Integrand size = 20, antiderivative size = 171 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {5 b d^4 x}{3 c^2}+\frac {88 b d^4 x^2}{105 c}+\frac {5}{9} b d^4 x^3+\frac {47}{140} b c d^4 x^4+\frac {2}{15} b c^2 d^4 x^5+\frac {1}{42} b c^3 d^4 x^6+\frac {d^4 (1+c x)^5 (a+b \text {arctanh}(c x))}{5 c^3}-\frac {d^4 (1+c x)^6 (a+b \text {arctanh}(c x))}{3 c^3}+\frac {d^4 (1+c x)^7 (a+b \text {arctanh}(c x))}{7 c^3}+\frac {176 b d^4 \log (1-c x)}{105 c^3} \] Output:
5/3*b*d^4*x/c^2+88/105*b*d^4*x^2/c+5/9*b*d^4*x^3+47/140*b*c*d^4*x^4+2/15*b *c^2*d^4*x^5+1/42*b*c^3*d^4*x^6+1/5*d^4*(c*x+1)^5*(a+b*arctanh(c*x))/c^3-1 /3*d^4*(c*x+1)^6*(a+b*arctanh(c*x))/c^3+1/7*d^4*(c*x+1)^7*(a+b*arctanh(c*x ))/c^3+176/105*b*d^4*ln(-c*x+1)/c^3
Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.98 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^4 \left (2100 b c x+1056 b c^2 x^2+420 a c^3 x^3+700 b c^3 x^3+1260 a c^4 x^4+423 b c^4 x^4+1512 a c^5 x^5+168 b c^5 x^5+840 a c^6 x^6+30 b c^6 x^6+180 a c^7 x^7+12 b c^3 x^3 \left (35+105 c x+126 c^2 x^2+70 c^3 x^3+15 c^4 x^4\right ) \text {arctanh}(c x)+2106 b \log (1-c x)+6 b \log (1+c x)\right )}{1260 c^3} \] Input:
Integrate[x^2*(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]
Output:
(d^4*(2100*b*c*x + 1056*b*c^2*x^2 + 420*a*c^3*x^3 + 700*b*c^3*x^3 + 1260*a *c^4*x^4 + 423*b*c^4*x^4 + 1512*a*c^5*x^5 + 168*b*c^5*x^5 + 840*a*c^6*x^6 + 30*b*c^6*x^6 + 180*a*c^7*x^7 + 12*b*c^3*x^3*(35 + 105*c*x + 126*c^2*x^2 + 70*c^3*x^3 + 15*c^4*x^4)*ArcTanh[c*x] + 2106*b*Log[1 - c*x] + 6*b*Log[1 + c*x]))/(1260*c^3)
Time = 0.41 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6498, 27, 1195, 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.
| \(\displaystyle \int x^2 (c d x+d)^4 (a+b \text {arctanh}(c x)) \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int \frac {d^4 (c x+1)^4 \left (15 c^2 x^2-5 c x+1\right )}{105 c^3 (1-c x)}dx+\frac {d^4 (c x+1)^7 (a+b \text {arctanh}(c x))}{7 c^3}-\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{3 c^3}+\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b d^4 \int \frac {(c x+1)^4 \left (15 c^2 x^2-5 c x+1\right )}{1-c x}dx}{105 c^2}+\frac {d^4 (c x+1)^7 (a+b \text {arctanh}(c x))}{7 c^3}-\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{3 c^3}+\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^3}\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle -\frac {b d^4 \int \left (-15 c^5 x^5-70 c^4 x^4-141 c^3 x^3-175 c^2 x^2-176 c x-\frac {176}{c x-1}-175\right )dx}{105 c^2}+\frac {d^4 (c x+1)^7 (a+b \text {arctanh}(c x))}{7 c^3}-\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{3 c^3}+\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 (c x+1)^7 (a+b \text {arctanh}(c x))}{7 c^3}-\frac {d^4 (c x+1)^6 (a+b \text {arctanh}(c x))}{3 c^3}+\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c^3}-\frac {b d^4 \left (-\frac {5}{2} c^5 x^6-14 c^4 x^5-\frac {141 c^3 x^4}{4}-\frac {175 c^2 x^3}{3}-88 c x^2-\frac {176 \log (1-c x)}{c}-175 x\right )}{105 c^2}\) |
Input:
Int[x^2*(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]
Output:
(d^4*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^3) - (d^4*(1 + c*x)^6*(a + b*A rcTanh[c*x]))/(3*c^3) + (d^4*(1 + c*x)^7*(a + b*ArcTanh[c*x]))/(7*c^3) - ( b*d^4*(-175*x - 88*c*x^2 - (175*c^2*x^3)/3 - (141*c^3*x^4)/4 - 14*c^4*x^5 - (5*c^5*x^6)/2 - (176*Log[1 - c*x])/c))/(105*c^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
