4)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.

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

\(\Big \downarrow \) 6498

\(\displaystyle -b c \int \frac {d^2 x^4 \left (10 c^2 x^2+24 c x+15\right )}{60 \left (1-c^2 x^2\right )}dx+\frac {1}{6} c^2 d^2 x^6 (a+b \text {arctanh}(c x))+\frac {2}{5} c d^2 x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d^2 x^4 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{60} b c d^2 \int \frac {x^4 \left (10 c^2 x^2+24 c x+15\right )}{1-c^2 x^2}dx+\frac {1}{6} c^2 d^2 x^6 (a+b \text {arctanh}(c x))+\frac {2}{5} c d^2 x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d^2 x^4 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 2333

\(\displaystyle -\frac {1}{60} b c d^2 \int \left (-10 x^4-\frac {24 x^3}{c}-\frac {25 x^2}{c^2}-\frac {24 x}{c^3}+\frac {24 c x+25}{c^4 \left (1-c^2 x^2\right )}-\frac {25}{c^4}\right )dx+\frac {1}{6} c^2 d^2 x^6 (a+b \text {arctanh}(c x))+\frac {2}{5} c d^2 x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d^2 x^4 (a+b \text {arctanh}(c x))\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{6} c^2 d^2 x^6 (a+b \text {arctanh}(c x))+\frac {2}{5} c d^2 x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d^2 x^4 (a+b \text {arctanh}(c x))-\frac {1}{60} b c d^2 \left (\frac {25 \text {arctanh}(c x)}{c^5}-\frac {25 x}{c^4}-\frac {12 x^2}{c^3}-\frac {25 x^3}{3 c^2}-\frac {12 \log \left (1-c^2 x^2\right )}{c^5}-\frac {6 x^4}{c}-2 x^5\right )\)

Input:

Int[x^3*(d + c*d*x)^2*(a + b*ArcTanh[c*x]),x]
 

Output:

(d^2*x^4*(a + b*ArcTanh[c*x]))/4 + (2*c*d^2*x^5*(a + b*ArcTanh[c*x]))/5 + 
(c^2*d^2*x^6*(a + b*ArcTanh[c*x]))/6 - (b*c*d^2*((-25*x)/c^4 - (12*x^2)/c^ 
3 - (25*x^3)/(3*c^2) - (6*x^4)/c - 2*x^5 + (25*ArcTanh[c*x])/c^5 - (12*Log 
[1 - c^2*x^2])/c^5))/60
 

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 = 0.41 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79

method	result	size
parts	\(d^{2} a \left (\frac {1}{6} c^{2} x^{6}+\frac {2}{5} c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{10}+\frac {5 x^{3} c^{3}}{36}+\frac {c^{2} x^{2}}{5}+\frac {5 c x}{12}+\frac {49 \ln \left (c x -1\right )}{120}-\frac {\ln \left (c x +1\right )}{120}\right )}{c^{4}}\)	\(124\)
derivativedivides	\(\frac {d^{2} a \left (\frac {1}{6} c^{6} x^{6}+\frac {2}{5} c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{10}+\frac {5 x^{3} c^{3}}{36}+\frac {c^{2} x^{2}}{5}+\frac {5 c x}{12}+\frac {49 \ln \left (c x -1\right )}{120}-\frac {\ln \left (c x +1\right )}{120}\right )}{c^{4}}\)	\(130\)
default	\(\frac {d^{2} a \left (\frac {1}{6} c^{6} x^{6}+\frac {2}{5} c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{10}+\frac {5 x^{3} c^{3}}{36}+\frac {c^{2} x^{2}}{5}+\frac {5 c x}{12}+\frac {49 \ln \left (c x -1\right )}{120}-\frac {\ln \left (c x +1\right )}{120}\right )}{c^{4}}\)	\(130\)
parallelrisch	\(\frac {30 b \,c^{6} d^{2} \operatorname {arctanh}\left (c x \right ) x^{6}+30 c^{6} d^{2} x^{6} a +72 b \,c^{5} d^{2} \operatorname {arctanh}\left (c x \right ) x^{5}+72 a \,c^{5} d^{2} x^{5}+6 c^{5} d^{2} x^{5} b +45 d^{2} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+45 a \,c^{4} d^{2} x^{4}+18 b \,c^{4} d^{2} x^{4}+25 b \,c^{3} d^{2} x^{3}+36 b \,c^{2} d^{2} x^{2}+75 b c \,d^{2} x +72 \ln \left (c x -1\right ) b \,d^{2}-3 b \,d^{2} \operatorname {arctanh}\left (c x \right )+36 d^{2} b}{180 c^{4}}\)	\(175\)
risch	\(\frac {d^{2} b \,x^{4} \left (10 c^{2} x^{2}+24 c x +15\right ) \ln \left (c x +1\right )}{120}-\frac {d^{2} c^{2} b \,x^{6} \ln \left (-c x +1\right )}{12}+\frac {d^{2} c^{2} a \,x^{6}}{6}-\frac {d^{2} c b \,x^{5} \ln \left (-c x +1\right )}{5}+\frac {2 d^{2} c a \,x^{5}}{5}+\frac {b c \,d^{2} x^{5}}{30}-\frac {d^{2} b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {d^{2} a \,x^{4}}{4}+\frac {b \,d^{2} x^{4}}{10}+\frac {5 b \,d^{2} x^{3}}{36 c}+\frac {b \,d^{2} x^{2}}{5 c^{2}}+\frac {5 b \,d^{2} x}{12 c^{3}}-\frac {b \,d^{2} \ln \left (c x +1\right )}{120 c^{4}}+\frac {49 b \,d^{2} \ln \left (-c x +1\right )}{120 c^{4}}\)	\(198\)

