Integrand size = 20, antiderivative size = 191 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=d (a+b \text {arctanh}(c x))^2+c d x (a+b \text {arctanh}(c x))^2+2 d (a+b \text {arctanh}(c x))^2 \text {arctanh}\left (1-\frac {2}{1-c x}\right )-2 b d (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )-b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )-b d (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+b d (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-1+\frac {2}{1-c x}\right ) \]
d*(a+b*arctanh(c*x))^2+c*d*x*(a+b*arctanh(c*x))^2-2*d*(a+b*arctanh(c*x))^2 *arctanh(-1+2/(-c*x+1))-2*b*d*(a+b*arctanh(c*x))*ln(2/(-c*x+1))-b^2*d*poly log(2,1-2/(-c*x+1))-b*d*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))+b*d*(a+ b*arctanh(c*x))*polylog(2,-1+2/(-c*x+1))+1/2*b^2*d*polylog(3,1-2/(-c*x+1)) -1/2*b^2*d*polylog(3,-1+2/(-c*x+1))
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.19 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=d \left (a^2 c x+a^2 \log (c x)+a b \left (2 c x \text {arctanh}(c x)+\log \left (1-c^2 x^2\right )\right )+b^2 \left (\text {arctanh}(c x) \left ((-1+c x) \text {arctanh}(c x)-2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )\right )+a b (-\operatorname {PolyLog}(2,-c x)+\operatorname {PolyLog}(2,c x))+b^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} \text {arctanh}(c x)^3-\text {arctanh}(c x)^2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+\text {arctanh}(c x)^2 \log \left (1-e^{2 \text {arctanh}(c x)}\right )+\text {arctanh}(c x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+\text {arctanh}(c x) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}(c x)}\right )\right )\right ) \]
d*(a^2*c*x + a^2*Log[c*x] + a*b*(2*c*x*ArcTanh[c*x] + Log[1 - c^2*x^2]) + b^2*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c*x] - 2*Log[1 + E^(-2*ArcTanh[c*x]) ]) + PolyLog[2, -E^(-2*ArcTanh[c*x])]) + a*b*(-PolyLog[2, -(c*x)] + PolyLo g[2, c*x]) + b^2*((I/24)*Pi^3 - (2*ArcTanh[c*x]^3)/3 - ArcTanh[c*x]^2*Log[ 1 + E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + Ar cTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]*PolyLog[2, E^(2 *ArcTanh[c*x])] + PolyLog[3, -E^(-2*ArcTanh[c*x])]/2 - PolyLog[3, E^(2*Arc Tanh[c*x])]/2))
Time = 0.67 (sec) , antiderivative size = 191, 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.
| \(\displaystyle \int \frac {(c d x+d) (a+b \text {arctanh}(c x))^2}{x} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (c d (a+b \text {arctanh}(c x))^2+\frac {d (a+b \text {arctanh}(c x))^2}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -b d \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))+b d \operatorname {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) (a+b \text {arctanh}(c x))+d (a+b \text {arctanh}(c x))^2+c d x (a+b \text {arctanh}(c x))^2+2 d \text {arctanh}\left (1-\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))^2-2 b d \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))+b^2 (-d) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,\frac {2}{1-c x}-1\right )\) |
d*(a + b*ArcTanh[c*x])^2 + c*d*x*(a + b*ArcTanh[c*x])^2 + 2*d*(a + b*ArcTa nh[c*x])^2*ArcTanh[1 - 2/(1 - c*x)] - 2*b*d*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)] - b^2*d*PolyLog[2, 1 - 2/(1 - c*x)] - b*d*(a + b*ArcTanh[c*x])*Pol yLog[2, 1 - 2/(1 - c*x)] + b*d*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)] + (b^2*d*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (b^2*d*PolyLog[3, -1 + 2/ (1 - c*x)])/2
3.1.72.3.1 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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 9.73 (sec) , antiderivative size = 1835, normalized size of antiderivative = 9.61
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1835\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1837\) |
| default | \(\text {Expression too large to display}\) | \(1837\) |
a^2*d*(c*x+ln(x))+b^2*d*(c*x*arctanh(c*x)^2-arctanh(c*x)*ln(1+(c*x+1)^2/(- c^2*x^2+1))-dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-dilog(1-I*(c*x+1)/(-c^2* x^2+1)^(1/2))+arctanh(c*x)^2-1/2*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-arctan h(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-arctanh(c*x)*ln(1-I*(c*x+1)/(-c^ 2*x^2+1)^(1/2))+1/2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-2*polylog(3,-(c*x+1 )/(-c^2*x^2+1)^(1/2))-2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))+ln(c*x)*arct anh(c*x)^2-arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-arctanh(c*x)^2* ln((c*x+1)^2/(-c^2*x^2+1)-1)+arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2 ))+2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c*x)^2*ln (1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+ 1)^(1/2))+1/4*I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(-(c*x+1)^2/(c ^2*x^2-1)-1))*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1))) *(2*arctanh(c*x)^2-2*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))-polylog(2,- (c*x+1)^2/(-c^2*x^2+1)))+1/2*I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I *(-(c*x+1)^2/(c^2*x^2-1)-1))*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^ 2/(c^2*x^2-1)))*(arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c *x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2 ))+dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2)))+1/4*I*Pi*csgn(I*(-(c*x+1)^2/(c^2 *x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3*(2*arctanh(c*x)^2-2*arctanh(c*x)*l n(1+(c*x+1)^2/(-c^2*x^2+1))-polylog(2,-(c*x+1)^2/(-c^2*x^2+1)))+1/2*I*P...
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
integral((a^2*c*d*x + a^2*d + (b^2*c*d*x + b^2*d)*arctanh(c*x)^2 + 2*(a*b* c*d*x + a*b*d)*arctanh(c*x))/x, x)
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=d \left (\int a^{2} c\, dx + \int \frac {a^{2}}{x}\, dx + \int b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x}\, dx + \int 2 a b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]
d*(Integral(a**2*c, x) + Integral(a**2/x, x) + Integral(b**2*c*atanh(c*x)* *2, x) + Integral(b**2*atanh(c*x)**2/x, x) + Integral(2*a*b*c*atanh(c*x), x) + Integral(2*a*b*atanh(c*x)/x, x))
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
1/4*b^2*c*d*x*log(-c*x + 1)^2 + a^2*c*d*x + (2*c*x*arctanh(c*x) + log(-c^2 *x^2 + 1))*a*b*d + a^2*d*log(x) - integrate(-1/4*((b^2*c^2*d*x^2 - b^2*d)* log(c*x + 1)^2 + 4*(a*b*c*d*x - a*b*d)*log(c*x + 1) - 2*(b^2*c^2*d*x^2 + 2 *a*b*c*d*x - 2*a*b*d + (b^2*c^2*d*x^2 - b^2*d)*log(c*x + 1))*log(-c*x + 1) )/(c*x^2 - x), x)
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
Timed out. \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\right )}{x} \,d x \]