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 = 0.72 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99
| method | result | size |
| parts | \(d^{4} a \left (\frac {1}{7} c^{4} x^{7}+\frac {2}{3} c^{3} x^{6}+\frac {6}{5} c^{2} x^{5}+c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{7} x^{7}}{7}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{3}+\frac {6 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{6} x^{6}}{42}+\frac {2 c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}+\frac {5 x^{3} c^{3}}{9}+\frac {88 c^{2} x^{2}}{105}+\frac {5 c x}{3}+\frac {117 \ln \left (c x -1\right )}{70}+\frac {\ln \left (c x +1\right )}{210}\right )}{c^{3}}\) | \(170\) |
| derivativedivides | \(\frac {d^{4} a \left (\frac {1}{7} c^{7} x^{7}+\frac {2}{3} c^{6} x^{6}+\frac {6}{5} c^{5} x^{5}+c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{7} x^{7}}{7}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{3}+\frac {6 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{6} x^{6}}{42}+\frac {2 c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}+\frac {5 x^{3} c^{3}}{9}+\frac {88 c^{2} x^{2}}{105}+\frac {5 c x}{3}+\frac {117 \ln \left (c x -1\right )}{70}+\frac {\ln \left (c x +1\right )}{210}\right )}{c^{3}}\) | \(176\) |
| default | \(\frac {d^{4} a \left (\frac {1}{7} c^{7} x^{7}+\frac {2}{3} c^{6} x^{6}+\frac {6}{5} c^{5} x^{5}+c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{7} x^{7}}{7}+\frac {2 \,\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{3}+\frac {6 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{6} x^{6}}{42}+\frac {2 c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}+\frac {5 x^{3} c^{3}}{9}+\frac {88 c^{2} x^{2}}{105}+\frac {5 c x}{3}+\frac {117 \ln \left (c x -1\right )}{70}+\frac {\ln \left (c x +1\right )}{210}\right )}{c^{3}}\) | \(176\) |
| parallelrisch | \(\frac {180 b \,c^{7} d^{4} \operatorname {arctanh}\left (c x \right ) x^{7}+180 a \,c^{7} d^{4} x^{7}+840 b \,c^{6} d^{4} \operatorname {arctanh}\left (c x \right ) x^{6}+840 a \,c^{6} d^{4} x^{6}+30 b \,c^{6} d^{4} x^{6}+1512 b \,c^{5} d^{4} \operatorname {arctanh}\left (c x \right ) x^{5}+1512 a \,c^{5} d^{4} x^{5}+168 b \,c^{5} d^{4} x^{5}+1260 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+1260 a \,c^{4} d^{4} x^{4}+423 b \,c^{4} d^{4} x^{4}+420 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+420 a \,c^{3} d^{4} x^{3}+700 b \,c^{3} d^{4} x^{3}+1056 b \,c^{2} d^{4} x^{2}+2100 b c \,d^{4} x +2112 \ln \left (c x -1\right ) b \,d^{4}+12 b \,d^{4} \operatorname {arctanh}\left (c x \right )}{1260 c^{3}}\) | \(237\) |
| risch | \(\frac {d^{4} b \,x^{3} \left (15 c^{4} x^{4}+70 x^{3} c^{3}+126 c^{2} x^{2}+105 c x +35\right ) \ln \left (c x +1\right )}{210}-\frac {d^{4} c^{4} b \,x^{7} \ln \left (-c x +1\right )}{14}+\frac {d^{4} c^{4} a \,x^{7}}{7}-\frac {d^{4} c^{3} b \,x^{6} \ln \left (-c x +1\right )}{3}+\frac {2 d^{4} c^{3} a \,x^{6}}{3}+\frac {b \,c^{3} d^{4} x^{6}}{42}-\frac {3 d^{4} c^{2} b \,x^{5} \ln \left (-c x +1\right )}{5}+\frac {6 d^{4} c^{2} a \,x^{5}}{5}+\frac {2 b \,c^{2} d^{4} x^{5}}{15}-\frac {d^{4} c b \,x^{4} \ln \left (-c x +1\right )}{2}+d^{4} c a \,x^{4}+\frac {47 b c \,d^{4} x^{4}}{140}-\frac {d^{4} b \,x^{3} \ln \left (-c x +1\right )}{6}+\frac {d^{4} a \,x^{3}}{3}+\frac {5 b \,d^{4} x^{3}}{9}+\frac {88 b \,d^{4} x^{2}}{105 c}+\frac {5 b \,d^{4} x}{3 c^{2}}+\frac {d^{4} b \ln \left (c x +1\right )}{210 c^{3}}+\frac {117 b \,d^{4} \ln \left (-c x +1\right )}{70 c^{3}}\) | \(287\) |
