Integrand size = 20, antiderivative size = 178 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {11 b d^3 x}{12 c^2}+\frac {7 b d^3 x^2}{15 c}+\frac {11}{36} b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} b c^2 d^3 x^5+\frac {1}{3} d^3 x^3 (a+b \text {arctanh}(c x))+\frac {3}{4} c d^3 x^4 (a+b \text {arctanh}(c x))+\frac {3}{5} c^2 d^3 x^5 (a+b \text {arctanh}(c x))+\frac {1}{6} c^3 d^3 x^6 (a+b \text {arctanh}(c x))+\frac {37 b d^3 \log (1-c x)}{40 c^3}+\frac {b d^3 \log (1+c x)}{120 c^3} \] Output:
11/12*b*d^3*x/c^2+7/15*b*d^3*x^2/c+11/36*b*d^3*x^3+3/20*b*c*d^3*x^4+1/30*b *c^2*d^3*x^5+1/3*d^3*x^3*(a+b*arctanh(c*x))+3/4*c*d^3*x^4*(a+b*arctanh(c*x ))+3/5*c^2*d^3*x^5*(a+b*arctanh(c*x))+1/6*c^3*d^3*x^6*(a+b*arctanh(c*x))+3 7/40*b*d^3*ln(-c*x+1)/c^3+1/120*b*d^3*ln(c*x+1)/c^3
Time = 0.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.80 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3 \left (330 b c x+168 b c^2 x^2+120 a c^3 x^3+110 b c^3 x^3+270 a c^4 x^4+54 b c^4 x^4+216 a c^5 x^5+12 b c^5 x^5+60 a c^6 x^6+6 b c^3 x^3 \left (20+45 c x+36 c^2 x^2+10 c^3 x^3\right ) \text {arctanh}(c x)+333 b \log (1-c x)+3 b \log (1+c x)\right )}{360 c^3} \] Input:
Integrate[x^2*(d + c*d*x)^3*(a + b*ArcTanh[c*x]),x]
Output:
(d^3*(330*b*c*x + 168*b*c^2*x^2 + 120*a*c^3*x^3 + 110*b*c^3*x^3 + 270*a*c^ 4*x^4 + 54*b*c^4*x^4 + 216*a*c^5*x^5 + 12*b*c^5*x^5 + 60*a*c^6*x^6 + 6*b*c ^3*x^3*(20 + 45*c*x + 36*c^2*x^2 + 10*c^3*x^3)*ArcTanh[c*x] + 333*b*Log[1 - c*x] + 3*b*Log[1 + c*x]))/(360*c^3)
Time = 0.55 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.84, 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.
\(\displaystyle \int x^2 (c d x+d)^3 (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle -b c \int \frac {d^3 x^3 \left (10 c^3 x^3+36 c^2 x^2+45 c x+20\right )}{60 \left (1-c^2 x^2\right )}dx+\frac {1}{6} c^3 d^3 x^6 (a+b \text {arctanh}(c x))+\frac {3}{5} c^2 d^3 x^5 (a+b \text {arctanh}(c x))+\frac {3}{4} c d^3 x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d^3 x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{60} b c d^3 \int \frac {x^3 \left (10 c^3 x^3+36 c^2 x^2+45 c x+20\right )}{1-c^2 x^2}dx+\frac {1}{6} c^3 d^3 x^6 (a+b \text {arctanh}(c x))+\frac {3}{5} c^2 d^3 x^5 (a+b \text {arctanh}(c x))+\frac {3}{4} c d^3 x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d^3 x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle -\frac {1}{60} b c d^3 \int \left (-10 c x^4-36 x^3-\frac {55 x^2}{c}-\frac {56 x}{c^2}+\frac {56 c x+55}{c^3 \left (1-c^2 x^2\right )}-\frac {55}{c^3}\right )dx+\frac {1}{6} c^3 d^3 x^6 (a+b \text {arctanh}(c x))+\frac {3}{5} c^2 d^3 x^5 (a+b \text {arctanh}(c x))+\frac {3}{4} c d^3 x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d^3 x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} c^3 d^3 x^6 (a+b \text {arctanh}(c x))+\frac {3}{5} c^2 d^3 x^5 (a+b \text {arctanh}(c x))+\frac {3}{4} c d^3 x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d^3 x^3 (a+b \text {arctanh}(c x))-\frac {1}{60} b c d^3 \left (\frac {55 \text {arctanh}(c x)}{c^4}-\frac {55 x}{c^3}-\frac {28 x^2}{c^2}-\frac {28 \log \left (1-c^2 x^2\right )}{c^4}-2 c x^5-\frac {55 x^3}{3 c}-9 x^4\right )\) |
Input:
Int[x^2*(d + c*d*x)^3*(a + b*ArcTanh[c*x]),x]
Output:
