Integrand size = 23, antiderivative size = 257 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\frac {2 (a+b \text {arctanh}(c+d x))^3 \text {arctanh}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {3 b (a+b \text {arctanh}(c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b (a+b \text {arctanh}(c+d x))^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1-c-d x}\right )}{2 d e}-\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (3,-1+\frac {2}{1-c-d x}\right )}{2 d e}-\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1-c-d x}\right )}{4 d e}+\frac {3 b^3 \operatorname {PolyLog}\left (4,-1+\frac {2}{1-c-d x}\right )}{4 d e} \]
-2*(a+b*arctanh(d*x+c))^3*arctanh(-1+2/(-d*x-c+1))/d/e-3/2*b*(a+b*arctanh( d*x+c))^2*polylog(2,1-2/(-d*x-c+1))/d/e+3/2*b*(a+b*arctanh(d*x+c))^2*polyl og(2,-1+2/(-d*x-c+1))/d/e+3/2*b^2*(a+b*arctanh(d*x+c))*polylog(3,1-2/(-d*x -c+1))/d/e-3/2*b^2*(a+b*arctanh(d*x+c))*polylog(3,-1+2/(-d*x-c+1))/d/e-3/4 *b^3*polylog(4,1-2/(-d*x-c+1))/d/e+3/4*b^3*polylog(4,-1+2/(-d*x-c+1))/d/e
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.26 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\frac {4 a^3 \log (c+d x)+12 a^2 b \text {arctanh}(c+d x) \left (-\log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+\log \left (\frac {i (c+d x)}{\sqrt {1-(c+d x)^2}}\right )\right )-\frac {3}{2} a^2 b \left (\pi ^2-4 i \pi \text {arctanh}(c+d x)-8 \text {arctanh}(c+d x)^2-8 \text {arctanh}(c+d x) \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )+4 i \pi \log \left (1+e^{2 \text {arctanh}(c+d x)}\right )+8 \text {arctanh}(c+d x) \log \left (1+e^{2 \text {arctanh}(c+d x)}\right )-4 i \pi \log \left (\frac {2}{\sqrt {1-(c+d x)^2}}\right )-8 \text {arctanh}(c+d x) \log \left (\frac {2}{\sqrt {1-(c+d x)^2}}\right )+8 \text {arctanh}(c+d x) \log \left (\frac {2 i (c+d x)}{\sqrt {1-(c+d x)^2}}\right )+4 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )+4 \operatorname {PolyLog}\left (2,-e^{2 \text {arctanh}(c+d x)}\right )\right )+6 a b^2 \left (2 \text {arctanh}(c+d x)^2 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )-2 \text {arctanh}(c+d x)^2 \log \left (1+e^{-2 \text {arctanh}(c+d x)}\right )+2 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c+d x)}\right )-2 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )+\operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c+d x)}\right )-\operatorname {PolyLog}\left (3,e^{-2 \text {arctanh}(c+d x)}\right )\right )+b^3 \left (4 \text {arctanh}(c+d x)^3 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )-4 \text {arctanh}(c+d x)^3 \log \left (1+e^{-2 \text {arctanh}(c+d x)}\right )+6 \text {arctanh}(c+d x)^2 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c+d x)}\right )-6 \text {arctanh}(c+d x)^2 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )+6 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c+d x)}\right )-6 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (3,e^{-2 \text {arctanh}(c+d x)}\right )+3 \operatorname {PolyLog}\left (4,-e^{-2 \text {arctanh}(c+d x)}\right )-3 \operatorname {PolyLog}\left (4,e^{-2 \text {arctanh}(c+d x)}\right )\right )}{4 d e} \]
