Integrand size = 20, antiderivative size = 214 \[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=-\frac {(a+b \text {arctanh}(c+d x))^2 \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {(a+b \text {arctanh}(c+d x))^2 \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {b (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{f}-\frac {b (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f} \]
-(a+b*arctanh(d*x+c))^2*ln(2/(d*x+c+1))/f+(a+b*arctanh(d*x+c))^2*ln(2*d*(f *x+e)/(-c*f+d*e+f)/(d*x+c+1))/f+b*(a+b*arctanh(d*x+c))*polylog(2,1-2/(d*x+ c+1))/f-b*(a+b*arctanh(d*x+c))*polylog(2,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c +1))/f+1/2*b^2*polylog(3,1-2/(d*x+c+1))/f-1/2*b^2*polylog(3,1-2*d*(f*x+e)/ (-c*f+d*e+f)/(d*x+c+1))/f
Result contains complex when optimal does not.
Time = 102.08 (sec) , antiderivative size = 3806, normalized size of antiderivative = 17.79 \[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\text {Result too large to show} \]
(a^2*Log[e + f*x])/f - ((2*I)*a*b*(I*ArcTanh[c + d*x]*(-Log[1/Sqrt[1 - (c + d*x)^2]] + Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]]) + ((- I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x])^2 - (I/4)*(Pi - (2*I)*A rcTanh[c + d*x])^2 + 2*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x])*Log [1 - E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))] + (Pi - (2 *I)*ArcTanh[c + d*x])*Log[1 - E^(I*(Pi - (2*I)*ArcTanh[c + d*x]))] - (Pi - (2*I)*ArcTanh[c + d*x])*Log[2*Sin[(Pi - (2*I)*ArcTanh[c + d*x])/2]] - 2*( I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x])*Log[(2*I)*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]] - I*PolyLog[2, E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))] - I*PolyLog[2, E^(I*(Pi - (2*I)*ArcTanh[c + d*x]))])/2))/f + (b^2*(d*e - c*f + f*(c + d*x))*((Sqrt[1 - (c + d*x)^2]* ((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x)) /Sqrt[1 - (c + d*x)^2])*ArcTanh[c + d*x]^3)/(3*(d*e - c*f)*(d*e - c*f + f* (c + d*x))) - (Sqrt[1 - (c + d*x)^2]*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/ Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])*(ArcTanh[c + d*x]^3/3 + ArcTanh[c + d*x]^2*Log[1 + E^(-2*ArcTanh[c + d*x])] - ArcTanh[c + d*x]*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] - PolyLog[3, -E^(-2*ArcTanh[c + d*x])]/2))/(f*(d*e - c*f + f*(c + d*x))) - (-6*d*e*ArcTanh[c + d*x]^3 + 2*f*ArcTanh[c + d*x]^3 + 6*c*f*ArcTanh[c + d*x]^3 - 4*E^ArcTanh[c - (d*e) /f]*Sqrt[1 - c^2 - (d^2*e^2)/f^2 + (2*c*d*e)/f]*f*ArcTanh[c + d*x]^3 - ...
Time = 0.42 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6661, 27, 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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx\) |
\(\Big \downarrow \) 6661 |
\(\displaystyle \frac {\int \frac {d (a+b \text {arctanh}(c+d x))^2}{d \left (e-\frac {c f}{d}\right )+f (c+d x)}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^2}{f (c+d x)-c f+d e}d(c+d x)\) |
\(\Big \downarrow \) 6474 |
\(\displaystyle -\frac {b (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{f}+\frac {(a+b \text {arctanh}(c+d x))^2 \log \left (\frac {2 (f (c+d x)-c f+d e)}{(c+d x+1) (-c f+d e+f)}\right )}{f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^2}{f}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c+d x+1}\right )}{2 f}\) |
-(((a + b*ArcTanh[c + d*x])^2*Log[2/(1 + c + d*x)])/f) + ((a + b*ArcTanh[c + d*x])^2*Log[(2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x ))])/f + (b*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - 2/(1 + c + d*x)])/f - (b*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - (2*(d*e - c*f + f*(c + d*x)))/( (d*e + f - c*f)*(1 + c + d*x))])/f + (b^2*PolyLog[3, 1 - 2/(1 + c + d*x)]) /(2*f) - (b^2*PolyLog[3, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f )*(1 + c + d*x))])/(2*f)
3.1.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IG tQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.36 (sec) , antiderivative size = 1700, normalized size of antiderivative = 7.94
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1700\) |
default | \(\text {Expression too large to display}\) | \(1700\) |
parts | \(\text {Expression too large to display}\) | \(1800\) |
1/d*(a^2*d*ln(c*f-d*e-f*(d*x+c))/f-b^2*d*(-ln(c*f-d*e-f*(d*x+c))/f*arctanh (d*x+c)^2+2/f*(1/2*arctanh(d*x+c)^2*ln(f*c*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+( -(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(-(d*x+c+1)^2/(1-(d*x+c)^2)+1)*f)-1/4*I* Pi*csgn(I*(f*c*(1-(d*x+c+1)^2/((d*x+c)^2-1))+((d*x+c+1)^2/((d*x+c)^2-1)-1) *e*d+((d*x+c+1)^2/((d*x+c)^2-1)+1)*f)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*(csgn (I*(f*c*(1-(d*x+c+1)^2/((d*x+c)^2-1))+((d*x+c+1)^2/((d*x+c)^2-1)-1)*e*d+(( d*x+c+1)^2/((d*x+c)^2-1)+1)*f))*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))-csgn (I*(f*c*(1-(d*x+c+1)^2/((d*x+c)^2-1))+((d*x+c+1)^2/((d*x+c)^2-1)-1)*e*d+(( d*x+c+1)^2/((d*x+c)^2-1)+1)*f)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I/(1-(d *x+c+1)^2/((d*x+c)^2-1)))-csgn(I*(f*c*(1-(d*x+c+1)^2/((d*x+c)^2-1))+((d*x+ c+1)^2/((d*x+c)^2-1)-1)*e*d+((d*x+c+1)^2/((d*x+c)^2-1)+1)*f))*csgn(I*(f*c* (1-(d*x+c+1)^2/((d*x+c)^2-1))+((d*x+c+1)^2/((d*x+c)^2-1)-1)*e*d+((d*x+c+1) ^2/((d*x+c)^2-1)+1)*f)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))+csgn(I*(f*c*(1-(d*x+ c+1)^2/((d*x+c)^2-1))+((d*x+c+1)^2/((d*x+c)^2-1)-1)*e*d+((d*x+c+1)^2/((d*x +c)^2-1)+1)*f)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2)*arctanh(d*x+c)^2+1/2*arct anh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-1/4*polylog(3,-(d*x+c+1)^ 2/(1-(d*x+c)^2))-1/2/(c*f-d*e-f)*f*c*arctanh(d*x+c)^2*ln(1-(c*f-d*e-f)*(d* x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-1/2/(c*f-d*e-f)*f*c*arctanh(d*x+c)*po lylog(2,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+1/4/(c*f-d*e-f )*f*c*polylog(3,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+1/2...
\[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{f x + e} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{2}}{e + f x}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{f x + e} \,d x } \]
a^2*log(f*x + e)/f + integrate(1/4*b^2*(log(d*x + c + 1) - log(-d*x - c + 1))^2/(f*x + e) + a*b*(log(d*x + c + 1) - log(-d*x - c + 1))/(f*x + e), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{f x + e} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c+d x))^2}{e+f x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2}{e+f\,x} \,d x \]