(d^3*x^3*(a + b*ArcTanh[c*x]))/3 + (3*c*d^3*x^4*(a + b*ArcTanh[c*x]))/4 + (3*c^2*d^3*x^5*(a + b*ArcTanh[c*x]))/5 + (c^3*d^3*x^6*(a + b*ArcTanh[c*x]) )/6 - (b*c*d^3*((-55*x)/c^3 - (28*x^2)/c^2 - (55*x^3)/(3*c) - 9*x^4 - 2*c* x^5 + (55*ArcTanh[c*x])/c^4 - (28*Log[1 - c^2*x^2])/c^4))/60
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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 = 0.51 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.81
method | result | size |
parts | \(d^{3} a \left (\frac {1}{6} c^{3} x^{6}+\frac {3}{5} c^{2} x^{5}+\frac {3}{4} c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}+\frac {11 x^{3} c^{3}}{36}+\frac {7 c^{2} x^{2}}{15}+\frac {11 c x}{12}+\frac {37 \ln \left (c x -1\right )}{40}+\frac {\ln \left (c x +1\right )}{120}\right )}{c^{3}}\) | \(144\) |
derivativedivides | \(\frac {d^{3} a \left (\frac {1}{6} c^{6} x^{6}+\frac {3}{5} c^{5} x^{5}+\frac {3}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}+\frac {11 x^{3} c^{3}}{36}+\frac {7 c^{2} x^{2}}{15}+\frac {11 c x}{12}+\frac {37 \ln \left (c x -1\right )}{40}+\frac {\ln \left (c x +1\right )}{120}\right )}{c^{3}}\) | \(150\) |
default | \(\frac {d^{3} a \left (\frac {1}{6} c^{6} x^{6}+\frac {3}{5} c^{5} x^{5}+\frac {3}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d^{3} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\frac {3 \,\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}+\frac {11 x^{3} c^{3}}{36}+\frac {7 c^{2} x^{2}}{15}+\frac {11 c x}{12}+\frac {37 \ln \left (c x -1\right )}{40}+\frac {\ln \left (c x +1\right )}{120}\right )}{c^{3}}\) | \(150\) |
parallelrisch | \(\frac {30 b \,c^{6} d^{3} \operatorname {arctanh}\left (c x \right ) x^{6}+30 a \,c^{6} d^{3} x^{6}+108 b \,c^{5} d^{3} \operatorname {arctanh}\left (c x \right ) x^{5}+108 a \,c^{5} d^{3} x^{5}+6 b \,c^{5} d^{3} x^{5}+135 d^{3} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+135 a \,c^{4} d^{3} x^{4}+27 b \,c^{4} d^{3} x^{4}+60 d^{3} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+60 a \,c^{3} d^{3} x^{3}+55 b \,c^{3} d^{3} x^{3}+84 b \,c^{2} d^{3} x^{2}+165 b c \,d^{3} x +168 \ln \left (c x -1\right ) b \,d^{3}+3 b \,d^{3} \operatorname {arctanh}\left (c x \right )}{180 c^{3}}\) | \(197\) |
risch | \(\frac {d^{3} b \,x^{3} \left (10 x^{3} c^{3}+36 c^{2} x^{2}+45 c x +20\right ) \ln \left (c x +1\right )}{120}-\frac {d^{3} c^{3} b \,x^{6} \ln \left (-c x +1\right )}{12}+\frac {d^{3} c^{3} a \,x^{6}}{6}-\frac {3 d^{3} c^{2} b \,x^{5} \ln \left (-c x +1\right )}{10}+\frac {3 d^{3} c^{2} a \,x^{5}}{5}+\frac {b \,c^{2} d^{3} x^{5}}{30}-\frac {3 d^{3} c b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {3 d^{3} c a \,x^{4}}{4}+\frac {3 b c \,d^{3} x^{4}}{20}-\frac {d^{3} b \,x^{3} \ln \left (-c x +1\right )}{6}+\frac {d^{3} a \,x^{3}}{3}+\frac {11 b \,d^{3} x^{3}}{36}+\frac {7 b \,d^{3} x^{2}}{15 c}+\frac {11 b \,d^{3} x}{12 c^{2}}+\frac {b \,d^{3} \ln \left (c x +1\right )}{120 c^{3}}+\frac {37 b \,d^{3} \ln \left (-c x +1\right )}{40 c^{3}}\) | \(237\) |
Input:
int(x^2*(c*d*x+d)^3*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
d^3*a*(1/6*c^3*x^6+3/5*c^2*x^5+3/4*c*x^4+1/3*x^3)+d^3*b/c^3*(1/6*arctanh(c *x)*c^6*x^6+3/5*arctanh(c*x)*c^5*x^5+3/4*arctanh(c*x)*c^4*x^4+1/3*arctanh( c*x)*c^3*x^3+1/30*c^5*x^5+3/20*c^4*x^4+11/36*x^3*c^3+7/15*c^2*x^2+11/12*c* x+37/40*ln(c*x-1)+1/120*ln(c*x+1))
Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {60 \, a c^{6} d^{3} x^{6} + 12 \, {\left (18 \, a + b\right )} c^{5} d^{3} x^{5} + 54 \, {\left (5 \, a + b\right )} c^{4} d^{3} x^{4} + 10 \, {\left (12 \, a + 11 \, b\right )} c^{3} d^{3} x^{3} + 168 \, b c^{2} d^{3} x^{2} + 330 \, b c d^{3} x + 3 \, b d^{3} \log \left (c x + 1\right ) + 333 \, b d^{3} \log \left (c x - 1\right ) + 3 \, {\left (10 \, b c^{6} d^{3} x^{6} + 36 \, b c^{5} d^{3} x^{5} + 45 \, b c^{4} d^{3} x^{4} + 20 \, b c^{3} d^{3} x^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{360 \, c^{3}} \] Input:
integrate(x^2*(c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/360*(60*a*c^6*d^3*x^6 + 12*(18*a + b)*c^5*d^3*x^5 + 54*(5*a + b)*c^4*d^3 *x^4 + 10*(12*a + 11*b)*c^3*d^3*x^3 + 168*b*c^2*d^3*x^2 + 330*b*c*d^3*x + 3*b*d^3*log(c*x + 1) + 333*b*d^3*log(c*x - 1) + 3*(10*b*c^6*d^3*x^6 + 36*b *c^5*d^3*x^5 + 45*b*c^4*d^3*x^4 + 20*b*c^3*d^3*x^3)*log(-(c*x + 1)/(c*x - 1)))/c^3
Time = 0.49 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.32 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{3} d^{3} x^{6}}{6} + \frac {3 a c^{2} d^{3} x^{5}}{5} + \frac {3 a c d^{3} x^{4}}{4} + \frac {a d^{3} x^{3}}{3} + \frac {b c^{3} d^{3} x^{6} \operatorname {atanh}{\left (c x \right )}}{6} + \frac {3 b c^{2} d^{3} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b c^{2} d^{3} x^{5}}{30} + \frac {3 b c d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {3 b c d^{3} x^{4}}{20} + \frac {b d^{3} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {11 b d^{3} x^{3}}{36} + \frac {7 b d^{3} x^{2}}{15 c} + \frac {11 b d^{3} x}{12 c^{2}} + \frac {14 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{15 c^{3}} + \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{60 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d^{3} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(c*d*x+d)**3*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*c**3*d**3*x**6/6 + 3*a*c**2*d**3*x**5/5 + 3*a*c*d**3*x**4/4 + a*d**3*x**3/3 + b*c**3*d**3*x**6*atanh(c*x)/6 + 3*b*c**2*d**3*x**5*atanh( c*x)/5 + b*c**2*d**3*x**5/30 + 3*b*c*d**3*x**4*atanh(c*x)/4 + 3*b*c*d**3*x **4/20 + b*d**3*x**3*atanh(c*x)/3 + 11*b*d**3*x**3/36 + 7*b*d**3*x**2/(15* c) + 11*b*d**3*x/(12*c**2) + 14*b*d**3*log(x - 1/c)/(15*c**3) + b*d**3*ata nh(c*x)/(60*c**3), Ne(c, 0)), (a*d**3*x**3/3, True))
Time = 0.03 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.49 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{6} \, a c^{3} d^{3} x^{6} + \frac {3}{5} \, a c^{2} d^{3} x^{5} + \frac {3}{4} \, a c d^{3} x^{4} + \frac {1}{180} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} b c^{3} d^{3} + \frac {3}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{3} + \frac {1}{3} \, a d^{3} x^{3} + \frac {1}{8} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c d^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d^{3} \] Input:
integrate(x^2*(c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/6*a*c^3*d^3*x^6 + 3/5*a*c^2*d^3*x^5 + 3/4*a*c*d^3*x^4 + 1/180*(30*x^6*ar ctanh(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^3 + 3/20*(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^2*d^3 + 1/3*a*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*lo g(c*x - 1)/c^5))*b*c*d^3 + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2* x^2 - 1)/c^4))*b*d^3
Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (156) = 312\).