Input:

int(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 

Output:

d^2*a*(1/6*c^2*x^6+2/5*c*x^5+1/4*x^4)+d^2*b/c^4*(1/6*arctanh(c*x)*c^6*x^6+ 
2/5*arctanh(c*x)*c^5*x^5+1/4*arctanh(c*x)*c^4*x^4+1/30*c^5*x^5+1/10*c^4*x^ 
4+5/36*x^3*c^3+1/5*c^2*x^2+5/12*c*x+49/120*ln(c*x-1)-1/120*ln(c*x+1))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {60 \, a c^{6} d^{2} x^{6} + 12 \, {\left (12 \, a + b\right )} c^{5} d^{2} x^{5} + 18 \, {\left (5 \, a + 2 \, b\right )} c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} + 72 \, b c^{2} d^{2} x^{2} + 150 \, b c d^{2} x - 3 \, b d^{2} \log \left (c x + 1\right ) + 147 \, b d^{2} \log \left (c x - 1\right ) + 3 \, {\left (10 \, b c^{6} d^{2} x^{6} + 24 \, b c^{5} d^{2} x^{5} + 15 \, b c^{4} d^{2} x^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{360 \, c^{4}} \] Input:

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="fricas")
 

Output:

1/360*(60*a*c^6*d^2*x^6 + 12*(12*a + b)*c^5*d^2*x^5 + 18*(5*a + 2*b)*c^4*d 
^2*x^4 + 50*b*c^3*d^2*x^3 + 72*b*c^2*d^2*x^2 + 150*b*c*d^2*x - 3*b*d^2*log 
(c*x + 1) + 147*b*d^2*log(c*x - 1) + 3*(10*b*c^6*d^2*x^6 + 24*b*c^5*d^2*x^ 
5 + 15*b*c^4*d^2*x^4)*log(-(c*x + 1)/(c*x - 1)))/c^4
 

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.25 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{2} d^{2} x^{6}}{6} + \frac {2 a c d^{2} x^{5}}{5} + \frac {a d^{2} x^{4}}{4} + \frac {b c^{2} d^{2} x^{6} \operatorname {atanh}{\left (c x \right )}}{6} + \frac {2 b c d^{2} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b c d^{2} x^{5}}{30} + \frac {b d^{2} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d^{2} x^{4}}{10} + \frac {5 b d^{2} x^{3}}{36 c} + \frac {b d^{2} x^{2}}{5 c^{2}} + \frac {5 b d^{2} x}{12 c^{3}} + \frac {2 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{5 c^{4}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{60 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

integrate(x**3*(c*d*x+d)**2*(a+b*atanh(c*x)),x)
 

Output:

Piecewise((a*c**2*d**2*x**6/6 + 2*a*c*d**2*x**5/5 + a*d**2*x**4/4 + b*c**2 
*d**2*x**6*atanh(c*x)/6 + 2*b*c*d**2*x**5*atanh(c*x)/5 + b*c*d**2*x**5/30 
+ b*d**2*x**4*atanh(c*x)/4 + b*d**2*x**4/10 + 5*b*d**2*x**3/(36*c) + b*d** 
2*x**2/(5*c**2) + 5*b*d**2*x/(12*c**3) + 2*b*d**2*log(x - 1/c)/(5*c**4) - 
b*d**2*atanh(c*x)/(60*c**4), Ne(c, 0)), (a*d**2*x**4/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.34 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{6} \, a c^{2} d^{2} x^{6} + \frac {2}{5} \, a c d^{2} x^{5} + \frac {1}{4} \, a d^{2} x^{4} + \frac {1}{180} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{2} d^{2} + \frac {1}{10} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c d^{2} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d^{2} \] Input:

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="maxima")
 

Output:

1/6*a*c^2*d^2*x^6 + 2/5*a*c*d^2*x^5 + 1/4*a*d^2*x^4 + 1/180*(30*x^6*arctan 
h(c*x) + c*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^6 - 15*log(c*x + 1)/c^7 + 1 
5*log(c*x - 1)/c^7))*b*c^2*d^2 + 1/10*(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*d^2 + 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*d^ 
2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (137) = 274\).

