Integrand size = 17, antiderivative size = 107 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {8}{5} b d^4 x+\frac {2 b d^4 (1+c x)^2}{5 c}+\frac {2 b d^4 (1+c x)^3}{15 c}+\frac {b d^4 (1+c x)^4}{20 c}+\frac {d^4 (1+c x)^5 (a+b \text {arctanh}(c x))}{5 c}+\frac {16 b d^4 \log (1-c x)}{5 c} \] Output:
8/5*b*d^4*x+2/5*b*d^4*(c*x+1)^2/c+2/15*b*d^4*(c*x+1)^3/c+1/20*b*d^4*(c*x+1 )^4/c+1/5*d^4*(c*x+1)^5*(a+b*arctanh(c*x))/c+16/5*b*d^4*ln(-c*x+1)/c
Time = 0.06 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.36 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^4 \left (60 a c x+180 b c x+120 a c^2 x^2+66 b c^2 x^2+120 a c^3 x^3+20 b c^3 x^3+60 a c^4 x^4+3 b c^4 x^4+12 a c^5 x^5+12 b c x \left (5+10 c x+10 c^2 x^2+5 c^3 x^3+c^4 x^4\right ) \text {arctanh}(c x)+180 b \log (1-c x)+6 b \log \left (1-c^2 x^2\right )\right )}{60 c} \] Input:
Integrate[(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]
Output:
(d^4*(60*a*c*x + 180*b*c*x + 120*a*c^2*x^2 + 66*b*c^2*x^2 + 120*a*c^3*x^3 + 20*b*c^3*x^3 + 60*a*c^4*x^4 + 3*b*c^4*x^4 + 12*a*c^5*x^5 + 12*b*c*x*(5 + 10*c*x + 10*c^2*x^2 + 5*c^3*x^3 + c^4*x^4)*ArcTanh[c*x] + 180*b*Log[1 - c *x] + 6*b*Log[1 - c^2*x^2]))/(60*c)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6478, 27, 456, 49, 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.
| \(\displaystyle \int (c d x+d)^4 (a+b \text {arctanh}(c x)) \, dx\) |
| \(\Big \downarrow \) 6478 |
| \(\displaystyle \frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c}-\frac {b \int \frac {d^5 (c x+1)^5}{1-c^2 x^2}dx}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c}-\frac {1}{5} b d^4 \int \frac {(c x+1)^5}{1-c^2 x^2}dx\) |
| \(\Big \downarrow \) 456 |
| \(\displaystyle \frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c}-\frac {1}{5} b d^4 \int \frac {(c x+1)^4}{1-c x}dx\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c}-\frac {1}{5} b d^4 \int \left (-(c x+1)^3-2 (c x+1)^2-4 (c x+1)+\frac {16}{1-c x}-8\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 c}-\frac {1}{5} b d^4 \left (-\frac {(c x+1)^4}{4 c}-\frac {2 (c x+1)^3}{3 c}-\frac {2 (c x+1)^2}{c}-\frac {16 \log (1-c x)}{c}-8 x\right )\) |
Input:
Int[(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) - (b*d^4*(-8*x - (2*(1 + c*x) ^2)/c - (2*(1 + c*x)^3)/(3*c) - (1 + c*x)^4/(4*c) - (16*Log[1 - c*x])/c))/ 5
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol ] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])/(e*(q + 1))), x] - Simp[b *(c/(e*(q + 1))) Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]
Time = 0.39 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {\frac {d^{4} a \left (c x +1\right )^{5}}{5}+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {x^{3} c^{3}}{3}+\frac {11 c^{2} x^{2}}{10}+3 c x +\frac {16 \ln \left (c x -1\right )}{5}\right )}{c}\) | \(121\) |
| default | \(\frac {\frac {d^{4} a \left (c x +1\right )^{5}}{5}+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {x^{3} c^{3}}{3}+\frac {11 c^{2} x^{2}}{10}+3 c x +\frac {16 \ln \left (c x -1\right )}{5}\right )}{c}\) | \(121\) |
