Integrand size = 18, antiderivative size = 153 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {16 b d^4 x}{15 c}+\frac {4 b d^4 (1+c x)^2}{15 c^2}+\frac {4 b d^4 (1+c x)^3}{45 c^2}+\frac {b d^4 (1+c x)^4}{30 c^2}+\frac {b d^4 (1+c x)^5}{30 c^2}-\frac {d^4 (1+c x)^5 (a+b \text {arctanh}(c x))}{5 c^2}+\frac {d^4 (1+c x)^6 (a+b \text {arctanh}(c x))}{6 c^2}+\frac {32 b d^4 \log (1-c x)}{15 c^2} \] Output:
16/15*b*d^4*x/c+4/15*b*d^4*(c*x+1)^2/c^2+4/45*b*d^4*(c*x+1)^3/c^2+1/30*b*d ^4*(c*x+1)^4/c^2+1/30*b*d^4*(c*x+1)^5/c^2-1/5*d^4*(c*x+1)^5*(a+b*arctanh(c *x))/c^2+1/6*d^4*(c*x+1)^6*(a+b*arctanh(c*x))/c^2+32/15*b*d^4*ln(-c*x+1)/c ^2
Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.04 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^4 \left (390 b c x+90 a c^2 x^2+192 b c^2 x^2+240 a c^3 x^3+100 b c^3 x^3+270 a c^4 x^4+36 b c^4 x^4+144 a c^5 x^5+6 b c^5 x^5+30 a c^6 x^6+6 b c^2 x^2 \left (15+40 c x+45 c^2 x^2+24 c^3 x^3+5 c^4 x^4\right ) \text {arctanh}(c x)+387 b \log (1-c x)-3 b \log (1+c x)\right )}{180 c^2} \] Input:
Integrate[x*(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]
Output:
(d^4*(390*b*c*x + 90*a*c^2*x^2 + 192*b*c^2*x^2 + 240*a*c^3*x^3 + 100*b*c^3 *x^3 + 270*a*c^4*x^4 + 36*b*c^4*x^4 + 144*a*c^5*x^5 + 6*b*c^5*x^5 + 30*a*c ^6*x^6 + 6*b*c^2*x^2*(15 + 40*c*x + 45*c^2*x^2 + 24*c^3*x^3 + 5*c^4*x^4)*A rcTanh[c*x] + 387*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(180*c^2)
Time = 0.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82, 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.
| \(\displaystyle \int x (c d x+d)^4 (a+b \text {arctanh}(c x)) \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int -\frac {d^4 (1-5 c x) (c x+1)^4}{30 c^2 (1-c x)}dx+\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{6 c^2}-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b d^4 \int \frac {(1-5 c x) (c x+1)^4}{1-c x}dx}{30 c}+\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{6 c^2}-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {b d^4 \int \left (5 (c x+1)^4+4 (c x+1)^3+8 (c x+1)^2+16 (c x+1)+\frac {64}{c x-1}+32\right )dx}{30 c}+\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{6 c^2}-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{6 c^2}-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^2}+\frac {b d^4 \left (\frac {(c x+1)^5}{c}+\frac {(c x+1)^4}{c}+\frac {8 (c x+1)^3}{3 c}+\frac {8 (c x+1)^2}{c}+\frac {64 \log (1-c x)}{c}+32 x\right )}{30 c}\) |
Input:
Int[x*(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]
Output:
-1/5*(d^4*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/c^2 + (d^4*(1 + c*x)^6*(a + b* ArcTanh[c*x]))/(6*c^2) + (b*d^4*(32*x + (8*(1 + c*x)^2)/c + (8*(1 + c*x)^3 )/(3*c) + (1 + c*x)^4/c + (1 + c*x)^5/c + (64*Log[1 - c*x])/c))/(30*c)
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 = 0.64 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07
| method | result | size |