Input:
int(x^2*(c*d*x+d)^4*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
d^4*a*(1/7*c^4*x^7+2/3*c^3*x^6+6/5*c^2*x^5+c*x^4+1/3*x^3)+d^4*b/c^3*(1/7*a rctanh(c*x)*c^7*x^7+2/3*arctanh(c*x)*c^6*x^6+6/5*arctanh(c*x)*c^5*x^5+arct anh(c*x)*c^4*x^4+1/3*arctanh(c*x)*c^3*x^3+1/42*c^6*x^6+2/15*c^5*x^5+47/140 *c^4*x^4+5/9*x^3*c^3+88/105*c^2*x^2+5/3*c*x+117/70*ln(c*x-1)+1/210*ln(c*x+ 1))
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.22 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {180 \, a c^{7} d^{4} x^{7} + 30 \, {\left (28 \, a + b\right )} c^{6} d^{4} x^{6} + 168 \, {\left (9 \, a + b\right )} c^{5} d^{4} x^{5} + 9 \, {\left (140 \, a + 47 \, b\right )} c^{4} d^{4} x^{4} + 140 \, {\left (3 \, a + 5 \, b\right )} c^{3} d^{4} x^{3} + 1056 \, b c^{2} d^{4} x^{2} + 2100 \, b c d^{4} x + 6 \, b d^{4} \log \left (c x + 1\right ) + 2106 \, b d^{4} \log \left (c x - 1\right ) + 6 \, {\left (15 \, b c^{7} d^{4} x^{7} + 70 \, b c^{6} d^{4} x^{6} + 126 \, b c^{5} d^{4} x^{5} + 105 \, b c^{4} d^{4} x^{4} + 35 \, b c^{3} d^{4} x^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{1260 \, c^{3}} \] Input:
integrate(x^2*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/1260*(180*a*c^7*d^4*x^7 + 30*(28*a + b)*c^6*d^4*x^6 + 168*(9*a + b)*c^5* d^4*x^5 + 9*(140*a + 47*b)*c^4*d^4*x^4 + 140*(3*a + 5*b)*c^3*d^4*x^3 + 105 6*b*c^2*d^4*x^2 + 2100*b*c*d^4*x + 6*b*d^4*log(c*x + 1) + 2106*b*d^4*log(c *x - 1) + 6*(15*b*c^7*d^4*x^7 + 70*b*c^6*d^4*x^6 + 126*b*c^5*d^4*x^5 + 105 *b*c^4*d^4*x^4 + 35*b*c^3*d^4*x^3)*log(-(c*x + 1)/(c*x - 1)))/c^3
Time = 0.74 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.63 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{4} d^{4} x^{7}}{7} + \frac {2 a c^{3} d^{4} x^{6}}{3} + \frac {6 a c^{2} d^{4} x^{5}}{5} + a c d^{4} x^{4} + \frac {a d^{4} x^{3}}{3} + \frac {b c^{4} d^{4} x^{7} \operatorname {atanh}{\left (c x \right )}}{7} + \frac {2 b c^{3} d^{4} x^{6} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b c^{3} d^{4} x^{6}}{42} + \frac {6 b c^{2} d^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {2 b c^{2} d^{4} x^{5}}{15} + b c d^{4} x^{4} \operatorname {atanh}{\left (c x \right )} + \frac {47 b c d^{4} x^{4}}{140} + \frac {b d^{4} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {5 b d^{4} x^{3}}{9} + \frac {88 b d^{4} x^{2}}{105 c} + \frac {5 b d^{4} x}{3 c^{2}} + \frac {176 b d^{4} \log {\left (x - \frac {1}{c} \right )}}{105 c^{3}} + \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{105 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d^{4} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(c*d*x+d)**4*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*c**4*d**4*x**7/7 + 2*a*c**3*d**4*x**6/3 + 6*a*c**2*d**4*x**5/ 5 + a*c*d**4*x**4 + a*d**4*x**3/3 + b*c**4*d**4*x**7*atanh(c*x)/7 + 2*b*c* *3*d**4*x**6*atanh(c*x)/3 + b*c**3*d**4*x**6/42 + 6*b*c**2*d**4*x**5*atanh (c*x)/5 + 2*b*c**2*d**4*x**5/15 + b*c*d**4*x**4*atanh(c*x) + 47*b*c*d**4*x **4/140 + b*d**4*x**3*atanh(c*x)/3 + 5*b*d**4*x**3/9 + 88*b*d**4*x**2/(105 *c) + 5*b*d**4*x/(3*c**2) + 176*b*d**4*log(x - 1/c)/(105*c**3) + b*d**4*at anh(c*x)/(105*c**3), Ne(c, 0)), (a*d**4*x**3/3, True))
Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (151) = 302\).
