Integrand size = 20, antiderivative size = 109 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=-\frac {b c d^4}{20 x^4}-\frac {b c^2 d^4}{3 x^3}-\frac {11 b c^3 d^4}{10 x^2}-\frac {3 b c^4 d^4}{x}-\frac {d^4 (1+c x)^5 (a+b \text {arctanh}(c x))}{5 x^5}+\frac {16}{5} b c^5 d^4 \log (x)-\frac {16}{5} b c^5 d^4 \log (1-c x) \] Output:
-1/20*b*c*d^4/x^4-1/3*b*c^2*d^4/x^3-11/10*b*c^3*d^4/x^2-3*b*c^4*d^4/x-1/5* d^4*(c*x+1)^5*(a+b*arctanh(c*x))/x^5+16/5*b*c^5*d^4*ln(x)-16/5*b*c^5*d^4*l n(-c*x+1)
Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.44 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=-\frac {d^4 \left (12 a+60 a c x+3 b c x+120 a c^2 x^2+20 b c^2 x^2+120 a c^3 x^3+66 b c^3 x^3+60 a c^4 x^4+180 b c^4 x^4+12 b \left (1+5 c x+10 c^2 x^2+10 c^3 x^3+5 c^4 x^4\right ) \text {arctanh}(c x)-192 b c^5 x^5 \log (x)+186 b c^5 x^5 \log (1-c x)+6 b c^5 x^5 \log (1+c x)\right )}{60 x^5} \] Input:
Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^6,x]
Output:
-1/60*(d^4*(12*a + 60*a*c*x + 3*b*c*x + 120*a*c^2*x^2 + 20*b*c^2*x^2 + 120 *a*c^3*x^3 + 66*b*c^3*x^3 + 60*a*c^4*x^4 + 180*b*c^4*x^4 + 12*b*(1 + 5*c*x + 10*c^2*x^2 + 10*c^3*x^3 + 5*c^4*x^4)*ArcTanh[c*x] - 192*b*c^5*x^5*Log[x ] + 186*b*c^5*x^5*Log[1 - c*x] + 6*b*c^5*x^5*Log[1 + c*x]))/x^5
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6498, 27, 99, 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.
| \(\displaystyle \int \frac {(c d x+d)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int -\frac {d^4 (c x+1)^4}{5 x^5 (1-c x)}dx-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} b c d^4 \int \frac {(c x+1)^4}{x^5 (1-c x)}dx-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 x^5}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{5} b c d^4 \int \left (-\frac {16 c^5}{c x-1}+\frac {16 c^4}{x}+\frac {15 c^3}{x^2}+\frac {11 c^2}{x^3}+\frac {5 c}{x^4}+\frac {1}{x^5}\right )dx-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 x^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} b c d^4 \left (16 c^4 \log (x)-16 c^4 \log (1-c x)-\frac {15 c^3}{x}-\frac {11 c^2}{2 x^2}-\frac {5 c}{3 x^3}-\frac {1}{4 x^4}\right )-\frac {d^4 (c x+1)^5 (a+b \text {arctanh}(c x))}{5 x^5}\) |
Input:
Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^6,x]
Output:
-1/5*(d^4*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/x^5 + (b*c*d^4*(-1/4*1/x^4 - ( 5*c)/(3*x^3) - (11*c^2)/(2*x^2) - (15*c^3)/x + 16*c^4*Log[x] - 16*c^4*Log[ 1 - c*x]))/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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.57 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.52
| method | result | size |
| parts | \(d^{4} a \left (-\frac {c}{x^{4}}-\frac {c^{4}}{x}-\frac {1}{5 x^{5}}-\frac {2 c^{3}}{x^{2}}-\frac {2 c^{2}}{x^{3}}\right )+d^{4} b \,c^{5} \left (-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {31 \ln \left (c x -1\right )}{10}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{3 c^{3} x^{3}}-\frac {11}{10 c^{2} x^{2}}-\frac {3}{c x}+\frac {16 \ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}\right )\) | \(166\) |
| derivativedivides | \(c^{5} \left (d^{4} a \left (-\frac {2}{c^{3} x^{3}}-\frac {2}{c^{2} x^{2}}-\frac {1}{c x}-\frac {1}{c^{4} x^{4}}-\frac {1}{5 c^{5} x^{5}}\right )+d^{4} b \left (-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {31 \ln \left (c x -1\right )}{10}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{3 c^{3} x^{3}}-\frac {11}{10 c^{2} x^{2}}-\frac {3}{c x}+\frac {16 \ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}\right )\right )\) | \(172\) |
