Integrand size = 18, antiderivative size = 188 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=-\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{e}+\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{e}-\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e} \]
-(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/e+(a+b*arctanh(c*x))^2*ln(2*c*(e*x+d)/ (c*d+e)/(c*x+1))/e+b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/e-b*(a+b*ar ctanh(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/e+1/2*b^2*polylog(3,1 -2/(c*x+1))/e-1/2*b^2*polylog(3,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/e
Result contains complex when optimal does not.
Time = 10.32 (sec) , antiderivative size = 1055, normalized size of antiderivative = 5.61 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx =\text {Too large to display} \]
(6*a^2*Log[d + e*x] + 6*a*b*ArcTanh[c*x]*(Log[1 - c^2*x^2] + 2*Log[I*Sinh[ ArcTanh[(c*d)/e] + ArcTanh[c*x]]]) - (6*I)*a*b*((-1/4*I)*(Pi - (2*I)*ArcTa nh[c*x])^2 + I*(ArcTanh[(c*d)/e] + ArcTanh[c*x])^2 + (Pi - (2*I)*ArcTanh[c *x])*Log[1 + E^(2*ArcTanh[c*x])] + (2*I)*(ArcTanh[(c*d)/e] + ArcTanh[c*x]) *Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - (Pi - (2*I)*ArcTanh[c *x])*Log[2/Sqrt[1 - c^2*x^2]] - (2*I)*(ArcTanh[(c*d)/e] + ArcTanh[c*x])*Lo g[(2*I)*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - I*PolyLog[2, -E^(2*ArcTan h[c*x])] - I*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]) + (b^2* (-8*c*d*ArcTanh[c*x]^3 + 4*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*c*d*ArcTanh[c*x]^2*Log[1 + E^(-2*A rcTanh[c*x])] - (6*I)*c*d*Pi*ArcTanh[c*x]*Log[(E^(-ArcTanh[c*x]) + E^ArcTa nh[c*x])/2] - 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 + (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] + Arc Tanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*(ArcTanh[(c*d)/e] + ArcTan h[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*ArcTan h[c*x])))/(2*E^ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/...
Time = 0.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6474}
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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx\) |
\(\Big \downarrow \) 6474 |
\(\displaystyle -\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{e}+\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{e}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 e}\) |
-(((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/e) + ((a + b*ArcTanh[c*x])^2*L og[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e + (b*(a + b*ArcTanh[c*x])*Pol yLog[2, 1 - 2/(1 + c*x)])/e - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c* (d + e*x))/((c*d + e)*(1 + c*x))])/e + (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/( 2*e) - (b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e)
3.1.12.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^2)*(Log[2/(1 + c*x)]/e), x] + (Simp[(a + b*Arc Tanh[c*x])^2*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[b*(a + b*ArcTanh[c*x])*(PolyLog[2, 1 - 2/(1 + c*x)]/e), x] - Simp[b*(a + b*ArcT anh[c*x])*(PolyLog[2, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + S imp[b^2*(PolyLog[3, 1 - 2/(1 + c*x)]/(2*e)), x] - Simp[b^2*(PolyLog[3, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/(2*e)), x]) /; FreeQ[{a, b, c, d, e} , x] && NeQ[c^2*d^2 - e^2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 35.05 (sec) , antiderivative size = 1087, normalized size of antiderivative = 5.78
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1087\) |
default | \(\text {Expression too large to display}\) | \(1087\) |
parts | \(\text {Expression too large to display}\) | \(1090\) |
1/c*(a^2*c*ln(c*e*x+c*d)/e+b^2*c*(ln(c*e*x+c*d)/e*arctanh(c*x)^2-2/e*(1/2* arctanh(c*x)^2*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*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)))*(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)))-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)))*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)))*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)))+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+1/2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2 +1))-1/4*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/2*e/(c*d+e)*arctanh(c*x)^2*l n(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*e/(c*d+e)*arctanh(c*x)*po lylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+1/4*e/(c*d+e)*polylog(3,( c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*d*c/(c*d+e)*arctanh(c*x)^2*ln( 1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*d*c/(c*d+e)*arctanh(c*x)*po lylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+1/4*d*c/(c*d+e)*polylog(3 ,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))))+2*a*b*c*(ln(c*e*x+c*d)/e*arcta nh(c*x)-1/e^2*(1/2*e*(dilog((c*e*x+e)/(-c*d+e))+ln(c*e*x+c*d)*ln((c*e*x+e) /(-c*d+e)))-1/2*e*(dilog((c*e*x-e)/(-c*d-e))+ln(c*e*x+c*d)*ln((c*e*x-e)...
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{e x + d} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{e x + d} \,d x } \]
a^2*log(e*x + d)/e + integrate(1/4*b^2*(log(c*x + 1) - log(-c*x + 1))^2/(e *x + d) + a*b*(log(c*x + 1) - log(-c*x + 1))/(e*x + d), x)
\[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^2}{d+e x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]