Integrand size = 20, antiderivative size = 185 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=4 a c d^4 x+\frac {13}{4} b c d^4 x+\frac {2}{3} b c^2 d^4 x^2+\frac {1}{12} b c^3 d^4 x^3-\frac {13}{4} b d^4 \text {arctanh}(c x)+4 b c d^4 x \text {arctanh}(c x)+3 c^2 d^4 x^2 (a+b \text {arctanh}(c x))+\frac {4}{3} c^3 d^4 x^3 (a+b \text {arctanh}(c x))+\frac {1}{4} c^4 d^4 x^4 (a+b \text {arctanh}(c x))+a d^4 \log (x)+\frac {8}{3} b d^4 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^4 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b d^4 \operatorname {PolyLog}(2,c x) \]
4*a*c*d^4*x+13/4*b*c*d^4*x+2/3*b*c^2*d^4*x^2+1/12*b*c^3*d^4*x^3-13/4*b*d^4 *arctanh(c*x)+4*b*c*d^4*x*arctanh(c*x)+3*c^2*d^4*x^2*(a+b*arctanh(c*x))+4/ 3*c^3*d^4*x^3*(a+b*arctanh(c*x))+1/4*c^4*d^4*x^4*(a+b*arctanh(c*x))+a*d^4* ln(x)+8/3*b*d^4*ln(-c^2*x^2+1)-1/2*b*d^4*polylog(2,-c*x)+1/2*b*d^4*polylog (2,c*x)
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.97 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=\frac {1}{24} d^4 \left (96 a c x+78 b c x+72 a c^2 x^2+16 b c^2 x^2+32 a c^3 x^3+2 b c^3 x^3+6 a c^4 x^4+96 b c x \text {arctanh}(c x)+72 b c^2 x^2 \text {arctanh}(c x)+32 b c^3 x^3 \text {arctanh}(c x)+6 b c^4 x^4 \text {arctanh}(c x)+24 a \log (x)+39 b \log (1-c x)-39 b \log (1+c x)+48 b \log \left (1-c^2 x^2\right )+16 b \log \left (-1+c^2 x^2\right )-12 b \operatorname {PolyLog}(2,-c x)+12 b \operatorname {PolyLog}(2,c x)\right ) \]
(d^4*(96*a*c*x + 78*b*c*x + 72*a*c^2*x^2 + 16*b*c^2*x^2 + 32*a*c^3*x^3 + 2 *b*c^3*x^3 + 6*a*c^4*x^4 + 96*b*c*x*ArcTanh[c*x] + 72*b*c^2*x^2*ArcTanh[c* x] + 32*b*c^3*x^3*ArcTanh[c*x] + 6*b*c^4*x^4*ArcTanh[c*x] + 24*a*Log[x] + 39*b*Log[1 - c*x] - 39*b*Log[1 + c*x] + 48*b*Log[1 - c^2*x^2] + 16*b*Log[- 1 + c^2*x^2] - 12*b*PolyLog[2, -(c*x)] + 12*b*PolyLog[2, c*x]))/24
Time = 0.43 (sec) , antiderivative size = 185, 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.