Time = 0.04 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.98 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{7} \, a c^{4} d^{4} x^{7} + \frac {2}{3} \, a c^{3} d^{4} x^{6} + \frac {6}{5} \, a c^{2} d^{4} x^{5} + \frac {1}{84} \, {\left (12 \, x^{7} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, c^{4} x^{6} + 3 \, c^{2} x^{4} + 6 \, x^{2}}{c^{6}} + \frac {6 \, \log \left (c^{2} x^{2} - 1\right )}{c^{8}}\right )}\right )} b c^{4} d^{4} + a c d^{4} x^{4} + \frac {1}{45} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{3} d^{4} + \frac {3}{10} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{4} + \frac {1}{3} \, a d^{4} x^{3} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c d^{4} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d^{4} \] Input:
integrate(x^2*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/7*a*c^4*d^4*x^7 + 2/3*a*c^3*d^4*x^6 + 6/5*a*c^2*d^4*x^5 + 1/84*(12*x^7*a rctanh(c*x) + c*((2*c^4*x^6 + 3*c^2*x^4 + 6*x^2)/c^6 + 6*log(c^2*x^2 - 1)/ c^8))*b*c^4*d^4 + a*c*d^4*x^4 + 1/45*(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^3*d^4 + 3/10*(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^2*d^4 + 1/3*a*d^4*x^3 + 1/6*(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*d^4 + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d^4
Leaf count of result is larger than twice the leaf count of optimal. 723 vs. \(2 (151) = 302\).
Time = 0.13 (sec) , antiderivative size = 723, normalized size of antiderivative = 4.23 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx =\text {Too large to display} \] Input:
integrate(x^2*(c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
-4/315*(132*b*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/c^4 - 132*b*d^4*log(-(c*x + 1)/(c*x - 1))/c^4 - 12*(105*(c*x + 1)^6*b*d^4/(c*x - 1)^6 - 210*(c*x + 1 )^5*b*d^4/(c*x - 1)^5 + 385*(c*x + 1)^4*b*d^4/(c*x - 1)^4 - 385*(c*x + 1)^ 3*b*d^4/(c*x - 1)^3 + 231*(c*x + 1)^2*b*d^4/(c*x - 1)^2 - 77*(c*x + 1)*b*d ^4/(c*x - 1) + 11*b*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^7*c^4/(c*x - 1)^7 - 7*(c*x + 1)^6*c^4/(c*x - 1)^6 + 21*(c*x + 1)^5*c^4/(c*x - 1)^5 - 3 5*(c*x + 1)^4*c^4/(c*x - 1)^4 + 35*(c*x + 1)^3*c^4/(c*x - 1)^3 - 21*(c*x + 1)^2*c^4/(c*x - 1)^2 + 7*(c*x + 1)*c^4/(c*x - 1) - c^4) - (2520*(c*x + 1) ^6*a*d^4/(c*x - 1)^6 - 5040*(c*x + 1)^5*a*d^4/(c*x - 1)^5 + 9240*(c*x + 1) ^4*a*d^4/(c*x - 1)^4 - 9240*(c*x + 1)^3*a*d^4/(c*x - 1)^3 + 5544*(c*x + 1) ^2*a*d^4/(c*x - 1)^2 - 1848*(c*x + 1)*a*d^4/(c*x - 1) + 264*a*d^4 + 1128*( c*x + 1)^6*b*d^4/(c*x - 1)^6 - 4812*(c*x + 1)^5*b*d^4/(c*x - 1)^5 + 9476*( c*x + 1)^4*b*d^4/(c*x - 1)^4 - 10631*(c*x + 1)^3*b*d^4/(c*x - 1)^3 + 6933* (c*x + 1)^2*b*d^4/(c*x - 1)^2 - 2465*(c*x + 1)*b*d^4/(c*x - 1) + 371*b*d^4 )/((c*x + 1)^7*c^4/(c*x - 1)^7 - 7*(c*x + 1)^6*c^4/(c*x - 1)^6 + 21*(c*x + 1)^5*c^4/(c*x - 1)^5 - 35*(c*x + 1)^4*c^4/(c*x - 1)^4 + 35*(c*x + 1)^3*c^ 4/(c*x - 1)^3 - 21*(c*x + 1)^2*c^4/(c*x - 1)^2 + 7*(c*x + 1)*c^4/(c*x - 1) - c^4))*c