Time = 0.13 (sec) , antiderivative size = 620, normalized size of antiderivative = 3.95 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx =\text {Too large to display} \] Input:

integrate(x^3*(c*d*x+d)^2*(a+b*arctanh(c*x)),x, algorithm="giac")
 

Output:

1/45*c*(6*(30*(c*x + 1)^5*b*d^2/(c*x - 1)^5 - 30*(c*x + 1)^4*b*d^2/(c*x - 
1)^4 + 70*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 45*(c*x + 1)^2*b*d^2/(c*x - 1)^2 
 + 18*(c*x + 1)*b*d^2/(c*x - 1) - 3*b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x 
 + 1)^6*c^5/(c*x - 1)^6 - 6*(c*x + 1)^5*c^5/(c*x - 1)^5 + 15*(c*x + 1)^4*c 
^5/(c*x - 1)^4 - 20*(c*x + 1)^3*c^5/(c*x - 1)^3 + 15*(c*x + 1)^2*c^5/(c*x 
- 1)^2 - 6*(c*x + 1)*c^5/(c*x - 1) + c^5) + (360*(c*x + 1)^5*a*d^2/(c*x - 
1)^5 - 360*(c*x + 1)^4*a*d^2/(c*x - 1)^4 + 840*(c*x + 1)^3*a*d^2/(c*x - 1) 
^3 - 540*(c*x + 1)^2*a*d^2/(c*x - 1)^2 + 216*(c*x + 1)*a*d^2/(c*x - 1) - 3 
6*a*d^2 + 162*(c*x + 1)^5*b*d^2/(c*x - 1)^5 - 531*(c*x + 1)^4*b*d^2/(c*x - 
 1)^4 + 818*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 696*(c*x + 1)^2*b*d^2/(c*x - 1 
)^2 + 300*(c*x + 1)*b*d^2/(c*x - 1) - 53*b*d^2)/((c*x + 1)^6*c^5/(c*x - 1) 
^6 - 6*(c*x + 1)^5*c^5/(c*x - 1)^5 + 15*(c*x + 1)^4*c^5/(c*x - 1)^4 - 20*( 
c*x + 1)^3*c^5/(c*x - 1)^3 + 15*(c*x + 1)^2*c^5/(c*x - 1)^2 - 6*(c*x + 1)* 
c^5/(c*x - 1) + c^5) - 18*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 18*b*d 
^2*log(-(c*x + 1)/(c*x - 1))/c^5)
 

Mupad [B] (verification not implemented)

Time = 3.64 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.93 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {\frac {b\,c^2\,d^2\,x^2}{5}-\frac {d^2\,\left (75\,b\,\mathrm {atanh}\left (c\,x\right )-36\,b\,\ln \left (c^2\,x^2-1\right )\right )}{180}+\frac {5\,b\,c^3\,d^2\,x^3}{36}+\frac {5\,b\,c\,d^2\,x}{12}}{c^4}+\frac {d^2\,\left (45\,a\,x^4+18\,b\,x^4+45\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c^2\,d^2\,\left (30\,a\,x^6+30\,b\,x^6\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c\,d^2\,\left (72\,a\,x^5+6\,b\,x^5+72\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{180} \] Input:

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

Output:

((b*c^2*d^2*x^2)/5 - (d^2*(75*b*atanh(c*x) - 36*b*log(c^2*x^2 - 1)))/180 + 
 (5*b*c^3*d^2*x^3)/36 + (5*b*c*d^2*x)/12)/c^4 + (d^2*(45*a*x^4 + 18*b*x^4 
+ 45*b*x^4*atanh(c*x)))/180 + (c^2*d^2*(30*a*x^6 + 30*b*x^6*atanh(c*x)))/1 
80 + (c*d^2*(72*a*x^5 + 6*b*x^5 + 72*b*x^5*atanh(c*x)))/180
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int x^3 (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{2} \left (30 \mathit {atanh} \left (c x \right ) b \,c^{6} x^{6}+72 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}+45 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-3 \mathit {atanh} \left (c x \right ) b +72 \,\mathrm {log}\left (c^{2} x -c \right ) b +30 a \,c^{6} x^{6}+72 a \,c^{5} x^{5}+45 a \,c^{4} x^{4}+6 b \,c^{5} x^{5}+18 b \,c^{4} x^{4}+25 b \,c^{3} x^{3}+36 b \,c^{2} x^{2}+75 b c x \right )}{180 c^{4}} \] Input:

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

Output:

(d**2*(30*atanh(c*x)*b*c**6*x**6 + 72*atanh(c*x)*b*c**5*x**5 + 45*atanh(c* 
x)*b*c**4*x**4 - 3*atanh(c*x)*b + 72*log(c**2*x - c)*b + 30*a*c**6*x**6 + 
72*a*c**5*x**5 + 45*a*c**4*x**4 + 6*b*c**5*x**5 + 18*b*c**4*x**4 + 25*b*c* 
*3*x**3 + 36*b*c**2*x**2 + 75*b*c*x))/(180*c**4)