Integrand size = 20, antiderivative size = 177 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {a x}{c^3 d}-\frac {b x}{2 c^3 d}+\frac {b x^2}{6 c^2 d}+\frac {b \text {arctanh}(c x)}{2 c^4 d}+\frac {b x \text {arctanh}(c x)}{c^3 d}-\frac {x^2 (a+b \text {arctanh}(c x))}{2 c^2 d}+\frac {x^3 (a+b \text {arctanh}(c x))}{3 c d}+\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^4 d}+\frac {2 b \log \left (1-c^2 x^2\right )}{3 c^4 d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^4 d} \]
a*x/c^3/d-1/2*b*x/c^3/d+1/6*b*x^2/c^2/d+1/2*b*arctanh(c*x)/c^4/d+b*x*arcta nh(c*x)/c^3/d-1/2*x^2*(a+b*arctanh(c*x))/c^2/d+1/3*x^3*(a+b*arctanh(c*x))/ c/d+(a+b*arctanh(c*x))*ln(2/(c*x+1))/c^4/d+2/3*b*ln(-c^2*x^2+1)/c^4/d-1/2* b*polylog(2,1-2/(c*x+1))/c^4/d
Time = 0.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {-b+6 a c x-3 b c x-3 a c^2 x^2+b c^2 x^2+2 a c^3 x^3+b \text {arctanh}(c x) \left (3+6 c x-3 c^2 x^2+2 c^3 x^3+6 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )-6 a \log (1+c x)+4 b \log \left (1-c^2 x^2\right )-3 b \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )}{6 c^4 d} \]
(-b + 6*a*c*x - 3*b*c*x - 3*a*c^2*x^2 + b*c^2*x^2 + 2*a*c^3*x^3 + b*ArcTan h[c*x]*(3 + 6*c*x - 3*c^2*x^2 + 2*c^3*x^3 + 6*Log[1 + E^(-2*ArcTanh[c*x])] ) - 6*a*Log[1 + c*x] + 4*b*Log[1 - c^2*x^2] - 3*b*PolyLog[2, -E^(-2*ArcTan h[c*x])])/(6*c^4*d)
Time = 1.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6492, 27, 6452, 243, 49, 2009, 6492, 6452, 262, 219, 6492, 2009, 6470, 2849, 2752}
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.
| \(\displaystyle \int \frac {x^3 (a+b \text {arctanh}(c x))}{c d x+d} \, dx\) |
| \(\Big \downarrow \) 6492 |
| \(\displaystyle \frac {\int x^2 (a+b \text {arctanh}(c x))dx}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{d (c x+1)}dx}{c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int x^2 (a+b \text {arctanh}(c x))dx}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{c x+1}dx}{c d}\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{3} b c \int \frac {x^3}{1-c^2 x^2}dx}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{c x+1}dx}{c d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \int \frac {x^2}{1-c^2 x^2}dx^2}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{c x+1}dx}{c d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2-1\right )}\right )dx^2}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{c x+1}dx}{c d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{c x+1}dx}{c d}\) |
| \(\Big \downarrow \) 6492 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\int x (a+b \text {arctanh}(c x))dx}{c}-\frac {\int \frac {x (a+b \text {arctanh}(c x))}{c x+1}dx}{c}}{c d}\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \int \frac {x^2}{1-c^2 x^2}dx}{c}-\frac {\int \frac {x (a+b \text {arctanh}(c x))}{c x+1}dx}{c}}{c d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\int \frac {1}{1-c^2 x^2}dx}{c^2}-\frac {x}{c^2}\right )}{c}-\frac {\int \frac {x (a+b \text {arctanh}(c x))}{c x+1}dx}{c}}{c d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\int \frac {x (a+b \text {arctanh}(c x))}{c x+1}dx}{c}}{c d}\) |
| \(\Big \downarrow \) 6492 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\frac {\int (a+b \text {arctanh}(c x))dx}{c}-\frac {\int \frac {a+b \text {arctanh}(c x)}{c x+1}dx}{c}}{c}}{c d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c}-\frac {\int \frac {a+b \text {arctanh}(c x)}{c x+1}dx}{c}}{c}}{c d}\) |
| \(\Big \downarrow \) 6470 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c}-\frac {b \int \frac {\log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c}}{c}}{c}}{c d}\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c}-\frac {\frac {b \int \frac {\log \left (\frac {2}{c x+1}\right )}{1-\frac {2}{c x+1}}d\frac {1}{c x+1}}{c}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c}}{c}}{c}}{c d}\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c d}-\frac {\frac {\frac {1}{2} x^2 (a+b \text {arctanh}(c x))-\frac {1}{2} b c \left (\frac {\text {arctanh}(c x)}{c^3}-\frac {x}{c^2}\right )}{c}-\frac {\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c}-\frac {\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c}}{c}}{c}}{c d}\) |
((x^3*(a + b*ArcTanh[c*x]))/3 - (b*c*(-(x^2/c^2) - Log[1 - c^2*x^2]/c^4))/ 6)/(c*d) - (((x^2*(a + b*ArcTanh[c*x]))/2 - (b*c*(-(x/c^2) + ArcTanh[c*x]/ c^3))/2)/c - ((a*x + b*x*ArcTanh[c*x] + (b*Log[1 - c^2*x^2])/(2*c))/c - (- (((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/c) + (b*PolyLog[2, 1 - 2/(1 + c*x )])/(2*c))/c)/c)/(c*d)
