Integrand size = 18, antiderivative size = 106 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\frac {b}{2 c^2 d^2 (1+c x)}-\frac {b \text {arctanh}(c x)}{2 c^2 d^2}+\frac {a+b \text {arctanh}(c x)}{c^2 d^2 (1+c x)}-\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^2 d^2}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^2 d^2} \]
1/2*b/c^2/d^2/(c*x+1)-1/2*b*arctanh(c*x)/c^2/d^2+(a+b*arctanh(c*x))/c^2/d^ 2/(c*x+1)-(a+b*arctanh(c*x))*ln(2/(c*x+1))/c^2/d^2+1/2*b*polylog(2,1-2/(c* x+1))/c^2/d^2
Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\frac {\frac {4 a}{1+c x}+4 a \log (1+c x)+b \left (\cosh (2 \text {arctanh}(c x))+2 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+2 \text {arctanh}(c x) \left (\cosh (2 \text {arctanh}(c x))-2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-\sinh (2 \text {arctanh}(c x))\right )-\sinh (2 \text {arctanh}(c x))\right )}{4 c^2 d^2} \]
((4*a)/(1 + c*x) + 4*a*Log[1 + c*x] + b*(Cosh[2*ArcTanh[c*x]] + 2*PolyLog[ 2, -E^(-2*ArcTanh[c*x])] + 2*ArcTanh[c*x]*(Cosh[2*ArcTanh[c*x]] - 2*Log[1 + E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[c*x]]))/(4 *c^2*d^2)
Time = 0.36 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
| \(\displaystyle \int \frac {x (a+b \text {arctanh}(c x))}{(c d x+d)^2} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a+b \text {arctanh}(c x)}{c^2 d^2 (c x+1)}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^2 d^2}-\frac {b \text {arctanh}(c x)}{2 c^2 d^2}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^2 d^2}+\frac {b}{2 c^2 d^2 (c x+1)}\) |
b/(2*c^2*d^2*(1 + c*x)) - (b*ArcTanh[c*x])/(2*c^2*d^2) + (a + b*ArcTanh[c* x])/(c^2*d^2*(1 + c*x)) - ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^2*d^2 ) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^2*d^2)
3.1.53.3.1 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 = 1.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.16
| method | result | size |
| derivativedivides | \(\frac {\frac {a \left (\ln \left (c x +1\right )+\frac {1}{c x +1}\right )}{d^{2}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+\frac {\ln \left (c x -1\right )}{4}+\frac {1}{2 c x +2}-\frac {\ln \left (c x +1\right )}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d^{2}}}{c^{2}}\) | \(123\) |
| default | \(\frac {\frac {a \left (\ln \left (c x +1\right )+\frac {1}{c x +1}\right )}{d^{2}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+\frac {\ln \left (c x -1\right )}{4}+\frac {1}{2 c x +2}-\frac {\ln \left (c x +1\right )}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d^{2}}}{c^{2}}\) | \(123\) |
| parts | \(\frac {a \left (\frac {\ln \left (c x +1\right )}{c^{2}}+\frac {1}{c^{2} \left (c x +1\right )}\right )}{d^{2}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+\frac {\ln \left (c x -1\right )}{4}+\frac {1}{2 c x +2}-\frac {\ln \left (c x +1\right )}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d^{2} c^{2}}\) | \(130\) |
| risch | \(\frac {b \ln \left (c x +1\right )^{2}}{4 c^{2} d^{2}}+\frac {b \ln \left (c x +1\right )}{2 c^{2} d^{2} \left (c x +1\right )}+\frac {b}{2 c^{2} d^{2} \left (c x +1\right )}+\frac {a \ln \left (-c x -1\right )}{c^{2} d^{2}}-\frac {a}{c^{2} d^{2} \left (-c x -1\right )}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 c^{2} d^{2}}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 c^{2} d^{2}}+\frac {b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 c^{2} d^{2}}-\frac {b \ln \left (-c x -1\right )}{4 c^{2} d^{2}}-\frac {b \ln \left (-c x +1\right ) x}{4 c \,d^{2} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right )}{4 c^{2} d^{2} \left (-c x -1\right )}\) | \(216\) |
1/c^2*(a/d^2*(ln(c*x+1)+1/(c*x+1))+b/d^2*(arctanh(c*x)*ln(c*x+1)+1/(c*x+1) *arctanh(c*x)+1/4*ln(c*x-1)+1/2/(c*x+1)-1/4*ln(c*x+1)+1/2*(ln(c*x+1)-ln(1/ 2*c*x+1/2))*ln(-1/2*c*x+1/2)-1/2*dilog(1/2*c*x+1/2)-1/4*ln(c*x+1)^2))
\[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{{\left (c d x + d\right )}^{2}} \,d x } \]
\[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\frac {\int \frac {a x}{c^{2} x^{2} + 2 c x + 1}\, dx + \int \frac {b x \operatorname {atanh}{\left (c x \right )}}{c^{2} x^{2} + 2 c x + 1}\, dx}{d^{2}} \]
(Integral(a*x/(c**2*x**2 + 2*c*x + 1), x) + Integral(b*x*atanh(c*x)/(c**2* x**2 + 2*c*x + 1), x))/d**2
\[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{{\left (c d x + d\right )}^{2}} \,d x } \]
1/8*(8*c^2*integrate(x^2*log(c*x + 1)/(c^4*d^2*x^3 + c^3*d^2*x^2 - c^2*d^2 *x - c*d^2), x) - c*(2/(c^4*d^2*x + c^3*d^2) + log(c*x + 1)/(c^3*d^2) - lo g(c*x - 1)/(c^3*d^2)) + 4*c*integrate(x*log(c*x + 1)/(c^4*d^2*x^3 + c^3*d^ 2*x^2 - c^2*d^2*x - c*d^2), x) - 4*((c*x + 1)*log(c*x + 1) + 1)*log(-c*x + 1)/(c^3*d^2*x + c^2*d^2) + 2/(c^3*d^2*x + c^2*d^2) - log(c*x + 1)/(c^2*d^ 2) + log(c*x - 1)/(c^2*d^2) + 4*integrate(log(c*x + 1)/(c^4*d^2*x^3 + c^3* d^2*x^2 - c^2*d^2*x - c*d^2), x))*b + a*(1/(c^3*d^2*x + c^2*d^2) + log(c*x + 1)/(c^2*d^2))
\[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{{\left (c d x + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\int \frac {x\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\right )}^2} \,d x \]