Integrand size = 20, antiderivative size = 229 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=-\frac {b c d^4}{42 x^6}-\frac {2 b c^2 d^4}{15 x^5}-\frac {47 b c^3 d^4}{140 x^4}-\frac {5 b c^4 d^4}{9 x^3}-\frac {88 b c^5 d^4}{105 x^2}-\frac {5 b c^6 d^4}{3 x}-\frac {d^4 (a+b \text {arctanh}(c x))}{7 x^7}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{3 x^6}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{5 x^5}-\frac {c^3 d^4 (a+b \text {arctanh}(c x))}{x^4}-\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{3 x^3}+\frac {176}{105} b c^7 d^4 \log (x)-\frac {117}{70} b c^7 d^4 \log (1-c x)-\frac {1}{210} b c^7 d^4 \log (1+c x) \] Output:
-1/42*b*c*d^4/x^6-2/15*b*c^2*d^4/x^5-47/140*b*c^3*d^4/x^4-5/9*b*c^4*d^4/x^ 3-88/105*b*c^5*d^4/x^2-5/3*b*c^6*d^4/x-1/7*d^4*(a+b*arctanh(c*x))/x^7-2/3* c*d^4*(a+b*arctanh(c*x))/x^6-6/5*c^2*d^4*(a+b*arctanh(c*x))/x^5-c^3*d^4*(a +b*arctanh(c*x))/x^4-1/3*c^4*d^4*(a+b*arctanh(c*x))/x^3+176/105*b*c^7*d^4* ln(x)-117/70*b*c^7*d^4*ln(-c*x+1)-1/210*b*c^7*d^4*ln(c*x+1)
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=-\frac {d^4 \left (180 a+840 a c x+30 b c x+1512 a c^2 x^2+168 b c^2 x^2+1260 a c^3 x^3+423 b c^3 x^3+420 a c^4 x^4+700 b c^4 x^4+1056 b c^5 x^5+2100 b c^6 x^6+12 b \left (15+70 c x+126 c^2 x^2+105 c^3 x^3+35 c^4 x^4\right ) \text {arctanh}(c x)-2112 b c^7 x^7 \log (x)+2106 b c^7 x^7 \log (1-c x)+6 b c^7 x^7 \log (1+c x)\right )}{1260 x^7} \] Input:
Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^8,x]
Output:
-1/1260*(d^4*(180*a + 840*a*c*x + 30*b*c*x + 1512*a*c^2*x^2 + 168*b*c^2*x^ 2 + 1260*a*c^3*x^3 + 423*b*c^3*x^3 + 420*a*c^4*x^4 + 700*b*c^4*x^4 + 1056* b*c^5*x^5 + 2100*b*c^6*x^6 + 12*b*(15 + 70*c*x + 126*c^2*x^2 + 105*c^3*x^3 + 35*c^4*x^4)*ArcTanh[c*x] - 2112*b*c^7*x^7*Log[x] + 2106*b*c^7*x^7*Log[1 - c*x] + 6*b*c^7*x^7*Log[1 + c*x]))/x^7
Time = 0.54 (sec) , antiderivative size = 192, 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.
| \(\displaystyle \int \frac {(c d x+d)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int -\frac {d^4 \left (35 c^4 x^4+105 c^3 x^3+126 c^2 x^2+70 c x+15\right )}{105 x^7 \left (1-c^2 x^2\right )}dx-\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c^3 d^4 (a+b \text {arctanh}(c x))}{x^4}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{5 x^5}-\frac {d^4 (a+b \text {arctanh}(c x))}{7 x^7}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{3 x^6}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{105} b c d^4 \int \frac {35 c^4 x^4+105 c^3 x^3+126 c^2 x^2+70 c x+15}{x^7 \left (1-c^2 x^2\right )}dx-\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c^3 d^4 (a+b \text {arctanh}(c x))}{x^4}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{5 x^5}-\frac {d^4 (a+b \text {arctanh}(c x))}{7 x^7}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{3 x^6}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {1}{105} b c d^4 \int \left (-\frac {351 c^7}{2 (c x-1)}-\frac {c^7}{2 (c x+1)}+\frac {176 c^6}{x}+\frac {175 c^5}{x^2}+\frac {176 c^4}{x^3}+\frac {175 c^3}{x^4}+\frac {141 c^2}{x^5}+\frac {70 c}{x^6}+\frac {15}{x^7}\right )dx-\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c^3 d^4 (a+b \text {arctanh}(c x))}{x^4}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{5 x^5}-\frac {d^4 (a+b \text {arctanh}(c x))}{7 x^7}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{3 x^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c^3 d^4 (a+b \text {arctanh}(c x))}{x^4}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{5 x^5}-\frac {d^4 (a+b \text {arctanh}(c x))}{7 x^7}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{3 x^6}+\frac {1}{105} b c d^4 \left (176 c^6 \log (x)-\frac {351}{2} c^6 \log (1-c x)-\frac {1}{2} c^6 \log (c x+1)-\frac {175 c^5}{x}-\frac {88 c^4}{x^2}-\frac {175 c^3}{3 x^3}-\frac {141 c^2}{4 x^4}-\frac {14 c}{x^5}-\frac {5}{2 x^6}\right )\) |
Input:
Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^8,x]
Output:
-1/7*(d^4*(a + b*ArcTanh[c*x]))/x^7 - (2*c*d^4*(a + b*ArcTanh[c*x]))/(3*x^ 6) - (6*c^2*d^4*(a + b*ArcTanh[c*x]))/(5*x^5) - (c^3*d^4*(a + b*ArcTanh[c* x]))/x^4 - (c^4*d^4*(a + b*ArcTanh[c*x]))/(3*x^3) + (b*c*d^4*(-5/(2*x^6) - (14*c)/x^5 - (141*c^2)/(4*x^4) - (175*c^3)/(3*x^3) - (88*c^4)/x^2 - (175* c^5)/x + 176*c^6*Log[x] - (351*c^6*Log[1 - c*x])/2 - (c^6*Log[1 + c*x])/2) )/105
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 = 0.58 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.79
| method | result | size |
| parts | \(d^{4} a \left (-\frac {2 c}{3 x^{6}}-\frac {c^{3}}{x^{4}}-\frac {1}{7 x^{7}}-\frac {6 c^{2}}{5 x^{5}}-\frac {c^{4}}{3 x^{3}}\right )+d^{4} b \,c^{7} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{7 c^{7} x^{7}}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{6} x^{6}}-\frac {117 \ln \left (c x -1\right )}{70}-\frac {1}{42 c^{6} x^{6}}-\frac {2}{15 c^{5} x^{5}}-\frac {47}{140 c^{4} x^{4}}-\frac {5}{9 c^{3} x^{3}}-\frac {88}{105 c^{2} x^{2}}-\frac {5}{3 c x}+\frac {176 \ln \left (c x \right )}{105}-\frac {\ln \left (c x +1\right )}{210}\right )\) | \(182\) |
| derivativedivides | \(c^{7} \left (d^{4} a \left (-\frac {1}{3 c^{3} x^{3}}-\frac {1}{7 c^{7} x^{7}}-\frac {1}{c^{4} x^{4}}-\frac {6}{5 c^{5} x^{5}}-\frac {2}{3 c^{6} x^{6}}\right )+d^{4} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{7 c^{7} x^{7}}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{6} x^{6}}-\frac {117 \ln \left (c x -1\right )}{70}-\frac {1}{42 c^{6} x^{6}}-\frac {2}{15 c^{5} x^{5}}-\frac {47}{140 c^{4} x^{4}}-\frac {5}{9 c^{3} x^{3}}-\frac {88}{105 c^{2} x^{2}}-\frac {5}{3 c x}+\frac {176 \ln \left (c x \right )}{105}-\frac {\ln \left (c x +1\right )}{210}\right )\right )\) | \(188\) |
| default | \(c^{7} \left (d^{4} a \left (-\frac {1}{3 c^{3} x^{3}}-\frac {1}{7 c^{7} x^{7}}-\frac {1}{c^{4} x^{4}}-\frac {6}{5 c^{5} x^{5}}-\frac {2}{3 c^{6} x^{6}}\right )+d^{4} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{7 c^{7} x^{7}}-\frac {\operatorname {arctanh}\left (c x \right )}{c^{4} x^{4}}-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{6} x^{6}}-\frac {117 \ln \left (c x -1\right )}{70}-\frac {1}{42 c^{6} x^{6}}-\frac {2}{15 c^{5} x^{5}}-\frac {47}{140 c^{4} x^{4}}-\frac {5}{9 c^{3} x^{3}}-\frac {88}{105 c^{2} x^{2}}-\frac {5}{3 c x}+\frac {176 \ln \left (c x \right )}{105}-\frac {\ln \left (c x +1\right )}{210}\right )\right )\) | \(188\) |
