Integrand size = 17, antiderivative size = 275 \[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\frac {2 \text {arctanh}(c x)^2 \text {arctanh}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\text {arctanh}(c x)^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {\text {arctanh}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{d}+\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-c x}\right )}{d}-\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{d}+\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}+\frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d}-\frac {\operatorname {PolyLog}\left (3,-1+\frac {2}{1-c x}\right )}{2 d}-\frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 d}+\frac {\operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d} \]
-2*arctanh(c*x)^2*arctanh(-1+2/(-c*x+1))/d+arctanh(c*x)^2*ln(2/(c*x+1))/d- arctanh(c*x)^2*ln(2*c*(e*x+d)/(c*d+e)/(c*x+1))/d-arctanh(c*x)*polylog(2,1- 2/(-c*x+1))/d+arctanh(c*x)*polylog(2,-1+2/(-c*x+1))/d-arctanh(c*x)*polylog (2,1-2/(c*x+1))/d+arctanh(c*x)*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/d+ 1/2*polylog(3,1-2/(-c*x+1))/d-1/2*polylog(3,-1+2/(-c*x+1))/d-1/2*polylog(3 ,1-2/(c*x+1))/d+1/2*polylog(3,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/d
Result contains complex when optimal does not.
Time = 7.16 (sec) , antiderivative size = 850, normalized size of antiderivative = 3.09 \[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx =\text {Too large to display} \]
(I*c*d*Pi^3 - 8*c*d*ArcTanh[c*x]^3 - 8*e*ArcTanh[c*x]^3 + 24*c*d*ArcTanh[c *x]^2*Log[1 - E^(2*ArcTanh[c*x])] + 24*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*Ar cTanh[c*x])] - 12*c*d*PolyLog[3, E^(2*ArcTanh[c*x])] - (24*(c*d - e)*(c*d + e)*(-6*c*d*ArcTanh[c*x]^3 + 2*e*ArcTanh[c*x]^3 - (4*Sqrt[1 - (c^2*d^2)/e ^2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - (6*I)*c*d*Pi*ArcTanh[c*x]*Log[( E^(-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] - 6*c*d*ArcTanh[c*x]^2*Log[1 - (Sqr t[c*d + e]*E^ArcTanh[c*x])/Sqrt[-(c*d) + e]] - 6*c*d*ArcTanh[c*x]^2*Log[1 + (Sqrt[c*d + e]*E^ArcTanh[c*x])/Sqrt[-(c*d) + e]] + 6*c*d*ArcTanh[c*x]^2* Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2 *(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x ]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTanh[(c*d) /e] + ArcTanh[c*x])))] + 6*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c* x])) + c*d*(1 + E^(2*ArcTanh[c*x])))/(2*E^ArcTanh[c*x])] - 6*c*d*ArcTanh[c *x]^2*Log[(c*(d + e*x))/Sqrt[1 - c^2*x^2]] - (3*I)*c*d*Pi*ArcTanh[c*x]*Log [1 - c^2*x^2] - 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(c *d)/e] + ArcTanh[c*x]]] - 12*c*d*ArcTanh[c*x]*PolyLog[2, -((Sqrt[c*d + e]* E^ArcTanh[c*x])/Sqrt[-(c*d) + e])] - 12*c*d*ArcTanh[c*x]*PolyLog[2, (Sqrt[ c*d + e]*E^ArcTanh[c*x])/Sqrt[-(c*d) + e]] + 12*c*d*ArcTanh[c*x]*PolyLog[2 , -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 12*c*d*ArcTanh[c*x]*PolyLog[2...
Time = 0.63 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.
