Integrand size = 20, antiderivative size = 209 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {b c d^4}{12 x^3}-\frac {2 b c^2 d^4}{3 x^2}-\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \text {arctanh}(c x)-\frac {d^4 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {4 c d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {3 c^2 d^4 (a+b \text {arctanh}(c x))}{x^2}-\frac {4 c^3 d^4 (a+b \text {arctanh}(c x))}{x}+a c^4 d^4 \log (x)+\frac {16}{3} b c^4 d^4 \log (x)-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^4 d^4 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b c^4 d^4 \operatorname {PolyLog}(2,c x) \]
-1/12*b*c*d^4/x^3-2/3*b*c^2*d^4/x^2-13/4*b*c^3*d^4/x+13/4*b*c^4*d^4*arctan h(c*x)-1/4*d^4*(a+b*arctanh(c*x))/x^4-4/3*c*d^4*(a+b*arctanh(c*x))/x^3-3*c ^2*d^4*(a+b*arctanh(c*x))/x^2-4*c^3*d^4*(a+b*arctanh(c*x))/x+a*c^4*d^4*ln( x)+16/3*b*c^4*d^4*ln(x)-8/3*b*c^4*d^4*ln(-c^2*x^2+1)-1/2*b*c^4*d^4*polylog (2,-c*x)+1/2*b*c^4*d^4*polylog(2,c*x)
Time = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.99 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {d^4 \left (-6 a-32 a c x-2 b c x-72 a c^2 x^2-16 b c^2 x^2-96 a c^3 x^3-78 b c^3 x^3-6 b \text {arctanh}(c x)-32 b c x \text {arctanh}(c x)-72 b c^2 x^2 \text {arctanh}(c x)-96 b c^3 x^3 \text {arctanh}(c x)+24 a c^4 x^4 \log (x)+128 b c^4 x^4 \log (c x)-39 b c^4 x^4 \log (1-c x)+39 b c^4 x^4 \log (1+c x)-64 b c^4 x^4 \log \left (1-c^2 x^2\right )-12 b c^4 x^4 \operatorname {PolyLog}(2,-c x)+12 b c^4 x^4 \operatorname {PolyLog}(2,c x)\right )}{24 x^4} \]
(d^4*(-6*a - 32*a*c*x - 2*b*c*x - 72*a*c^2*x^2 - 16*b*c^2*x^2 - 96*a*c^3*x ^3 - 78*b*c^3*x^3 - 6*b*ArcTanh[c*x] - 32*b*c*x*ArcTanh[c*x] - 72*b*c^2*x^ 2*ArcTanh[c*x] - 96*b*c^3*x^3*ArcTanh[c*x] + 24*a*c^4*x^4*Log[x] + 128*b*c ^4*x^4*Log[c*x] - 39*b*c^4*x^4*Log[1 - c*x] + 39*b*c^4*x^4*Log[1 + c*x] - 64*b*c^4*x^4*Log[1 - c^2*x^2] - 12*b*c^4*x^4*PolyLog[2, -(c*x)] + 12*b*c^4 *x^4*PolyLog[2, c*x]))/(24*x^4)
Time = 0.46 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6502, 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.
| \(\displaystyle \int \frac {(c d x+d)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (\frac {c^4 d^4 (a+b \text {arctanh}(c x))}{x}+\frac {4 c^3 d^4 (a+b \text {arctanh}(c x))}{x^2}+\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{x^3}+\frac {d^4 (a+b \text {arctanh}(c x))}{x^5}+\frac {4 c d^4 (a+b \text {arctanh}(c x))}{x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 c^3 d^4 (a+b \text {arctanh}(c x))}{x}-\frac {3 c^2 d^4 (a+b \text {arctanh}(c x))}{x^2}-\frac {d^4 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {4 c d^4 (a+b \text {arctanh}(c x))}{3 x^3}+a c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \text {arctanh}(c x)-\frac {1}{2} b c^4 d^4 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b c^4 d^4 \operatorname {PolyLog}(2,c x)+\frac {16}{3} b c^4 d^4 \log (x)-\frac {13 b c^3 d^4}{4 x}-\frac {2 b c^2 d^4}{3 x^2}-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {b c d^4}{12 x^3}\) |
-1/12*(b*c*d^4)/x^3 - (2*b*c^2*d^4)/(3*x^2) - (13*b*c^3*d^4)/(4*x) + (13*b *c^4*d^4*ArcTanh[c*x])/4 - (d^4*(a + b*ArcTanh[c*x]))/(4*x^4) - (4*c*d^4*( a + b*ArcTanh[c*x]))/(3*x^3) - (3*c^2*d^4*(a + b*ArcTanh[c*x]))/x^2 - (4*c ^3*d^4*(a + b*ArcTanh[c*x]))/x + a*c^4*d^4*Log[x] + (16*b*c^4*d^4*Log[x])/ 3 - (8*b*c^4*d^4*Log[1 - c^2*x^2])/3 - (b*c^4*d^4*PolyLog[2, -(c*x)])/2 + (b*c^4*d^4*PolyLog[2, c*x])/2
3.1.39.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p, ( f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Time = 1.59 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.86
| method | result | size |
| parts | \(d^{4} a \left (-\frac {4 c^{3}}{x}-\frac {1}{4 x^{4}}-\frac {4 c}{3 x^{3}}-\frac {3 c^{2}}{x^{2}}+c^{4} \ln \left (x \right )\right )+b \,d^{4} c^{4} \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {25 \ln \left (c x +1\right )}{24}-\frac {103 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {2}{3 c^{2} x^{2}}-\frac {13}{4 c x}+\frac {16 \ln \left (c x \right )}{3}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\) | \(179\) |