| parts | \(d^{4} a \left (\frac {1}{6} c^{4} x^{6}+\frac {4}{5} c^{3} x^{5}+\frac {3}{2} c^{2} x^{4}+\frac {4}{3} c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{2}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{5}+\frac {5 x^{3} c^{3}}{9}+\frac {16 c^{2} x^{2}}{15}+\frac {13 c x}{6}+\frac {43 \ln \left (c x -1\right )}{20}-\frac {\ln \left (c x +1\right )}{60}\right )}{c^{2}}\) | \(164\) |
| derivativedivides | \(\frac {d^{4} a \left (\frac {1}{6} c^{6} x^{6}+\frac {4}{5} c^{5} x^{5}+\frac {3}{2} c^{4} x^{4}+\frac {4}{3} x^{3} c^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{2}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{5}+\frac {5 x^{3} c^{3}}{9}+\frac {16 c^{2} x^{2}}{15}+\frac {13 c x}{6}+\frac {43 \ln \left (c x -1\right )}{20}-\frac {\ln \left (c x +1\right )}{60}\right )}{c^{2}}\) | \(170\) |
| default | \(\frac {d^{4} a \left (\frac {1}{6} c^{6} x^{6}+\frac {4}{5} c^{5} x^{5}+\frac {3}{2} c^{4} x^{4}+\frac {4}{3} x^{3} c^{3}+\frac {1}{2} c^{2} x^{2}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{2}+\frac {4 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+\frac {c^{5} x^{5}}{30}+\frac {c^{4} x^{4}}{5}+\frac {5 x^{3} c^{3}}{9}+\frac {16 c^{2} x^{2}}{15}+\frac {13 c x}{6}+\frac {43 \ln \left (c x -1\right )}{20}-\frac {\ln \left (c x +1\right )}{60}\right )}{c^{2}}\) | \(170\) |
| parallelrisch | \(\frac {15 b \,c^{6} d^{4} \operatorname {arctanh}\left (c x \right ) x^{6}+15 a \,c^{6} d^{4} x^{6}+72 b \,c^{5} d^{4} \operatorname {arctanh}\left (c x \right ) x^{5}+72 a \,c^{5} d^{4} x^{5}+3 b \,c^{5} d^{4} x^{5}+135 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+135 a \,c^{4} d^{4} x^{4}+18 b \,c^{4} d^{4} x^{4}+120 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+120 a \,c^{3} d^{4} x^{3}+50 b \,c^{3} d^{4} x^{3}+45 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{4}+45 a \,c^{2} d^{4} x^{2}+96 b \,c^{2} d^{4} x^{2}+195 b c \,d^{4} x +192 \ln \left (c x -1\right ) b \,d^{4}-3 b \,d^{4} \operatorname {arctanh}\left (c x \right )}{90 c^{2}}\) | \(225\) |
| risch | \(\frac {d^{4} b \,x^{2} \left (5 c^{4} x^{4}+24 x^{3} c^{3}+45 c^{2} x^{2}+40 c x +15\right ) \ln \left (c x +1\right )}{60}-\frac {d^{4} c^{4} b \,x^{6} \ln \left (-c x +1\right )}{12}+\frac {d^{4} c^{4} a \,x^{6}}{6}-\frac {2 d^{4} c^{3} b \,x^{5} \ln \left (-c x +1\right )}{5}+\frac {4 d^{4} c^{3} a \,x^{5}}{5}+\frac {d^{4} c^{3} b \,x^{5}}{30}-\frac {3 d^{4} c^{2} b \,x^{4} \ln \left (-c x +1\right )}{4}+\frac {3 d^{4} c^{2} a \,x^{4}}{2}+\frac {d^{4} c^{2} b \,x^{4}}{5}-\frac {2 d^{4} c b \,x^{3} \ln \left (-c x +1\right )}{3}+\frac {4 d^{4} c a \,x^{3}}{3}+\frac {5 d^{4} c b \,x^{3}}{9}-\frac {d^{4} b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {d^{4} a \,x^{2}}{2}+\frac {16 d^{4} b \,x^{2}}{15}+\frac {13 b \,d^{4} x}{6 c}-\frac {d^{4} b \ln \left (c x +1\right )}{60 c^{2}}+\frac {43 b \,d^{4} \ln \left (-c x +1\right )}{20 c^{2}}\) | \(276\) |
Input:
int(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