Time = 3.58 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.15 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {\frac {88\,b\,c^2\,d^4\,x^2}{105}-\frac {d^4\,\left (2100\,b\,\mathrm {atanh}\left (c\,x\right )-1056\,b\,\ln \left (c^2\,x^2-1\right )\right )}{1260}+\frac {5\,b\,c\,d^4\,x}{3}}{c^3}+\frac {d^4\,\left (420\,a\,x^3+700\,b\,x^3+420\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{1260}+\frac {c^4\,d^4\,\left (180\,a\,x^7+180\,b\,x^7\,\mathrm {atanh}\left (c\,x\right )\right )}{1260}+\frac {c\,d^4\,\left (1260\,a\,x^4+423\,b\,x^4+1260\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{1260}+\frac {c^3\,d^4\,\left (840\,a\,x^6+30\,b\,x^6+840\,b\,x^6\,\mathrm {atanh}\left (c\,x\right )\right )}{1260}+\frac {c^2\,d^4\,\left (1512\,a\,x^5+168\,b\,x^5+1512\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{1260} \] Input:
int(x^2*(a + b*atanh(c*x))*(d + c*d*x)^4,x)
Output:
((88*b*c^2*d^4*x^2)/105 - (d^4*(2100*b*atanh(c*x) - 1056*b*log(c^2*x^2 - 1 )))/1260 + (5*b*c*d^4*x)/3)/c^3 + (d^4*(420*a*x^3 + 700*b*x^3 + 420*b*x^3* atanh(c*x)))/1260 + (c^4*d^4*(180*a*x^7 + 180*b*x^7*atanh(c*x)))/1260 + (c *d^4*(1260*a*x^4 + 423*b*x^4 + 1260*b*x^4*atanh(c*x)))/1260 + (c^3*d^4*(84 0*a*x^6 + 30*b*x^6 + 840*b*x^6*atanh(c*x)))/1260 + (c^2*d^4*(1512*a*x^5 + 168*b*x^5 + 1512*b*x^5*atanh(c*x)))/1260
Time = 0.16 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.11 \[ \int x^2 (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{4} \left (180 \mathit {atanh} \left (c x \right ) b \,c^{7} x^{7}+840 \mathit {atanh} \left (c x \right ) b \,c^{6} x^{6}+1512 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}+1260 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+420 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+12 \mathit {atanh} \left (c x \right ) b +2112 \,\mathrm {log}\left (c^{2} x -c \right ) b +180 a \,c^{7} x^{7}+840 a \,c^{6} x^{6}+1512 a \,c^{5} x^{5}+1260 a \,c^{4} x^{4}+420 a \,c^{3} x^{3}+30 b \,c^{6} x^{6}+168 b \,c^{5} x^{5}+423 b \,c^{4} x^{4}+700 b \,c^{3} x^{3}+1056 b \,c^{2} x^{2}+2100 b c x \right )}{1260 c^{3}} \] Input:
int(x^2*(c*d*x+d)^4*(a+b*atanh(c*x)),x)
Output:
(d**4*(180*atanh(c*x)*b*c**7*x**7 + 840*atanh(c*x)*b*c**6*x**6 + 1512*atan h(c*x)*b*c**5*x**5 + 1260*atanh(c*x)*b*c**4*x**4 + 420*atanh(c*x)*b*c**3*x **3 + 12*atanh(c*x)*b + 2112*log(c**2*x - c)*b + 180*a*c**7*x**7 + 840*a*c **6*x**6 + 1512*a*c**5*x**5 + 1260*a*c**4*x**4 + 420*a*c**3*x**3 + 30*b*c* *6*x**6 + 168*b*c**5*x**5 + 423*b*c**4*x**4 + 700*b*c**3*x**3 + 1056*b*c** 2*x**2 + 2100*b*c*x))/(1260*c**3)