Time = 0.14 (sec) , antiderivative size = 621, normalized size of antiderivative = 3.49 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx =\text {Too large to display} \] Input:
integrate(x^2*(c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
-1/45*c*(42*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^4 - 6*(60*(c*x + 1)^5*b* d^3/(c*x - 1)^5 - 90*(c*x + 1)^4*b*d^3/(c*x - 1)^4 + 140*(c*x + 1)^3*b*d^3 /(c*x - 1)^3 - 105*(c*x + 1)^2*b*d^3/(c*x - 1)^2 + 42*(c*x + 1)*b*d^3/(c*x - 1) - 7*b*d^3)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^6*c^4/(c*x - 1)^6 - 6*(c*x + 1)^5*c^4/(c*x - 1)^5 + 15*(c*x + 1)^4*c^4/(c*x - 1)^4 - 20*(c*x + 1)^3*c^4/(c*x - 1)^3 + 15*(c*x + 1)^2*c^4/(c*x - 1)^2 - 6*(c*x + 1)*c^4/( c*x - 1) + c^4) - 42*b*d^3*log(-(c*x + 1)/(c*x - 1))/c^4 - (720*(c*x + 1)^ 5*a*d^3/(c*x - 1)^5 - 1080*(c*x + 1)^4*a*d^3/(c*x - 1)^4 + 1680*(c*x + 1)^ 3*a*d^3/(c*x - 1)^3 - 1260*(c*x + 1)^2*a*d^3/(c*x - 1)^2 + 504*(c*x + 1)*a *d^3/(c*x - 1) - 84*a*d^3 + 318*(c*x + 1)^5*b*d^3/(c*x - 1)^5 - 1119*(c*x + 1)^4*b*d^3/(c*x - 1)^4 + 1742*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 1464*(c*x + 1)^2*b*d^3/(c*x - 1)^2 + 636*(c*x + 1)*b*d^3/(c*x - 1) - 113*b*d^3)/((c* x + 1)^6*c^4/(c*x - 1)^6 - 6*(c*x + 1)^5*c^4/(c*x - 1)^5 + 15*(c*x + 1)^4* c^4/(c*x - 1)^4 - 20*(c*x + 1)^3*c^4/(c*x - 1)^3 + 15*(c*x + 1)^2*c^4/(c*x - 1)^2 - 6*(c*x + 1)*c^4/(c*x - 1) + c^4))
Time = 4.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.93 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {\frac {7\,b\,c^2\,d^3\,x^2}{15}-\frac {d^3\,\left (165\,b\,\mathrm {atanh}\left (c\,x\right )-84\,b\,\ln \left (c^2\,x^2-1\right )\right )}{180}+\frac {11\,b\,c\,d^3\,x}{12}}{c^3}+\frac {d^3\,\left (60\,a\,x^3+55\,b\,x^3+60\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c^3\,d^3\,\left (30\,a\,x^6+30\,b\,x^6\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c\,d^3\,\left (135\,a\,x^4+27\,b\,x^4+135\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{180}+\frac {c^2\,d^3\,\left (108\,a\,x^5+6\,b\,x^5+108\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{180} \] Input:
int(x^2*(a + b*atanh(c*x))*(d + c*d*x)^3,x)
Output:
((7*b*c^2*d^3*x^2)/15 - (d^3*(165*b*atanh(c*x) - 84*b*log(c^2*x^2 - 1)))/1 80 + (11*b*c*d^3*x)/12)/c^3 + (d^3*(60*a*x^3 + 55*b*x^3 + 60*b*x^3*atanh(c *x)))/180 + (c^3*d^3*(30*a*x^6 + 30*b*x^6*atanh(c*x)))/180 + (c*d^3*(135*a *x^4 + 27*b*x^4 + 135*b*x^4*atanh(c*x)))/180 + (c^2*d^3*(108*a*x^5 + 6*b*x ^5 + 108*b*x^5*atanh(c*x)))/180
Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.89 \[ \int x^2 (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{3} \left (30 \mathit {atanh} \left (c x \right ) b \,c^{6} x^{6}+108 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}+135 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+60 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+3 \mathit {atanh} \left (c x \right ) b +168 \,\mathrm {log}\left (c^{2} x -c \right ) b +30 a \,c^{6} x^{6}+108 a \,c^{5} x^{5}+135 a \,c^{4} x^{4}+60 a \,c^{3} x^{3}+6 b \,c^{5} x^{5}+27 b \,c^{4} x^{4}+55 b \,c^{3} x^{3}+84 b \,c^{2} x^{2}+165 b c x \right )}{180 c^{3}} \] Input:
int(x^2*(c*d*x+d)^3*(a+b*atanh(c*x)),x)
Output:
(d**3*(30*atanh(c*x)*b*c**6*x**6 + 108*atanh(c*x)*b*c**5*x**5 + 135*atanh( c*x)*b*c**4*x**4 + 60*atanh(c*x)*b*c**3*x**3 + 3*atanh(c*x)*b + 168*log(c* *2*x - c)*b + 30*a*c**6*x**6 + 108*a*c**5*x**5 + 135*a*c**4*x**4 + 60*a*c* *3*x**3 + 6*b*c**5*x**5 + 27*b*c**4*x**4 + 55*b*c**3*x**3 + 84*b*c**2*x**2 + 165*b*c*x))/(180*c**3)