| \(\displaystyle \int \frac {(c d x+d)^4 (a+b \text {arctanh}(c x))}{x} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (c^4 d^4 x^3 (a+b \text {arctanh}(c x))+4 c^3 d^4 x^2 (a+b \text {arctanh}(c x))+6 c^2 d^4 x (a+b \text {arctanh}(c x))+4 c d^4 (a+b \text {arctanh}(c x))+\frac {d^4 (a+b \text {arctanh}(c x))}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} c^4 d^4 x^4 (a+b \text {arctanh}(c x))+\frac {4}{3} c^3 d^4 x^3 (a+b \text {arctanh}(c x))+3 c^2 d^4 x^2 (a+b \text {arctanh}(c x))+4 a c d^4 x+a d^4 \log (x)-\frac {13}{4} b d^4 \text {arctanh}(c x)+4 b c d^4 x \text {arctanh}(c x)+\frac {1}{12} b c^3 d^4 x^3+\frac {2}{3} b c^2 d^4 x^2+\frac {8}{3} b d^4 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^4 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b d^4 \operatorname {PolyLog}(2,c x)+\frac {13}{4} b c d^4 x\) |
4*a*c*d^4*x + (13*b*c*d^4*x)/4 + (2*b*c^2*d^4*x^2)/3 + (b*c^3*d^4*x^3)/12 - (13*b*d^4*ArcTanh[c*x])/4 + 4*b*c*d^4*x*ArcTanh[c*x] + 3*c^2*d^4*x^2*(a + b*ArcTanh[c*x]) + (4*c^3*d^4*x^3*(a + b*ArcTanh[c*x]))/3 + (c^4*d^4*x^4* (a + b*ArcTanh[c*x]))/4 + a*d^4*Log[x] + (8*b*d^4*Log[1 - c^2*x^2])/3 - (b *d^4*PolyLog[2, -(c*x)])/2 + (b*d^4*PolyLog[2, c*x])/2
3.1.35.3.1 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 = 1.31 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.86
| method | result | size |
| parts | \(d^{4} a \left (\frac {c^{4} x^{4}}{4}+\frac {4 c^{3} x^{3}}{3}+3 c^{2} x^{2}+4 c x +\ln \left (x \right )\right )+b \,d^{4} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {4 c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+4 c x \,\operatorname {arctanh}\left (c x \right )+\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {c^{3} x^{3}}{12}+\frac {2 c^{2} x^{2}}{3}+\frac {13 c x}{4}+\frac {103 \ln \left (c x -1\right )}{24}+\frac {25 \ln \left (c x +1\right )}{24}\right )\) | \(159\) |
| derivativedivides | \(d^{4} a \left (\frac {c^{4} x^{4}}{4}+\frac {4 c^{3} x^{3}}{3}+3 c^{2} x^{2}+4 c x +\ln \left (c x \right )\right )+b \,d^{4} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {4 c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+4 c x \,\operatorname {arctanh}\left (c x \right )+\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {c^{3} x^{3}}{12}+\frac {2 c^{2} x^{2}}{3}+\frac {13 c x}{4}+\frac {103 \ln \left (c x -1\right )}{24}+\frac {25 \ln \left (c x +1\right )}{24}\right )\) | \(161\) |
| default | \(d^{4} a \left (\frac {c^{4} x^{4}}{4}+\frac {4 c^{3} x^{3}}{3}+3 c^{2} x^{2}+4 c x +\ln \left (c x \right )\right )+b \,d^{4} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {4 c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+4 c x \,\operatorname {arctanh}\left (c x \right )+\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {c^{3} x^{3}}{12}+\frac {2 c^{2} x^{2}}{3}+\frac {13 c x}{4}+\frac {103 \ln \left (c x -1\right )}{24}+\frac {25 \ln \left (c x +1\right )}{24}\right )\) | \(161\) |
| risch | \(-\frac {58 b \,d^{4}}{9}-\frac {103 d^{4} a}{12}+\frac {25 \ln \left (c x +1\right ) b \,d^{4}}{24}+\frac {103 \ln \left (-c x +1\right ) b \,d^{4}}{24}+\frac {13 b c \,d^{4} x}{4}+4 a c \,d^{4} x +\frac {2 b \,c^{2} d^{4} x^{2}}{3}+\frac {b \,c^{3} d^{4} x^{3}}{12}+\frac {a \,c^{4} d^{4} x^{4}}{4}+\frac {4 a \,c^{3} d^{4} x^{3}}{3}+3 a \,c^{2} d^{4} x^{2}-\frac {b \,d^{4} \operatorname {dilog}\left (c x +1\right )}{2}+d^{4} \ln \left (-c x \right ) a +\frac {d^{4} \operatorname {dilog}\left (-c x +1\right ) b}{2}-\frac {d^{4} b \,x^{4} \ln \left (-c x +1\right ) c^{4}}{8}-\frac {2 d^{4} b \,x^{3} \ln \left (-c x +1\right ) c^{3}}{3}-\frac {3 d^{4} b \,x^{2} \ln \left (-c x +1\right ) c^{2}}{2}-2 d^{4} b c x \ln \left (-c x +1\right )+\frac {b \,d^{4} \ln \left (c x +1\right ) c^{4} x^{4}}{8}+\frac {2 b \,d^{4} \ln \left (c x +1\right ) c^{3} x^{3}}{3}+\frac {3 b \,d^{4} \ln \left (c x +1\right ) c^{2} x^{2}}{2}+2 b \,d^{4} \ln \left (c x +1\right ) c x\) | \(290\) |