| default | \(c^{5} \left (d^{4} a \left (-\frac {2}{c^{3} x^{3}}-\frac {2}{c^{2} x^{2}}-\frac {1}{c x}-\frac {1}{c^{4} x^{4}}-\frac {1}{5 c^{5} x^{5}}\right )+d^{4} b \left (-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {31 \ln \left (c x -1\right )}{10}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{3 c^{3} x^{3}}-\frac {11}{10 c^{2} x^{2}}-\frac {3}{c x}+\frac {16 \ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}\right )\right )\) | \(172\) |
| risch | \(-\frac {d^{4} b \left (5 c^{4} x^{4}+10 x^{3} c^{3}+10 c^{2} x^{2}+5 c x +1\right ) \ln \left (c x +1\right )}{10 x^{5}}-\frac {d^{4} \left (6 b \,c^{5} \ln \left (c x +1\right ) x^{5}+186 b \,c^{5} x^{5} \ln \left (-c x +1\right )-192 b \,c^{5} \ln \left (-x \right ) x^{5}-30 b \,x^{4} \ln \left (-c x +1\right ) c^{4}+60 a \,c^{4} x^{4}+180 b \,c^{4} x^{4}-60 b \,x^{3} \ln \left (-c x +1\right ) c^{3}+120 a \,c^{3} x^{3}+66 b \,c^{3} x^{3}-60 b \,c^{2} x^{2} \ln \left (-c x +1\right )+120 a \,c^{2} x^{2}+20 b \,c^{2} x^{2}-30 b c x \ln \left (-c x +1\right )+60 a c x +3 b c x -6 b \ln \left (-c x +1\right )+12 a \right )}{60 x^{5}}\) | \(237\) |
| parallelrisch | \(\frac {192 b \,c^{5} d^{4} \ln \left (x \right ) x^{5}-192 \ln \left (c x -1\right ) x^{5} b \,c^{5} d^{4}-12 b \,c^{5} d^{4} \operatorname {arctanh}\left (c x \right ) x^{5}-120 a \,c^{5} d^{4} x^{5}-66 b \,c^{5} d^{4} x^{5}-60 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}-60 a \,c^{4} d^{4} x^{4}-180 b \,c^{4} d^{4} x^{4}-120 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-120 a \,c^{3} d^{4} x^{3}-66 b \,c^{3} d^{4} x^{3}-120 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{4}-120 a \,c^{2} d^{4} x^{2}-20 b \,c^{2} d^{4} x^{2}-60 b c \,d^{4} x \,\operatorname {arctanh}\left (c x \right )-60 a c \,d^{4} x -3 b c \,d^{4} x -12 b \,d^{4} \operatorname {arctanh}\left (c x \right )-12 d^{4} a}{60 x^{5}}\) | \(243\) |
Input:
int((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x,method=_RETURNVERBOSE)
Output:
d^4*a*(-c/x^4-c^4/x-1/5/x^5-2*c^3/x^2-2*c^2/x^3)+d^4*b*c^5*(-2*arctanh(c*x )/c^3/x^3-2*arctanh(c*x)/c^2/x^2-arctanh(c*x)/c/x-arctanh(c*x)/c^4/x^4-1/5 *arctanh(c*x)/c^5/x^5-31/10*ln(c*x-1)-1/20/c^4/x^4-1/3/c^3/x^3-11/10/c^2/x ^2-3/c/x+16/5*ln(c*x)-1/10*ln(c*x+1))
Time = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.75 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=-\frac {6 \, b c^{5} d^{4} x^{5} \log \left (c x + 1\right ) + 186 \, b c^{5} d^{4} x^{5} \log \left (c x - 1\right ) - 192 \, b c^{5} d^{4} x^{5} \log \left (x\right ) + 60 \, {\left (a + 3 \, b\right )} c^{4} d^{4} x^{4} + 6 \, {\left (20 \, a + 11 \, b\right )} c^{3} d^{4} x^{3} + 20 \, {\left (6 \, a + b\right )} c^{2} d^{4} x^{2} + 3 \, {\left (20 \, a + b\right )} c d^{4} x + 12 \, a d^{4} + 6 \, {\left (5 \, b c^{4} d^{4} x^{4} + 10 \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} + 5 \, b c d^{4} x + b d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, x^{5}} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="fricas")
Output:
-1/60*(6*b*c^5*d^4*x^5*log(c*x + 1) + 186*b*c^5*d^4*x^5*log(c*x - 1) - 192 *b*c^5*d^4*x^5*log(x) + 60*(a + 3*b)*c^4*d^4*x^4 + 6*(20*a + 11*b)*c^3*d^4 *x^3 + 20*(6*a + b)*c^2*d^4*x^2 + 3*(20*a + b)*c*d^4*x + 12*a*d^4 + 6*(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 + b*d^4)* log(-(c*x + 1)/(c*x - 1)))/x^5
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (109) = 218\).