d^4*a*(1/6*c^4*x^6+4/5*c^3*x^5+3/2*c^2*x^4+4/3*c*x^3+1/2*x^2)+d^4*b/c^2*(1 /6*arctanh(c*x)*c^6*x^6+4/5*arctanh(c*x)*c^5*x^5+3/2*arctanh(c*x)*c^4*x^4+ 4/3*arctanh(c*x)*c^3*x^3+1/2*arctanh(c*x)*c^2*x^2+1/30*c^5*x^5+1/5*c^4*x^4 +5/9*x^3*c^3+16/15*c^2*x^2+13/6*c*x+43/20*ln(c*x-1)-1/60*ln(c*x+1))
Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.29 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {30 \, a c^{6} d^{4} x^{6} + 6 \, {\left (24 \, a + b\right )} c^{5} d^{4} x^{5} + 18 \, {\left (15 \, a + 2 \, b\right )} c^{4} d^{4} x^{4} + 20 \, {\left (12 \, a + 5 \, b\right )} c^{3} d^{4} x^{3} + 6 \, {\left (15 \, a + 32 \, b\right )} c^{2} d^{4} x^{2} + 390 \, b c d^{4} x - 3 \, b d^{4} \log \left (c x + 1\right ) + 387 \, b d^{4} \log \left (c x - 1\right ) + 3 \, {\left (5 \, b c^{6} d^{4} x^{6} + 24 \, b c^{5} d^{4} x^{5} + 45 \, b c^{4} d^{4} x^{4} + 40 \, b c^{3} d^{4} x^{3} + 15 \, b c^{2} d^{4} x^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{180 \, c^{2}} \] Input:
integrate(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/180*(30*a*c^6*d^4*x^6 + 6*(24*a + b)*c^5*d^4*x^5 + 18*(15*a + 2*b)*c^4*d ^4*x^4 + 20*(12*a + 5*b)*c^3*d^4*x^3 + 6*(15*a + 32*b)*c^2*d^4*x^2 + 390*b *c*d^4*x - 3*b*d^4*log(c*x + 1) + 387*b*d^4*log(c*x - 1) + 3*(5*b*c^6*d^4* x^6 + 24*b*c^5*d^4*x^5 + 45*b*c^4*d^4*x^4 + 40*b*c^3*d^4*x^3 + 15*b*c^2*d^ 4*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2
Time = 0.53 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.76 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{4} d^{4} x^{6}}{6} + \frac {4 a c^{3} d^{4} x^{5}}{5} + \frac {3 a c^{2} d^{4} x^{4}}{2} + \frac {4 a c d^{4} x^{3}}{3} + \frac {a d^{4} x^{2}}{2} + \frac {b c^{4} d^{4} x^{6} \operatorname {atanh}{\left (c x \right )}}{6} + \frac {4 b c^{3} d^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b c^{3} d^{4} x^{5}}{30} + \frac {3 b c^{2} d^{4} x^{4} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b c^{2} d^{4} x^{4}}{5} + \frac {4 b c d^{4} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {5 b c d^{4} x^{3}}{9} + \frac {b d^{4} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {16 b d^{4} x^{2}}{15} + \frac {13 b d^{4} x}{6 c} + \frac {32 b d^{4} \log {\left (x - \frac {1}{c} \right )}}{15 c^{2}} - \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{30 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{4} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*(c*d*x+d)**4*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*c**4*d**4*x**6/6 + 4*a*c**3*d**4*x**5/5 + 3*a*c**2*d**4*x**4/ 2 + 4*a*c*d**4*x**3/3 + a*d**4*x**2/2 + b*c**4*d**4*x**6*atanh(c*x)/6 + 4* b*c**3*d**4*x**5*atanh(c*x)/5 + b*c**3*d**4*x**5/30 + 3*b*c**2*d**4*x**4*a tanh(c*x)/2 + b*c**2*d**4*x**4/5 + 4*b*c*d**4*x**3*atanh(c*x)/3 + 5*b*c*d* *4*x**3/9 + b*d**4*x**2*atanh(c*x)/2 + 16*b*d**4*x**2/15 + 13*b*d**4*x/(6* c) + 32*b*d**4*log(x - 1/c)/(15*c**2) - b*d**4*atanh(c*x)/(30*c**2), Ne(c, 0)), (a*d**4*x**2/2, True))
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (137) = 274\).