| risch | \(-\frac {d^{4} b \left (35 c^{4} x^{4}+105 x^{3} c^{3}+126 c^{2} x^{2}+70 c x +15\right ) \ln \left (c x +1\right )}{210 x^{7}}-\frac {d^{4} \left (6 b \,c^{7} \ln \left (c x +1\right ) x^{7}-2112 b \,c^{7} \ln \left (-x \right ) x^{7}+2106 b \,c^{7} x^{7} \ln \left (-c x +1\right )+2100 b \,c^{6} x^{6}+1056 b \,c^{5} x^{5}-210 b \,x^{4} \ln \left (-c x +1\right ) c^{4}+420 a \,c^{4} x^{4}+700 b \,c^{4} x^{4}-630 b \,x^{3} \ln \left (-c x +1\right ) c^{3}+1260 a \,c^{3} x^{3}+423 b \,c^{3} x^{3}-756 b \,c^{2} x^{2} \ln \left (-c x +1\right )+1512 a \,c^{2} x^{2}+168 b \,c^{2} x^{2}-420 b c x \ln \left (-c x +1\right )+840 a c x +30 b c x -90 b \ln \left (-c x +1\right )+180 a \right )}{1260 x^{7}}\) | \(255\) |
| parallelrisch | \(\frac {2112 b \,c^{7} d^{4} \ln \left (x \right ) x^{7}-2112 \ln \left (c x -1\right ) x^{7} b \,c^{7} d^{4}-12 b \,c^{7} d^{4} \operatorname {arctanh}\left (c x \right ) x^{7}-1056 c^{7} d^{4} x^{7} b -2100 b \,c^{6} d^{4} x^{6}-1056 b \,c^{5} d^{4} x^{5}-420 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}-420 a \,c^{4} d^{4} x^{4}-700 b \,c^{4} d^{4} x^{4}-1260 d^{4} b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-1260 a \,c^{3} d^{4} x^{3}-423 b \,c^{3} d^{4} x^{3}-1512 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{4}-1512 a \,c^{2} d^{4} x^{2}-168 b \,c^{2} d^{4} x^{2}-840 b c \,d^{4} x \,\operatorname {arctanh}\left (c x \right )-840 a c \,d^{4} x -30 b c \,d^{4} x -180 b \,d^{4} \operatorname {arctanh}\left (c x \right )-180 d^{4} a}{1260 x^{7}}\) | \(255\) |
Input:
int((c*d*x+d)^4*(a+b*arctanh(c*x))/x^8,x,method=_RETURNVERBOSE)
Output:
d^4*a*(-2/3*c/x^6-c^3/x^4-1/7/x^7-6/5*c^2/x^5-1/3*c^4/x^3)+d^4*b*c^7*(-1/3 *arctanh(c*x)/c^3/x^3-1/7*arctanh(c*x)/c^7/x^7-arctanh(c*x)/c^4/x^4-6/5*ar ctanh(c*x)/c^5/x^5-2/3*arctanh(c*x)/c^6/x^6-117/70*ln(c*x-1)-1/42/c^6/x^6- 2/15/c^5/x^5-47/140/c^4/x^4-5/9/c^3/x^3-88/105/c^2/x^2-5/3/c/x+176/105*ln( c*x)-1/210*ln(c*x+1))
Time = 0.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.95 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=-\frac {6 \, b c^{7} d^{4} x^{7} \log \left (c x + 1\right ) + 2106 \, b c^{7} d^{4} x^{7} \log \left (c x - 1\right ) - 2112 \, b c^{7} d^{4} x^{7} \log \left (x\right ) + 2100 \, b c^{6} d^{4} x^{6} + 1056 \, b c^{5} d^{4} x^{5} + 140 \, {\left (3 \, a + 5 \, b\right )} c^{4} d^{4} x^{4} + 9 \, {\left (140 \, a + 47 \, b\right )} c^{3} d^{4} x^{3} + 168 \, {\left (9 \, a + b\right )} c^{2} d^{4} x^{2} + 30 \, {\left (28 \, a + b\right )} c d^{4} x + 180 \, a d^{4} + 6 \, {\left (35 \, b c^{4} d^{4} x^{4} + 105 \, b c^{3} d^{4} x^{3} + 126 \, b c^{2} d^{4} x^{2} + 70 \, b c d^{4} x + 15 \, b d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{1260 \, x^{7}} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^8,x, algorithm="fricas")
Output:
-1/1260*(6*b*c^7*d^4*x^7*log(c*x + 1) + 2106*b*c^7*d^4*x^7*log(c*x - 1) - 2112*b*c^7*d^4*x^7*log(x) + 2100*b*c^6*d^4*x^6 + 1056*b*c^5*d^4*x^5 + 140* (3*a + 5*b)*c^4*d^4*x^4 + 9*(140*a + 47*b)*c^3*d^4*x^3 + 168*(9*a + b)*c^2 *d^4*x^2 + 30*(28*a + b)*c*d^4*x + 180*a*d^4 + 6*(35*b*c^4*d^4*x^4 + 105*b *c^3*d^4*x^3 + 126*b*c^2*d^4*x^2 + 70*b*c*d^4*x + 15*b*d^4)*log(-(c*x + 1) /(c*x - 1)))/x^7