| derivativedivides | \(c^{4} \left (d^{4} a \left (\ln \left (c x \right )-\frac {4}{c x}-\frac {3}{c^{2} x^{2}}-\frac {4}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+b \,d^{4} \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {25 \ln \left (c x +1\right )}{24}-\frac {103 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {2}{3 c^{2} x^{2}}-\frac {13}{4 c x}+\frac {16 \ln \left (c x \right )}{3}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\right )\) | \(183\) |
| default | \(c^{4} \left (d^{4} a \left (\ln \left (c x \right )-\frac {4}{c x}-\frac {3}{c^{2} x^{2}}-\frac {4}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+b \,d^{4} \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}-\frac {4 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {25 \ln \left (c x +1\right )}{24}-\frac {103 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {2}{3 c^{2} x^{2}}-\frac {13}{4 c x}+\frac {16 \ln \left (c x \right )}{3}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\right )\) | \(183\) |
| risch | \(-\frac {25 \ln \left (c x +1\right ) b \,c^{4} d^{4}}{24}-\frac {103 \ln \left (-c x +1\right ) b \,c^{4} d^{4}}{24}-\frac {b c \,d^{4}}{12 x^{3}}-\frac {2 b \,c^{2} d^{4}}{3 x^{2}}-\frac {13 b \,c^{3} d^{4}}{4 x}+\frac {2 c \,d^{4} b \ln \left (-c x +1\right )}{3 x^{3}}+\frac {3 c^{2} d^{4} b \ln \left (-c x +1\right )}{2 x^{2}}+\frac {2 c^{3} d^{4} b \ln \left (-c x +1\right )}{x}-\frac {2 b c \,d^{4} \ln \left (c x +1\right )}{3 x^{3}}-\frac {3 b \,c^{2} d^{4} \ln \left (c x +1\right )}{2 x^{2}}-\frac {2 b \,c^{3} d^{4} \ln \left (c x +1\right )}{x}+\frac {25 b \,c^{4} d^{4} \ln \left (c x \right )}{24}-\frac {b \,d^{4} \ln \left (c x +1\right )}{8 x^{4}}-\frac {b \,c^{4} d^{4} \operatorname {dilog}\left (c x +1\right )}{2}-\frac {4 c \,d^{4} a}{3 x^{3}}-\frac {3 c^{2} d^{4} a}{x^{2}}-\frac {4 c^{3} d^{4} a}{x}+c^{4} d^{4} \ln \left (-c x \right ) a +\frac {103 c^{4} d^{4} b \ln \left (-c x \right )}{24}+\frac {d^{4} b \ln \left (-c x +1\right )}{8 x^{4}}+\frac {c^{4} d^{4} b \operatorname {dilog}\left (-c x +1\right )}{2}-\frac {d^{4} a}{4 x^{4}}\) | \(319\) |
d^4*a*(-4*c^3/x-1/4/x^4-4/3*c/x^3-3*c^2/x^2+c^4*ln(x))+b*d^4*c^4*(ln(c*x)* arctanh(c*x)-4/c/x*arctanh(c*x)-3/c^2/x^2*arctanh(c*x)-4/3/c^3/x^3*arctanh (c*x)-1/4/c^4/x^4*arctanh(c*x)-25/24*ln(c*x+1)-103/24*ln(c*x-1)-1/12/c^3/x ^3-2/3/c^2/x^2-13/4/c/x+16/3*ln(c*x)-1/2*dilog(c*x+1)-1/2*ln(c*x)*ln(c*x+1 )-1/2*dilog(c*x))
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=\int { \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
integral((a*c^4*d^4*x^4 + 4*a*c^3*d^4*x^3 + 6*a*c^2*d^4*x^2 + 4*a*c*d^4*x + a*d^4 + (b*c^4*d^4*x^4 + 4*b*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 4*b*c*d^4*x + b*d^4)*arctanh(c*x))/x^5, x)
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=d^{4} \left (\int \frac {a}{x^{5}}\, dx + \int \frac {4 a c}{x^{4}}\, dx + \int \frac {6 a c^{2}}{x^{3}}\, dx + \int \frac {4 a c^{3}}{x^{2}}\, dx + \int \frac {a c^{4}}{x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac {4 b c \operatorname {atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {6 b c^{2} \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {4 b c^{3} \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b c^{4} \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]
d**4*(Integral(a/x**5, x) + Integral(4*a*c/x**4, x) + Integral(6*a*c**2/x* *3, x) + Integral(4*a*c**3/x**2, x) + Integral(a*c**4/x, x) + Integral(b*a tanh(c*x)/x**5, x) + Integral(4*b*c*atanh(c*x)/x**4, x) + Integral(6*b*c** 2*atanh(c*x)/x**3, x) + Integral(4*b*c**3*atanh(c*x)/x**2, x) + Integral(b *c**4*atanh(c*x)/x, x))
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=\int { \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
1/2*b*c^4*d^4*integrate((log(c*x + 1) - log(-c*x + 1))/x, x) + a*c^4*d^4*l og(x) - 2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c^3*d^4 + 3/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*c^ 2*d^4 - 2/3*((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*d^4 - 4*a*c^3*d^4/x + 1/24*((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*d^4 - 3*a*c^2*d ^4/x^2 - 4/3*a*c*d^4/x^3 - 1/4*a*d^4/x^4
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=\int { \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^5} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^4}{x^5} \,d x \]