Time = 0.65 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.32 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=\begin {cases} - \frac {a c^{4} d^{4}}{x} - \frac {2 a c^{3} d^{4}}{x^{2}} - \frac {2 a c^{2} d^{4}}{x^{3}} - \frac {a c d^{4}}{x^{4}} - \frac {a d^{4}}{5 x^{5}} + \frac {16 b c^{5} d^{4} \log {\left (x \right )}}{5} - \frac {16 b c^{5} d^{4} \log {\left (x - \frac {1}{c} \right )}}{5} - \frac {b c^{5} d^{4} \operatorname {atanh}{\left (c x \right )}}{5} - \frac {b c^{4} d^{4} \operatorname {atanh}{\left (c x \right )}}{x} - \frac {3 b c^{4} d^{4}}{x} - \frac {2 b c^{3} d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{2}} - \frac {11 b c^{3} d^{4}}{10 x^{2}} - \frac {2 b c^{2} d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{3}} - \frac {b c^{2} d^{4}}{3 x^{3}} - \frac {b c d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{4}} - \frac {b c d^{4}}{20 x^{4}} - \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a d^{4}}{5 x^{5}} & \text {otherwise} \end {cases} \] Input:
integrate((c*d*x+d)**4*(a+b*atanh(c*x))/x**6,x)
Output:
Piecewise((-a*c**4*d**4/x - 2*a*c**3*d**4/x**2 - 2*a*c**2*d**4/x**3 - a*c* d**4/x**4 - a*d**4/(5*x**5) + 16*b*c**5*d**4*log(x)/5 - 16*b*c**5*d**4*log (x - 1/c)/5 - b*c**5*d**4*atanh(c*x)/5 - b*c**4*d**4*atanh(c*x)/x - 3*b*c* *4*d**4/x - 2*b*c**3*d**4*atanh(c*x)/x**2 - 11*b*c**3*d**4/(10*x**2) - 2*b *c**2*d**4*atanh(c*x)/x**3 - b*c**2*d**4/(3*x**3) - b*c*d**4*atanh(c*x)/x* *4 - b*c*d**4/(20*x**4) - b*d**4*atanh(c*x)/(5*x**5), Ne(c, 0)), (-a*d**4/ (5*x**5), True))
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (97) = 194\).
Time = 0.03 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.74 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=-\frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c^{4} d^{4} + {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{3} d^{4} - {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c^{2} d^{4} - \frac {a c^{4} d^{4}}{x} + \frac {1}{6} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b c d^{4} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b d^{4} - \frac {2 \, a c^{3} d^{4}}{x^{2}} - \frac {2 \, a c^{2} d^{4}}{x^{3}} - \frac {a c d^{4}}{x^{4}} - \frac {a d^{4}}{5 \, x^{5}} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="maxima")
Output:
-1/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c^4*d^4 + ((c* log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*c^3*d^4 - ( (c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*b*c^ 2*d^4 - a*c^4*d^4/x + 1/6*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3 *c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b*c*d^4 - 1/20*((2*c^4*log(c^2* x^2 - 1) - 2*c^4*log(x^2) + (2*c^2*x^2 + 1)/x^4)*c + 4*arctanh(c*x)/x^5)*b *d^4 - 2*a*c^3*d^4/x^2 - 2*a*c^2*d^4/x^3 - a*c*d^4/x^4 - 1/5*a*d^4/x^5
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (97) = 194\).