Time = 0.87 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.31 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=\begin {cases} - \frac {a c^{4} d^{4}}{3 x^{3}} - \frac {a c^{3} d^{4}}{x^{4}} - \frac {6 a c^{2} d^{4}}{5 x^{5}} - \frac {2 a c d^{4}}{3 x^{6}} - \frac {a d^{4}}{7 x^{7}} + \frac {176 b c^{7} d^{4} \log {\left (x \right )}}{105} - \frac {176 b c^{7} d^{4} \log {\left (x - \frac {1}{c} \right )}}{105} - \frac {b c^{7} d^{4} \operatorname {atanh}{\left (c x \right )}}{105} - \frac {5 b c^{6} d^{4}}{3 x} - \frac {88 b c^{5} d^{4}}{105 x^{2}} - \frac {b c^{4} d^{4} \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} - \frac {5 b c^{4} d^{4}}{9 x^{3}} - \frac {b c^{3} d^{4} \operatorname {atanh}{\left (c x \right )}}{x^{4}} - \frac {47 b c^{3} d^{4}}{140 x^{4}} - \frac {6 b c^{2} d^{4} \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} - \frac {2 b c^{2} d^{4}}{15 x^{5}} - \frac {2 b c d^{4} \operatorname {atanh}{\left (c x \right )}}{3 x^{6}} - \frac {b c d^{4}}{42 x^{6}} - \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{7 x^{7}} & \text {for}\: c \neq 0 \\- \frac {a d^{4}}{7 x^{7}} & \text {otherwise} \end {cases} \] Input:
integrate((c*d*x+d)**4*(a+b*atanh(c*x))/x**8,x)
Output:
Piecewise((-a*c**4*d**4/(3*x**3) - a*c**3*d**4/x**4 - 6*a*c**2*d**4/(5*x** 5) - 2*a*c*d**4/(3*x**6) - a*d**4/(7*x**7) + 176*b*c**7*d**4*log(x)/105 - 176*b*c**7*d**4*log(x - 1/c)/105 - b*c**7*d**4*atanh(c*x)/105 - 5*b*c**6*d **4/(3*x) - 88*b*c**5*d**4/(105*x**2) - b*c**4*d**4*atanh(c*x)/(3*x**3) - 5*b*c**4*d**4/(9*x**3) - b*c**3*d**4*atanh(c*x)/x**4 - 47*b*c**3*d**4/(140 *x**4) - 6*b*c**2*d**4*atanh(c*x)/(5*x**5) - 2*b*c**2*d**4/(15*x**5) - 2*b *c*d**4*atanh(c*x)/(3*x**6) - b*c*d**4/(42*x**6) - b*d**4*atanh(c*x)/(7*x* *7), Ne(c, 0)), (-a*d**4/(7*x**7), True))
Time = 0.03 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.54 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c^{4} d^{4} + \frac {1}{6} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b c^{3} d^{4} - \frac {3}{10} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b c^{2} d^{4} + \frac {1}{45} \, {\left ({\left (15 \, c^{5} \log \left (c x + 1\right ) - 15 \, c^{5} \log \left (c x - 1\right ) - \frac {2 \, {\left (15 \, c^{4} x^{4} + 5 \, c^{2} x^{2} + 3\right )}}{x^{5}}\right )} c - \frac {30 \, \operatorname {artanh}\left (c x\right )}{x^{6}}\right )} b c d^{4} - \frac {1}{84} \, {\left ({\left (6 \, c^{6} \log \left (c^{2} x^{2} - 1\right ) - 6 \, c^{6} \log \left (x^{2}\right ) + \frac {6 \, c^{4} x^{4} + 3 \, c^{2} x^{2} + 2}{x^{6}}\right )} c + \frac {12 \, \operatorname {artanh}\left (c x\right )}{x^{7}}\right )} b d^{4} - \frac {a c^{4} d^{4}}{3 \, x^{3}} - \frac {a c^{3} d^{4}}{x^{4}} - \frac {6 \, a c^{2} d^{4}}{5 \, x^{5}} - \frac {2 \, a c d^{4}}{3 \, x^{6}} - \frac {a d^{4}}{7 \, x^{7}} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^8,x, algorithm="maxima")
Output:
-1/6*((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^4*d^4 + 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^3*d^4 - 3/10*((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*c^2*d ^4 + 1/45*((15*c^5*log(c*x + 1) - 15*c^5*log(c*x - 1) - 2*(15*c^4*x^4 + 5* c^2*x^2 + 3)/x^5)*c - 30*arctanh(c*x)/x^6)*b*c*d^4 - 1/84*((6*c^6*log(c^2* x^2 - 1) - 6*c^6*log(x^2) + (6*c^4*x^4 + 3*c^2*x^2 + 2)/x^6)*c + 12*arctan h(c*x)/x^7)*b*d^4 - 1/3*a*c^4*d^4/x^3 - a*c^3*d^4/x^4 - 6/5*a*c^2*d^4/x^5 - 2/3*a*c*d^4/x^6 - 1/7*a*d^4/x^7
Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (203) = 406\).