| parts | \(\frac {d^{4} a \left (c x +1\right )^{5}}{5 c}+\frac {d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}+2 \,\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {x^{3} c^{3}}{3}+\frac {11 c^{2} x^{2}}{10}+3 c x +\frac {16 \ln \left (c x -1\right )}{5}\right )}{c}\) | \(123\) |
| parallelrisch | \(\frac {12 b \,c^{5} d^{4} \operatorname {arctanh}\left (c x \right ) x^{5}+12 a \,c^{5} d^{4} x^{5}+60 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+60 a \,c^{4} d^{4} x^{4}+3 b \,c^{4} d^{4} x^{4}+120 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+120 a \,c^{3} d^{4} x^{3}+20 b \,c^{3} d^{4} x^{3}+120 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{4}+120 a \,c^{2} d^{4} x^{2}+66 b \,c^{2} d^{4} x^{2}+60 b c \,d^{4} x \,\operatorname {arctanh}\left (c x \right )+60 a c \,d^{4} x +180 b c \,d^{4} x +192 \ln \left (c x -1\right ) b \,d^{4}+12 b \,d^{4} \operatorname {arctanh}\left (c x \right )}{60 c}\) | \(205\) |
| risch | \(\frac {d^{4} \left (c x +1\right )^{5} b \ln \left (c x +1\right )}{10 c}-\frac {d^{4} c^{4} b \,x^{5} \ln \left (-c x +1\right )}{10}+\frac {d^{4} c^{4} a \,x^{5}}{5}-\frac {d^{4} c^{3} b \,x^{4} \ln \left (-c x +1\right )}{2}+d^{4} c^{3} a \,x^{4}+\frac {d^{4} c^{3} b \,x^{4}}{20}-d^{4} c^{2} b \,x^{3} \ln \left (-c x +1\right )+2 d^{4} c^{2} a \,x^{3}+\frac {d^{4} c^{2} b \,x^{3}}{3}-d^{4} c b \,x^{2} \ln \left (-c x +1\right )+2 d^{4} c a \,x^{2}+\frac {11 d^{4} c b \,x^{2}}{10}-\frac {b \,d^{4} x \ln \left (-c x +1\right )}{2}+a \,d^{4} x +3 b \,d^{4} x -\frac {\ln \left (-c x +1\right ) b \,d^{4}}{10 c}+\frac {16 d^{4} b \ln \left (c x -1\right )}{5 c}\) | \(235\) |
Input:
int((c*d*x+d)^4*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
1/c*(1/5*d^4*a*(c*x+1)^5+d^4*b*(1/5*arctanh(c*x)*c^5*x^5+arctanh(c*x)*c^4* x^4+2*arctanh(c*x)*c^3*x^3+2*arctanh(c*x)*c^2*x^2+arctanh(c*x)*c*x+1/5*arc tanh(c*x)+1/20*c^4*x^4+1/3*x^3*c^3+11/10*c^2*x^2+3*c*x+16/5*ln(c*x-1)))
Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.65 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {12 \, a c^{5} d^{4} x^{5} + 3 \, {\left (20 \, a + b\right )} c^{4} d^{4} x^{4} + 20 \, {\left (6 \, a + b\right )} c^{3} d^{4} x^{3} + 6 \, {\left (20 \, a + 11 \, b\right )} c^{2} d^{4} x^{2} + 60 \, {\left (a + 3 \, b\right )} c d^{4} x + 6 \, b d^{4} \log \left (c x + 1\right ) + 186 \, b d^{4} \log \left (c x - 1\right ) + 6 \, {\left (b c^{5} d^{4} x^{5} + 5 \, b c^{4} d^{4} x^{4} + 10 \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} + 5 \, b c d^{4} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, c} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/60*(12*a*c^5*d^4*x^5 + 3*(20*a + b)*c^4*d^4*x^4 + 20*(6*a + b)*c^3*d^4*x ^3 + 6*(20*a + 11*b)*c^2*d^4*x^2 + 60*(a + 3*b)*c*d^4*x + 6*b*d^4*log(c*x + 1) + 186*b*d^4*log(c*x - 1) + 6*(b*c^5*d^4*x^5 + 5*b*c^4*d^4*x^4 + 10*b* c^3*d^4*x^3 + 10*b*c^2*d^4*x^2 + 5*b*c*d^4*x)*log(-(c*x + 1)/(c*x - 1)))/c
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (97) = 194\).