Time = 0.13 (sec) , antiderivative size = 532, normalized size of antiderivative = 4.88 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=\frac {4}{15} \, {\left (12 \, b c^{4} d^{4} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 12 \, b c^{4} d^{4} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {12 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} b c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} b c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2} b c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b c^{4} d^{4}}{c x - 1} + b c^{4} d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {120 \, {\left (c x + 1\right )}^{4} a c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {240 \, {\left (c x + 1\right )}^{3} a c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {240 \, {\left (c x + 1\right )}^{2} a c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {120 \, {\left (c x + 1\right )} a c^{4} d^{4}}{c x - 1} + 24 \, a c^{4} d^{4} + \frac {48 \, {\left (c x + 1\right )}^{4} b c^{4} d^{4}}{{\left (c x - 1\right )}^{4}} + \frac {156 \, {\left (c x + 1\right )}^{3} b c^{4} d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {196 \, {\left (c x + 1\right )}^{2} b c^{4} d^{4}}{{\left (c x - 1\right )}^{2}} + \frac {113 \, {\left (c x + 1\right )} b c^{4} d^{4}}{c x - 1} + 25 \, b c^{4} d^{4}}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^6,x, algorithm="giac")
Output:
4/15*(12*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) - 1) - 12*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1)) + 12*(5*(c*x + 1)^4*b*c^4*d^4/(c*x - 1)^4 + 10*(c*x + 1)^3 *b*c^4*d^4/(c*x - 1)^3 + 10*(c*x + 1)^2*b*c^4*d^4/(c*x - 1)^2 + 5*(c*x + 1 )*b*c^4*d^4/(c*x - 1) + b*c^4*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5/ (c*x - 1)^5 + 5*(c*x + 1)^4/(c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10* (c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c*x - 1) + 1) + (120*(c*x + 1)^4*a* c^4*d^4/(c*x - 1)^4 + 240*(c*x + 1)^3*a*c^4*d^4/(c*x - 1)^3 + 240*(c*x + 1 )^2*a*c^4*d^4/(c*x - 1)^2 + 120*(c*x + 1)*a*c^4*d^4/(c*x - 1) + 24*a*c^4*d ^4 + 48*(c*x + 1)^4*b*c^4*d^4/(c*x - 1)^4 + 156*(c*x + 1)^3*b*c^4*d^4/(c*x - 1)^3 + 196*(c*x + 1)^2*b*c^4*d^4/(c*x - 1)^2 + 113*(c*x + 1)*b*c^4*d^4/ (c*x - 1) + 25*b*c^4*d^4)/((c*x + 1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4/(c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c*x - 1) + 1))*c
Time = 4.24 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.64 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=\frac {d^4\,\left (180\,b\,c^5\,\mathrm {atanh}\left (c\,x\right )-96\,b\,c^5\,\ln \left (c^2\,x^2-1\right )+192\,b\,c^5\,\ln \left (x\right )\right )}{60}-\frac {\frac {d^4\,\left (12\,a+12\,b\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x\,\left (60\,a\,c+3\,b\,c+60\,b\,c\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^2\,\left (120\,a\,c^2+20\,b\,c^2+120\,b\,c^2\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^4\,\left (60\,a\,c^4+180\,b\,c^4+60\,b\,c^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {d^4\,x^3\,\left (120\,a\,c^3+66\,b\,c^3+120\,b\,c^3\,\mathrm {atanh}\left (c\,x\right )\right )}{60}}{x^5} \] Input:
int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^6,x)
Output:
(d^4*(180*b*c^5*atanh(c*x) - 96*b*c^5*log(c^2*x^2 - 1) + 192*b*c^5*log(x)) )/60 - ((d^4*(12*a + 12*b*atanh(c*x)))/60 + (d^4*x*(60*a*c + 3*b*c + 60*b* c*atanh(c*x)))/60 + (d^4*x^2*(120*a*c^2 + 20*b*c^2 + 120*b*c^2*atanh(c*x)) )/60 + (d^4*x^4*(60*a*c^4 + 180*b*c^4 + 60*b*c^4*atanh(c*x)))/60 + (d^4*x^ 3*(120*a*c^3 + 66*b*c^3 + 120*b*c^3*atanh(c*x)))/60)/x^5
Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.60 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^6} \, dx=\frac {d^{4} \left (-12 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}-60 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-120 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}-120 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-60 \mathit {atanh} \left (c x \right ) b c x -12 \mathit {atanh} \left (c x \right ) b -192 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{5} x^{5}+192 \,\mathrm {log}\left (x \right ) b \,c^{5} x^{5}-60 a \,c^{4} x^{4}-120 a \,c^{3} x^{3}-120 a \,c^{2} x^{2}-60 a c x -12 a -180 b \,c^{4} x^{4}-66 b \,c^{3} x^{3}-20 b \,c^{2} x^{2}-3 b c x \right )}{60 x^{5}} \] Input:
int((c*d*x+d)^4*(a+b*atanh(c*x))/x^6,x)
Output:
(d**4*( - 12*atanh(c*x)*b*c**5*x**5 - 60*atanh(c*x)*b*c**4*x**4 - 120*atan h(c*x)*b*c**3*x**3 - 120*atanh(c*x)*b*c**2*x**2 - 60*atanh(c*x)*b*c*x - 12 *atanh(c*x)*b - 192*log(c**2*x - c)*b*c**5*x**5 + 192*log(x)*b*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 - 12*a - 18 0*b*c**4*x**4 - 66*b*c**3*x**3 - 20*b*c**2*x**2 - 3*b*c*x))/(60*x**5)