3.1.43.3.1 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_))^(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol ] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c *(p/e) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^2*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2 , 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[f/e Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x]) ^p, x], x] - Simp[d*(f/e) Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 0]
Time = 1.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\frac {a \left (\frac {c^{3} x^{3}}{3}-\frac {c^{2} x^{2}}{2}+c x -\ln \left (c x +1\right )\right )}{d}+\frac {b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}-\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\left (c x +1\right )^{2}}{6}-\frac {5 c x}{6}-\frac {5}{6}+\frac {5 \ln \left (c x -1\right )}{12}+\frac {11 \ln \left (c x +1\right )}{12}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d}}{c^{4}}\) | \(162\) |
| default | \(\frac {\frac {a \left (\frac {c^{3} x^{3}}{3}-\frac {c^{2} x^{2}}{2}+c x -\ln \left (c x +1\right )\right )}{d}+\frac {b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}-\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\left (c x +1\right )^{2}}{6}-\frac {5 c x}{6}-\frac {5}{6}+\frac {5 \ln \left (c x -1\right )}{12}+\frac {11 \ln \left (c x +1\right )}{12}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d}}{c^{4}}\) | \(162\) |
| parts | \(\frac {a \left (\frac {\frac {1}{3} x^{3} c^{2}-\frac {1}{2} c \,x^{2}+x}{c^{3}}-\frac {\ln \left (c x +1\right )}{c^{4}}\right )}{d}+\frac {b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}-\frac {c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\left (c x +1\right )^{2}}{6}-\frac {5 c x}{6}-\frac {5}{6}+\frac {5 \ln \left (c x -1\right )}{12}+\frac {11 \ln \left (c x +1\right )}{12}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d \,c^{4}}\) | \(165\) |
| risch | \(-\frac {b \ln \left (c x +1\right )^{2}}{4 d \,c^{4}}+\frac {b \left (\frac {1}{3} x^{3} c^{2}-\frac {1}{2} c \,x^{2}+x \right ) \ln \left (c x +1\right )}{2 d \,c^{3}}-\frac {b \,x^{3} \ln \left (-c x +1\right )}{6 d c}+\frac {b \,x^{2} \ln \left (-c x +1\right )}{4 d \,c^{2}}-\frac {b x \ln \left (-c x +1\right )}{2 d \,c^{3}}+\frac {5 b \ln \left (-c x +1\right )}{12 d \,c^{4}}+\frac {b \,x^{2}}{6 c^{2} d}-\frac {b x}{2 c^{3} d}-\frac {31 b}{72 d \,c^{4}}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d \,c^{4}}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d \,c^{4}}-\frac {b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d \,c^{4}}+\frac {x^{3} a}{3 d c}-\frac {a \,x^{2}}{2 d \,c^{2}}+\frac {a x}{c^{3} d}-\frac {5 a}{6 d \,c^{4}}-\frac {a \ln \left (-c x -1\right )}{d \,c^{4}}+\frac {11 b \ln \left (c x +1\right )}{12 d \,c^{4}}\) | \(287\) |
1/c^4*(a/d*(1/3*c^3*x^3-1/2*c^2*x^2+c*x-ln(c*x+1))+b/d*(1/3*c^3*x^3*arctan h(c*x)-1/2*c^2*x^2*arctanh(c*x)+c*x*arctanh(c*x)-arctanh(c*x)*ln(c*x+1)+1/ 6*(c*x+1)^2-5/6*c*x-5/6+5/12*ln(c*x-1)+11/12*ln(c*x+1)-1/2*(ln(c*x+1)-ln(1 /2*c*x+1/2))*ln(-1/2*c*x+1/2)+1/2*dilog(1/2*c*x+1/2)+1/4*ln(c*x+1)^2))
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{3}}{c d x + d} \,d x } \]
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {\int \frac {a x^{3}}{c x + 1}\, dx + \int \frac {b x^{3} \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{3}}{c d x + d} \,d x } \]
1/72*(2*c^4*(2*(c^2*x^3 + 3*x)/(c^7*d) - 3*log(c*x + 1)/(c^8*d) + 3*log(c* x - 1)/(c^8*d)) + 216*c^4*integrate(1/6*x^4*log(c*x + 1)/(c^5*d*x^2 - c^3* d), x) - 3*c^3*(x^2/(c^5*d) + log(c^2*x^2 - 1)/(c^7*d)) - 216*c^3*integrat e(1/6*x^3*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x) + 9*c^2*(2*x/(c^5*d) - log( c*x + 1)/(c^6*d) + log(c*x - 1)/(c^6*d)) - 216*c*integrate(1/6*x*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x) - 6*(2*c^3*x^3 - 3*c^2*x^2 + 6*c*x - 6*log(c*x + 1))*log(-c*x + 1)/(c^4*d) + 18*log(6*c^5*d*x^2 - 6*c^3*d)/(c^4*d) - 216 *integrate(1/6*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x))*b + 1/6*a*((2*c^2*x^3 - 3*c*x^2 + 6*x)/(c^3*d) - 6*log(c*x + 1)/(c^4*d))
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{3}}{c d x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{d+c\,d\,x} \,d x \]