Time = 0.40 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.11 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{4} d^{4} x^{5}}{5} + a c^{3} d^{4} x^{4} + 2 a c^{2} d^{4} x^{3} + 2 a c d^{4} x^{2} + a d^{4} x + \frac {b c^{4} d^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + b c^{3} d^{4} x^{4} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{3} d^{4} x^{4}}{20} + 2 b c^{2} d^{4} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{2} d^{4} x^{3}}{3} + 2 b c d^{4} x^{2} \operatorname {atanh}{\left (c x \right )} + \frac {11 b c d^{4} x^{2}}{10} + b d^{4} x \operatorname {atanh}{\left (c x \right )} + 3 b d^{4} x + \frac {16 b d^{4} \log {\left (x - \frac {1}{c} \right )}}{5 c} + \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{5 c} & \text {for}\: c \neq 0 \\a d^{4} x & \text {otherwise} \end {cases} \] Input:
integrate((c*d*x+d)**4*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*c**4*d**4*x**5/5 + a*c**3*d**4*x**4 + 2*a*c**2*d**4*x**3 + 2* a*c*d**4*x**2 + a*d**4*x + b*c**4*d**4*x**5*atanh(c*x)/5 + b*c**3*d**4*x** 4*atanh(c*x) + b*c**3*d**4*x**4/20 + 2*b*c**2*d**4*x**3*atanh(c*x) + b*c** 2*d**4*x**3/3 + 2*b*c*d**4*x**2*atanh(c*x) + 11*b*c*d**4*x**2/10 + b*d**4* x*atanh(c*x) + 3*b*d**4*x + 16*b*d**4*log(x - 1/c)/(5*c) + b*d**4*atanh(c* x)/(5*c), Ne(c, 0)), (a*d**4*x, True))
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (95) = 190\).
Time = 0.04 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.64 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, a c^{4} d^{4} x^{5} + a c^{3} d^{4} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{4} d^{4} + 2 \, a c^{2} d^{4} x^{3} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{3} d^{4} + {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{4} + 2 \, a c d^{4} x^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{4} + a d^{4} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/5*a*c^4*d^4*x^5 + a*c^3*d^4*x^4 + 1/20*(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^4*d^4 + 2*a*c^2*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*l og(c*x - 1)/c^5))*b*c^3*d^4 + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x ^2 - 1)/c^4))*b*c^2*d^4 + 2*a*c*d^4*x^2 + (2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*c*d^4 + a*d^4*x + 1/2*(2*c*x*ar ctanh(c*x) + log(-c^2*x^2 + 1))*b*d^4/c
Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (95) = 190\).
Time = 0.12 (sec) , antiderivative size = 526, normalized size of antiderivative = 4.92 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=-\frac {4}{15} \, {\left (\frac {12 \, b d^{4} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {12 \, b d^{4} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {12 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} b d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {10 \, {\left (c x + 1\right )}^{3} b d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2} b d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {5 \, {\left (c x + 1\right )} b d^{4}}{c x - 1} + b d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{2}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}} - \frac {\frac {120 \, {\left (c x + 1\right )}^{4} a d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {240 \, {\left (c x + 1\right )}^{3} a d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {240 \, {\left (c x + 1\right )}^{2} a d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {120 \, {\left (c x + 1\right )} a d^{4}}{c x - 1} + 24 \, a d^{4} + \frac {48 \, {\left (c x + 1\right )}^{4} b d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {156 \, {\left (c x + 1\right )}^{3} b d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {196 \, {\left (c x + 1\right )}^{2} b d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {113 \, {\left (c x + 1\right )} b d^{4}}{c x - 1} + 25 \, b d^{4}}{\frac {{\left (c x + 1\right )}^{5} c^{2}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}}\right )} c \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