\(\displaystyle \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 6502 |
\(\displaystyle \int \left (\frac {\text {arctanh}(c x)^2}{d x}-\frac {e \text {arctanh}(c x)^2}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{d}-\frac {\text {arctanh}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}-\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{d}+\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,\frac {2}{1-c x}-1\right )}{d}-\frac {\text {arctanh}(c x) \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{d}+\frac {2 \text {arctanh}\left (1-\frac {2}{1-c x}\right ) \text {arctanh}(c x)^2}{d}+\frac {\text {arctanh}(c x)^2 \log \left (\frac {2}{c x+1}\right )}{d}+\frac {\operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d}+\frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d}-\frac {\operatorname {PolyLog}\left (3,\frac {2}{1-c x}-1\right )}{2 d}-\frac {\operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 d}\) |
(2*ArcTanh[c*x]^2*ArcTanh[1 - 2/(1 - c*x)])/d + (ArcTanh[c*x]^2*Log[2/(1 + c*x)])/d - (ArcTanh[c*x]^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d - (ArcTanh[c*x]*PolyLog[2, 1 - 2/(1 - c*x)])/d + (ArcTanh[c*x]*PolyLog[2, -1 + 2/(1 - c*x)])/d - (ArcTanh[c*x]*PolyLog[2, 1 - 2/(1 + c*x)])/d + (Arc Tanh[c*x]*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d + PolyL og[3, 1 - 2/(1 - c*x)]/(2*d) - PolyLog[3, -1 + 2/(1 - c*x)]/(2*d) - PolyLo g[3, 1 - 2/(1 + c*x)]/(2*d) + PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))]/(2*d)
3.2.59.3.1 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 = 11.37 (sec) , antiderivative size = 1566, normalized size of antiderivative = 5.69
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1566\) |
default | \(\text {Expression too large to display}\) | \(1566\) |
parts | \(\text {Expression too large to display}\) | \(2521\) |
1/d*ln(c*x)*arctanh(c*x)^2-arctanh(c*x)^2/d*ln(c*e*x+c*d)-2*c*(-1/2*arctan h(c*x)^2/d/c*ln(d*c*(1+(c*x+1)^2/(-c^2*x^2+1))+e*((c*x+1)^2/(-c^2*x^2+1)-1 ))-1/4*I/d/c*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(d*c*(1-(c*x+1)^2 /(c^2*x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1))/(1-(c*x+1)^2/(c^2*x^2-1)))^2*a rctanh(c*x)^2-1/4*I/d/c*Pi*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/ (c^2*x^2-1)))^3*arctanh(c*x)^2+1/4*I/d/c*Pi*csgn(I*(d*c*(1-(c*x+1)^2/(c^2* x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1)))*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*c sgn(I*(d*c*(1-(c*x+1)^2/(c^2*x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1)-1))/(1-(c*x +1)^2/(c^2*x^2-1)))*arctanh(c*x)^2+1/4*I/d/c*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)/(1-(c*x+1)^2/(c^2*x^2-1)))^2*arc tanh(c*x)^2-1/4*I/d/c*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)^2+1/4*I/d/c*Pi*csgn(I*(d*c*(1-(c*x+1)^2/(c^2*x^2-1))+e *(-(c*x+1)^2/(c^2*x^2-1)-1))/(1-(c*x+1)^2/(c^2*x^2-1)))^3*arctanh(c*x)^2-1 /4*I/d/c*Pi*csgn(I*(d*c*(1-(c*x+1)^2/(c^2*x^2-1))+e*(-(c*x+1)^2/(c^2*x^2-1 )-1)))*csgn(I*(d*c*(1-(c*x+1)^2/(c^2*x^2-1))+e*(-(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+1/4*I/d/c*Pi*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+1/2/d/c*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1) -1/2/d/c*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-1/d/c*arctanh(...
\[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\int { \frac {\operatorname {artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x} \,d x } \]
\[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\int \frac {\operatorname {atanh}^{2}{\left (c x \right )}}{x \left (d + e x\right )}\, dx \]
\[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\int { \frac {\operatorname {artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x} \,d x } \]
\[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\int { \frac {\operatorname {artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}(c x)^2}{x (d+e x)} \, dx=\int \frac {{\mathrm {atanh}\left (c\,x\right )}^2}{x\,\left (d+e\,x\right )} \,d x \]