d^4*a*(1/4*c^4*x^4+4/3*c^3*x^3+3*c^2*x^2+4*c*x+ln(x))+b*d^4*(1/4*c^4*x^4*a rctanh(c*x)+4/3*c^3*x^3*arctanh(c*x)+3*c^2*x^2*arctanh(c*x)+4*c*x*arctanh( c*x)+ln(c*x)*arctanh(c*x)-1/2*dilog(c*x+1)-1/2*ln(c*x)*ln(c*x+1)-1/2*dilog (c*x)+1/12*c^3*x^3+2/3*c^2*x^2+13/4*c*x+103/24*ln(c*x-1)+25/24*ln(c*x+1))
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=\int { \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x} \,d x } \]
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, x)
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=d^{4} \left (\int 4 a c\, dx + \int \frac {a}{x}\, dx + \int 6 a c^{2} x\, dx + \int 4 a c^{3} x^{2}\, dx + \int a c^{4} x^{3}\, dx + \int 4 b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x}\, dx + \int 6 b c^{2} x \operatorname {atanh}{\left (c x \right )}\, dx + \int 4 b c^{3} x^{2} \operatorname {atanh}{\left (c x \right )}\, dx + \int b c^{4} x^{3} \operatorname {atanh}{\left (c x \right )}\, dx\right ) \]
d**4*(Integral(4*a*c, x) + Integral(a/x, x) + Integral(6*a*c**2*x, x) + In tegral(4*a*c**3*x**2, x) + Integral(a*c**4*x**3, x) + Integral(4*b*c*atanh (c*x), x) + Integral(b*atanh(c*x)/x, x) + Integral(6*b*c**2*x*atanh(c*x), x) + Integral(4*b*c**3*x**2*atanh(c*x), x) + Integral(b*c**4*x**3*atanh(c* x), x))
Time = 0.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.49 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=\frac {1}{4} \, a c^{4} d^{4} x^{4} + \frac {4}{3} \, a c^{3} d^{4} x^{3} + \frac {1}{12} \, b c^{3} d^{4} x^{3} + 3 \, a c^{2} d^{4} x^{2} + \frac {2}{3} \, b c^{2} d^{4} x^{2} + 4 \, a c d^{4} x + \frac {13}{4} \, b c d^{4} x + 2 \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4} - \frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b d^{4} + \frac {1}{2} \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b d^{4} - \frac {23}{24} \, b d^{4} \log \left (c x + 1\right ) + \frac {55}{24} \, b d^{4} \log \left (c x - 1\right ) + a d^{4} \log \left (x\right ) + \frac {1}{24} \, {\left (3 \, b c^{4} d^{4} x^{4} + 16 \, b c^{3} d^{4} x^{3} + 36 \, b c^{2} d^{4} x^{2}\right )} \log \left (c x + 1\right ) - \frac {1}{24} \, {\left (3 \, b c^{4} d^{4} x^{4} + 16 \, b c^{3} d^{4} x^{3} + 36 \, b c^{2} d^{4} x^{2}\right )} \log \left (-c x + 1\right ) \]
1/4*a*c^4*d^4*x^4 + 4/3*a*c^3*d^4*x^3 + 1/12*b*c^3*d^4*x^3 + 3*a*c^2*d^4*x ^2 + 2/3*b*c^2*d^4*x^2 + 4*a*c*d^4*x + 13/4*b*c*d^4*x + 2*(2*c*x*arctanh(c *x) + log(-c^2*x^2 + 1))*b*d^4 - 1/2*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*b*d^4 + 1/2*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b*d^4 - 23/24* b*d^4*log(c*x + 1) + 55/24*b*d^4*log(c*x - 1) + a*d^4*log(x) + 1/24*(3*b*c ^4*d^4*x^4 + 16*b*c^3*d^4*x^3 + 36*b*c^2*d^4*x^2)*log(c*x + 1) - 1/24*(3*b *c^4*d^4*x^4 + 16*b*c^3*d^4*x^3 + 36*b*c^2*d^4*x^2)*log(-c*x + 1)
\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=\int { \frac {{\left (c d x + d\right )}^{4} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^4}{x} \,d x \]