-4/15*(12*b*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 12*b*d^4*log(-(c*x + 1 )/(c*x - 1))/c^2 - 12*(5*(c*x + 1)^4*b*d^4/(c*x - 1)^4 - 10*(c*x + 1)^3*b* d^4/(c*x - 1)^3 + 10*(c*x + 1)^2*b*d^4/(c*x - 1)^2 - 5*(c*x + 1)*b*d^4/(c* x - 1) + b*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^2/(c*x - 1)^5 - 5 *(c*x + 1)^4*c^2/(c*x - 1)^4 + 10*(c*x + 1)^3*c^2/(c*x - 1)^3 - 10*(c*x + 1)^2*c^2/(c*x - 1)^2 + 5*(c*x + 1)*c^2/(c*x - 1) - c^2) - (120*(c*x + 1)^4 *a*d^4/(c*x - 1)^4 - 240*(c*x + 1)^3*a*d^4/(c*x - 1)^3 + 240*(c*x + 1)^2*a *d^4/(c*x - 1)^2 - 120*(c*x + 1)*a*d^4/(c*x - 1) + 24*a*d^4 + 48*(c*x + 1) ^4*b*d^4/(c*x - 1)^4 - 156*(c*x + 1)^3*b*d^4/(c*x - 1)^3 + 196*(c*x + 1)^2 *b*d^4/(c*x - 1)^2 - 113*(c*x + 1)*b*d^4/(c*x - 1) + 25*b*d^4)/((c*x + 1)^ 5*c^2/(c*x - 1)^5 - 5*(c*x + 1)^4*c^2/(c*x - 1)^4 + 10*(c*x + 1)^3*c^2/(c* x - 1)^3 - 10*(c*x + 1)^2*c^2/(c*x - 1)^2 + 5*(c*x + 1)*c^2/(c*x - 1) - c^ 2))*c
Time = 3.49 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.57 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^4\,\left (60\,a\,x+180\,b\,x+60\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^4\,d^4\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{60}-\frac {d^4\,\left (180\,b\,\mathrm {atanh}\left (c\,x\right )-96\,b\,\ln \left (c^2\,x^2-1\right )\right )}{60\,c}+\frac {c\,d^4\,\left (120\,a\,x^2+66\,b\,x^2+120\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^3\,d^4\,\left (60\,a\,x^4+3\,b\,x^4+60\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^2\,d^4\,\left (120\,a\,x^3+20\,b\,x^3+120\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{60} \] Input:
int((a + b*atanh(c*x))*(d + c*d*x)^4,x)
Output:
(d^4*(60*a*x + 180*b*x + 60*b*x*atanh(c*x)))/60 + (c^4*d^4*(12*a*x^5 + 12* b*x^5*atanh(c*x)))/60 - (d^4*(180*b*atanh(c*x) - 96*b*log(c^2*x^2 - 1)))/( 60*c) + (c*d^4*(120*a*x^2 + 66*b*x^2 + 120*b*x^2*atanh(c*x)))/60 + (c^3*d^ 4*(60*a*x^4 + 3*b*x^4 + 60*b*x^4*atanh(c*x)))/60 + (c^2*d^4*(120*a*x^3 + 2 0*b*x^3 + 120*b*x^3*atanh(c*x)))/60
Time = 0.16 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.52 \[ \int (d+c d x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {d^{4} \left (12 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}+60 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+120 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+120 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}+60 \mathit {atanh} \left (c x \right ) b c x +12 \mathit {atanh} \left (c x \right ) b +192 \,\mathrm {log}\left (c^{2} x -c \right ) b +12 a \,c^{5} x^{5}+60 a \,c^{4} x^{4}+120 a \,c^{3} x^{3}+120 a \,c^{2} x^{2}+60 a c x +3 b \,c^{4} x^{4}+20 b \,c^{3} x^{3}+66 b \,c^{2} x^{2}+180 b c x \right )}{60 c} \] Input:
int((c*d*x+d)^4*(a+b*atanh(c*x)),x)
Output:
(d**4*(12*atanh(c*x)*b*c**5*x**5 + 60*atanh(c*x)*b*c**4*x**4 + 120*atanh(c *x)*b*c**3*x**3 + 120*atanh(c*x)*b*c**2*x**2 + 60*atanh(c*x)*b*c*x + 12*at anh(c*x)*b + 192*log(c**2*x - c)*b + 12*a*c**5*x**5 + 60*a*c**4*x**4 + 120 *a*c**3*x**3 + 120*a*c**2*x**2 + 60*a*c*x + 3*b*c**4*x**4 + 20*b*c**3*x**3 + 66*b*c**2*x**2 + 180*b*c*x))/(60*c)