Integrand size = 20, antiderivative size = 308 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=-\frac {(a+b \text {arctanh}(c+d x))^3 \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {(a+b \text {arctanh}(c+d x))^3 \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {3 b (a+b \text {arctanh}(c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b (a+b \text {arctanh}(c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+c+d x}\right )}{4 f}-\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{4 f} \]
-(a+b*arctanh(d*x+c))^3*ln(2/(d*x+c+1))/f+(a+b*arctanh(d*x+c))^3*ln(2*d*(f *x+e)/(-c*f+d*e+f)/(d*x+c+1))/f+3/2*b*(a+b*arctanh(d*x+c))^2*polylog(2,1-2 /(d*x+c+1))/f-3/2*b*(a+b*arctanh(d*x+c))^2*polylog(2,1-2*d*(f*x+e)/(-c*f+d *e+f)/(d*x+c+1))/f+3/2*b^2*(a+b*arctanh(d*x+c))*polylog(3,1-2/(d*x+c+1))/f -3/2*b^2*(a+b*arctanh(d*x+c))*polylog(3,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+ 1))/f+3/4*b^3*polylog(4,1-2/(d*x+c+1))/f-3/4*b^3*polylog(4,1-2*d*(f*x+e)/( -c*f+d*e+f)/(d*x+c+1))/f
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx \]
Time = 0.47 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6661, 27, 6476}
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.
| \(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx\) |
| \(\Big \downarrow \) 6661 |
| \(\displaystyle \frac {\int \frac {d (a+b \text {arctanh}(c+d x))^3}{d \left (e-\frac {c f}{d}\right )+f (c+d x)}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^3}{f (c+d x)-c f+d e}d(c+d x)\) |
| \(\Big \downarrow \) 6476 |
| \(\displaystyle -\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))}{2 f}-\frac {3 b (a+b \text {arctanh}(c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {(a+b \text {arctanh}(c+d x))^3 \log \left (\frac {2 (f (c+d x)-c f+d e)}{(c+d x+1) (-c f+d e+f)}\right )}{f}+\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^2}{2 f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))^3}{f}-\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{4 f}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{c+d x+1}\right )}{4 f}\) |
-(((a + b*ArcTanh[c + d*x])^3*Log[2/(1 + c + d*x)])/f) + ((a + b*ArcTanh[c + d*x])^3*Log[(2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x ))])/f + (3*b*(a + b*ArcTanh[c + d*x])^2*PolyLog[2, 1 - 2/(1 + c + d*x)])/ (2*f) - (3*b*(a + b*ArcTanh[c + d*x])^2*PolyLog[2, 1 - (2*(d*e - c*f + f*( c + d*x)))/((d*e + f - c*f)*(1 + c + d*x))])/(2*f) + (3*b^2*(a + b*ArcTanh [c + d*x])*PolyLog[3, 1 - 2/(1 + c + d*x)])/(2*f) - (3*b^2*(a + b*ArcTanh[ c + d*x])*PolyLog[3, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x))])/(2*f) + (3*b^3*PolyLog[4, 1 - 2/(1 + c + d*x)])/(4*f) - (3* b^3*PolyLog[4, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x))])/(4*f)
3.1.48.3.1 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_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^3)*(Log[2/(1 + c*x)]/e), x] + (Simp[(a + b*Arc Tanh[c*x])^3*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[3*b* (a + b*ArcTanh[c*x])^2*(PolyLog[2, 1 - 2/(1 + c*x)]/(2*e)), x] - Simp[3*b*( a + b*ArcTanh[c*x])^2*(PolyLog[2, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x))) ]/(2*e)), x] + Simp[3*b^2*(a + b*ArcTanh[c*x])*(PolyLog[3, 1 - 2/(1 + c*x)] /(2*e)), x] - Simp[3*b^2*(a + b*ArcTanh[c*x])*(PolyLog[3, 1 - 2*c*((d + e*x )/((c*d + e)*(1 + c*x)))]/(2*e)), x] + Simp[3*b^3*(PolyLog[4, 1 - 2/(1 + c* x)]/(4*e)), x] - Simp[3*b^3*(PolyLog[4, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/(4*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 = 1.10 (sec) , antiderivative size = 3441, normalized size of antiderivative = 11.17
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(3441\) |
| default | \(\text {Expression too large to display}\) | \(3441\) |
| parts | \(\text {Expression too large to display}\) | \(3656\) |
1/d*(a^3*d*ln(c*f-d*e-f*(d*x+c))/f-b^3*d*(-ln(c*f-d*e-f*(d*x+c))/f*arctanh (d*x+c)^3+3/f*(1/3*arctanh(d*x+c)^3*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)-1/6*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)^3+1/2*arct anh(d*x+c)^2*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-1/2*arctanh(d*x+c)*poly log(3,-(d*x+c+1)^2/(1-(d*x+c)^2))+1/4*polylog(4,-(d*x+c+1)^2/(1-(d*x+c)^2) )-1/3*f*c/(c*f-d*e-f)*arctanh(d*x+c)^3*ln(1-(c*f-d*e-f)*(d*x+c+1)^2/(1-(d* x+c)^2)/(-c*f+d*e-f))-1/2*f*c/(c*f-d*e-f)*arctanh(d*x+c)^2*polylog(2,(c*f- d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+1/2*f*c/(c*f-d*e-f)*arct...
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \]
integral((b^3*arctanh(d*x + c)^3 + 3*a*b^2*arctanh(d*x + c)^2 + 3*a^2*b*ar ctanh(d*x + c) + a^3)/(f*x + e), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3}}{e + f x}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \]
a^3*log(f*x + e)/f + integrate(1/8*b^3*(log(d*x + c + 1) - log(-d*x - c + 1))^3/(f*x + e) + 3/4*a*b^2*(log(d*x + c + 1) - log(-d*x - c + 1))^2/(f*x + e) + 3/2*a^2*b*(log(d*x + c + 1) - log(-d*x - c + 1))/(f*x + e), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{e+f x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{e+f\,x} \,d x \]