Time = 0.03 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.13 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{6} \, a c^{4} d^{4} x^{6} + \frac {4}{5} \, a c^{3} d^{4} x^{5} + \frac {3}{2} \, a c^{2} d^{4} x^{4} + \frac {1}{180} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{4} d^{4} + \frac {1}{5} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{4} + \frac {4}{3} \, a c d^{4} x^{3} + \frac {1}{4} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{4} + \frac {2}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{4} + \frac {1}{2} \, a d^{4} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{4} \] Input:
integrate(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/6*a*c^4*d^4*x^6 + 4/5*a*c^3*d^4*x^5 + 3/2*a*c^2*d^4*x^4 + 1/180*(30*x^6* arctanh(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 + 15*log(c*x - 1)/c^7))*b*c^4*d^4 + 1/5*(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^4 + 4/3*a*c*d^4*x^3 + 1 /4*(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^4 + 2/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log (c^2*x^2 - 1)/c^4))*b*c*d^4 + 1/2*a*d^4*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^4
Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (137) = 274\).
Time = 0.13 (sec) , antiderivative size = 621, normalized size of antiderivative = 4.06 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx =\text {Too large to display} \] Input:
integrate(x*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
-8/45*(12*b*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/c^3 - 12*b*d^4*log(-(c*x + 1 )/(c*x - 1))/c^3 - 6*(15*(c*x + 1)^5*b*d^4/(c*x - 1)^5 - 30*(c*x + 1)^4*b* d^4/(c*x - 1)^4 + 40*(c*x + 1)^3*b*d^4/(c*x - 1)^3 - 30*(c*x + 1)^2*b*d^4/ (c*x - 1)^2 + 12*(c*x + 1)*b*d^4/(c*x - 1) - 2*b*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^6*c^3/(c*x - 1)^6 - 6*(c*x + 1)^5*c^3/(c*x - 1)^5 + 15*(c *x + 1)^4*c^3/(c*x - 1)^4 - 20*(c*x + 1)^3*c^3/(c*x - 1)^3 + 15*(c*x + 1)^ 2*c^3/(c*x - 1)^2 - 6*(c*x + 1)*c^3/(c*x - 1) + c^3) - (180*(c*x + 1)^5*a* d^4/(c*x - 1)^5 - 360*(c*x + 1)^4*a*d^4/(c*x - 1)^4 + 480*(c*x + 1)^3*a*d^ 4/(c*x - 1)^3 - 360*(c*x + 1)^2*a*d^4/(c*x - 1)^2 + 144*(c*x + 1)*a*d^4/(c *x - 1) - 24*a*d^4 + 78*(c*x + 1)^5*b*d^4/(c*x - 1)^5 - 294*(c*x + 1)^4*b* d^4/(c*x - 1)^4 + 472*(c*x + 1)^3*b*d^4/(c*x - 1)^3 - 399*(c*x + 1)^2*b*d^ 4/(c*x - 1)^2 + 174*(c*x + 1)*b*d^4/(c*x - 1) - 31*b*d^4)/((c*x + 1)^6*c^3 /(c*x - 1)^6 - 6*(c*x + 1)^5*c^3/(c*x - 1)^5 + 15*(c*x + 1)^4*c^3/(c*x - 1 )^4 - 20*(c*x + 1)^3*c^3/(c*x - 1)^3 + 15*(c*x + 1)^2*c^3/(c*x - 1)^2 - 6* (c*x + 1)*c^3/(c*x - 1) + c^3))*c