Time = 0.14 (sec) , antiderivative size = 735, normalized size of antiderivative = 3.21 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx =\text {Too large to display} \] Input:
integrate((c*d*x+d)^4*(a+b*arctanh(c*x))/x^8,x, algorithm="giac")
Output:
4/315*(132*b*c^6*d^4*log(-(c*x + 1)/(c*x - 1) - 1) - 132*b*c^6*d^4*log(-(c *x + 1)/(c*x - 1)) + 12*(105*(c*x + 1)^6*b*c^6*d^4/(c*x - 1)^6 + 210*(c*x + 1)^5*b*c^6*d^4/(c*x - 1)^5 + 385*(c*x + 1)^4*b*c^6*d^4/(c*x - 1)^4 + 385 *(c*x + 1)^3*b*c^6*d^4/(c*x - 1)^3 + 231*(c*x + 1)^2*b*c^6*d^4/(c*x - 1)^2 + 77*(c*x + 1)*b*c^6*d^4/(c*x - 1) + 11*b*c^6*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^7/(c*x - 1)^7 + 7*(c*x + 1)^6/(c*x - 1)^6 + 21*(c*x + 1)^5/ (c*x - 1)^5 + 35*(c*x + 1)^4/(c*x - 1)^4 + 35*(c*x + 1)^3/(c*x - 1)^3 + 21 *(c*x + 1)^2/(c*x - 1)^2 + 7*(c*x + 1)/(c*x - 1) + 1) + (2520*(c*x + 1)^6* a*c^6*d^4/(c*x - 1)^6 + 5040*(c*x + 1)^5*a*c^6*d^4/(c*x - 1)^5 + 9240*(c*x + 1)^4*a*c^6*d^4/(c*x - 1)^4 + 9240*(c*x + 1)^3*a*c^6*d^4/(c*x - 1)^3 + 5 544*(c*x + 1)^2*a*c^6*d^4/(c*x - 1)^2 + 1848*(c*x + 1)*a*c^6*d^4/(c*x - 1) + 264*a*c^6*d^4 + 1128*(c*x + 1)^6*b*c^6*d^4/(c*x - 1)^6 + 4812*(c*x + 1) ^5*b*c^6*d^4/(c*x - 1)^5 + 9476*(c*x + 1)^4*b*c^6*d^4/(c*x - 1)^4 + 10631* (c*x + 1)^3*b*c^6*d^4/(c*x - 1)^3 + 6933*(c*x + 1)^2*b*c^6*d^4/(c*x - 1)^2 + 2465*(c*x + 1)*b*c^6*d^4/(c*x - 1) + 371*b*c^6*d^4)/((c*x + 1)^7/(c*x - 1)^7 + 7*(c*x + 1)^6/(c*x - 1)^6 + 21*(c*x + 1)^5/(c*x - 1)^5 + 35*(c*x + 1)^4/(c*x - 1)^4 + 35*(c*x + 1)^3/(c*x - 1)^3 + 21*(c*x + 1)^2/(c*x - 1)^ 2 + 7*(c*x + 1)/(c*x - 1) + 1))*c
Time = 4.10 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.14 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=\frac {176\,b\,c^7\,d^4\,\ln \left (x\right )}{105}-\frac {88\,b\,c^7\,d^4\,\ln \left (c^2\,x^2-1\right )}{105}-\frac {6\,a\,c^2\,d^4}{5\,x^5}-\frac {a\,c^3\,d^4}{x^4}-\frac {a\,c^4\,d^4}{3\,x^3}-\frac {2\,b\,c^2\,d^4}{15\,x^5}-\frac {47\,b\,c^3\,d^4}{140\,x^4}-\frac {5\,b\,c^4\,d^4}{9\,x^3}-\frac {88\,b\,c^5\,d^4}{105\,x^2}-\frac {5\,b\,c^6\,d^4}{3\,x}-\frac {a\,d^4}{7\,x^7}-\frac {2\,a\,c\,d^4}{3\,x^6}-\frac {b\,c\,d^4}{42\,x^6}-\frac {b\,d^4\,\mathrm {atanh}\left (c\,x\right )}{7\,x^7}-\frac {5\,b\,c^8\,d^4\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{3\,\sqrt {-c^2}}-\frac {2\,b\,c\,d^4\,\mathrm {atanh}\left (c\,x\right )}{3\,x^6}-\frac {6\,b\,c^2\,d^4\,\mathrm {atanh}\left (c\,x\right )}{5\,x^5}-\frac {b\,c^3\,d^4\,\mathrm {atanh}\left (c\,x\right )}{x^4}-\frac {b\,c^4\,d^4\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3} \] Input:
int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^8,x)
Output:
(176*b*c^7*d^4*log(x))/105 - (88*b*c^7*d^4*log(c^2*x^2 - 1))/105 - (6*a*c^ 2*d^4)/(5*x^5) - (a*c^3*d^4)/x^4 - (a*c^4*d^4)/(3*x^3) - (2*b*c^2*d^4)/(15 *x^5) - (47*b*c^3*d^4)/(140*x^4) - (5*b*c^4*d^4)/(9*x^3) - (88*b*c^5*d^4)/ (105*x^2) - (5*b*c^6*d^4)/(3*x) - (a*d^4)/(7*x^7) - (2*a*c*d^4)/(3*x^6) - (b*c*d^4)/(42*x^6) - (b*d^4*atanh(c*x))/(7*x^7) - (5*b*c^8*d^4*atan((c^2*x )/(-c^2)^(1/2)))/(3*(-c^2)^(1/2)) - (2*b*c*d^4*atanh(c*x))/(3*x^6) - (6*b* c^2*d^4*atanh(c*x))/(5*x^5) - (b*c^3*d^4*atanh(c*x))/x^4 - (b*c^4*d^4*atan h(c*x))/(3*x^3)
Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.84 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^8} \, dx=\frac {d^{4} \left (-12 \mathit {atanh} \left (c x \right ) b \,c^{7} x^{7}-420 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-1260 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}-1512 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-840 \mathit {atanh} \left (c x \right ) b c x -180 \mathit {atanh} \left (c x \right ) b -2112 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{7} x^{7}+2112 \,\mathrm {log}\left (x \right ) b \,c^{7} x^{7}-420 a \,c^{4} x^{4}-1260 a \,c^{3} x^{3}-1512 a \,c^{2} x^{2}-840 a c x -180 a -2100 b \,c^{6} x^{6}-1056 b \,c^{5} x^{5}-700 b \,c^{4} x^{4}-423 b \,c^{3} x^{3}-168 b \,c^{2} x^{2}-30 b c x \right )}{1260 x^{7}} \] Input:
int((c*d*x+d)^4*(a+b*atanh(c*x))/x^8,x)
Output:
(d**4*( - 12*atanh(c*x)*b*c**7*x**7 - 420*atanh(c*x)*b*c**4*x**4 - 1260*at anh(c*x)*b*c**3*x**3 - 1512*atanh(c*x)*b*c**2*x**2 - 840*atanh(c*x)*b*c*x - 180*atanh(c*x)*b - 2112*log(c**2*x - c)*b*c**7*x**7 + 2112*log(x)*b*c**7 *x**7 - 420*a*c**4*x**4 - 1260*a*c**3*x**3 - 1512*a*c**2*x**2 - 840*a*c*x - 180*a - 2100*b*c**6*x**6 - 1056*b*c**5*x**5 - 700*b*c**4*x**4 - 423*b*c* *3*x**3 - 168*b*c**2*x**2 - 30*b*c*x))/(1260*x**7)