Time = 3.50 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.21 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^4\,\left (45\,a\,x^2+96\,b\,x^2+45\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{90}-\frac {\frac {d^4\,\left (195\,b\,\mathrm {atanh}\left (c\,x\right )-96\,b\,\ln \left (c^2\,x^2-1\right )\right )}{90}-\frac {13\,b\,c\,d^4\,x}{6}}{c^2}+\frac {c^4\,d^4\,\left (15\,a\,x^6+15\,b\,x^6\,\mathrm {atanh}\left (c\,x\right )\right )}{90}+\frac {c\,d^4\,\left (120\,a\,x^3+50\,b\,x^3+120\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{90}+\frac {c^3\,d^4\,\left (72\,a\,x^5+3\,b\,x^5+72\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{90}+\frac {c^2\,d^4\,\left (135\,a\,x^4+18\,b\,x^4+135\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{90} \] Input:
int(x*(a + b*atanh(c*x))*(d + c*d*x)^4,x)
Output:
(d^4*(45*a*x^2 + 96*b*x^2 + 45*b*x^2*atanh(c*x)))/90 - ((d^4*(195*b*atanh( c*x) - 96*b*log(c^2*x^2 - 1)))/90 - (13*b*c*d^4*x)/6)/c^2 + (c^4*d^4*(15*a *x^6 + 15*b*x^6*atanh(c*x)))/90 + (c*d^4*(120*a*x^3 + 50*b*x^3 + 120*b*x^3 *atanh(c*x)))/90 + (c^3*d^4*(72*a*x^5 + 3*b*x^5 + 72*b*x^5*atanh(c*x)))/90 + (c^2*d^4*(135*a*x^4 + 18*b*x^4 + 135*b*x^4*atanh(c*x)))/90
Time = 0.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.18 \[ \int x (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{4} \left (15 \mathit {atanh} \left (c x \right ) b \,c^{6} x^{6}+72 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}+135 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+120 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+45 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-3 \mathit {atanh} \left (c x \right ) b +192 \,\mathrm {log}\left (c^{2} x -c \right ) b +15 a \,c^{6} x^{6}+72 a \,c^{5} x^{5}+135 a \,c^{4} x^{4}+120 a \,c^{3} x^{3}+45 a \,c^{2} x^{2}+3 b \,c^{5} x^{5}+18 b \,c^{4} x^{4}+50 b \,c^{3} x^{3}+96 b \,c^{2} x^{2}+195 b c x \right )}{90 c^{2}} \] Input:
int(x*(c*d*x+d)^4*(a+b*atanh(c*x)),x)
Output:
(d**4*(15*atanh(c*x)*b*c**6*x**6 + 72*atanh(c*x)*b*c**5*x**5 + 135*atanh(c *x)*b*c**4*x**4 + 120*atanh(c*x)*b*c**3*x**3 + 45*atanh(c*x)*b*c**2*x**2 - 3*atanh(c*x)*b + 192*log(c**2*x - c)*b + 15*a*c**6*x**6 + 72*a*c**5*x**5 + 135*a*c**4*x**4 + 120*a*c**3*x**3 + 45*a*c**2*x**2 + 3*b*c**5*x**5 + 18* b*c**4*x**4 + 50*b*c**3*x**3 + 96*b*c**2*x**2 + 195*b*c*x))/(90*c**2)