(4*a^3*Log[c + d*x] + 12*a^2*b*ArcTanh[c + d*x]*(-Log[1/Sqrt[1 - (c + d*x) ^2]] + Log[(I*(c + d*x))/Sqrt[1 - (c + d*x)^2]]) - (3*a^2*b*(Pi^2 - (4*I)* Pi*ArcTanh[c + d*x] - 8*ArcTanh[c + d*x]^2 - 8*ArcTanh[c + d*x]*Log[1 - E^ (-2*ArcTanh[c + d*x])] + (4*I)*Pi*Log[1 + E^(2*ArcTanh[c + d*x])] + 8*ArcT anh[c + d*x]*Log[1 + E^(2*ArcTanh[c + d*x])] - (4*I)*Pi*Log[2/Sqrt[1 - (c + d*x)^2]] - 8*ArcTanh[c + d*x]*Log[2/Sqrt[1 - (c + d*x)^2]] + 8*ArcTanh[c + d*x]*Log[((2*I)*(c + d*x))/Sqrt[1 - (c + d*x)^2]] + 4*PolyLog[2, E^(-2* ArcTanh[c + d*x])] + 4*PolyLog[2, -E^(2*ArcTanh[c + d*x])]))/2 + 6*a*b^2*( 2*ArcTanh[c + d*x]^2*Log[1 - E^(-2*ArcTanh[c + d*x])] - 2*ArcTanh[c + d*x] ^2*Log[1 + E^(-2*ArcTanh[c + d*x])] + 2*ArcTanh[c + d*x]*PolyLog[2, -E^(-2 *ArcTanh[c + d*x])] - 2*ArcTanh[c + d*x]*PolyLog[2, E^(-2*ArcTanh[c + d*x] )] + PolyLog[3, -E^(-2*ArcTanh[c + d*x])] - PolyLog[3, E^(-2*ArcTanh[c + d *x])]) + b^3*(4*ArcTanh[c + d*x]^3*Log[1 - E^(-2*ArcTanh[c + d*x])] - 4*Ar cTanh[c + d*x]^3*Log[1 + E^(-2*ArcTanh[c + d*x])] + 6*ArcTanh[c + d*x]^2*P olyLog[2, -E^(-2*ArcTanh[c + d*x])] - 6*ArcTanh[c + d*x]^2*PolyLog[2, E^(- 2*ArcTanh[c + d*x])] + 6*ArcTanh[c + d*x]*PolyLog[3, -E^(-2*ArcTanh[c + d* x])] - 6*ArcTanh[c + d*x]*PolyLog[3, E^(-2*ArcTanh[c + d*x])] + 3*PolyLog[ 4, -E^(-2*ArcTanh[c + d*x])] - 3*PolyLog[4, E^(-2*ArcTanh[c + d*x])]))/(4* d*e)
Time = 1.14 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6657, 27, 6448, 6614, 6620, 6624, 7164}
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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx\) |
\(\Big \downarrow \) 6657 |
\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{e (c+d x)}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{c+d x}d(c+d x)}{d e}\) |
\(\Big \downarrow \) 6448 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^3-6 b \int \frac {(a+b \text {arctanh}(c+d x))^2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right )}{1-(c+d x)^2}d(c+d x)}{d e}\) |
\(\Big \downarrow \) 6614 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^3-6 b \left (\frac {1}{2} \int \frac {(a+b \text {arctanh}(c+d x))^2 \log \left (2-\frac {2}{-c-d x+1}\right )}{1-(c+d x)^2}d(c+d x)-\frac {1}{2} \int \frac {(a+b \text {arctanh}(c+d x))^2 \log \left (\frac {2}{-c-d x+1}\right )}{1-(c+d x)^2}d(c+d x)\right )}{d e}\) |
\(\Big \downarrow \) 6620 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^2-b \int \frac {(a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right )}{1-(c+d x)^2}d(c+d x)\right )+\frac {1}{2} \left (b \int \frac {(a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right )}{1-(c+d x)^2}d(c+d x)-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right ) (a+b \text {arctanh}(c+d x))^2\right )\right )}{d e}\) |
\(\Big \downarrow \) 6624 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^2-b \left (\frac {1}{2} \operatorname {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )}{1-(c+d x)^2}d(c+d x)\right )\right )+\frac {1}{2} \left (b \left (\frac {1}{2} \operatorname {PolyLog}\left (3,\frac {2}{-c-d x+1}-1\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (3,\frac {2}{-c-d x+1}-1\right )}{1-(c+d x)^2}d(c+d x)\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right ) (a+b \text {arctanh}(c+d x))^2\right )\right )}{d e}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {2 \text {arctanh}\left (1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^2-b \left (\frac {1}{2} \operatorname {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{4} b \operatorname {PolyLog}\left (4,1-\frac {2}{-c-d x+1}\right )\right )\right )+\frac {1}{2} \left (b \left (\frac {1}{2} \operatorname {PolyLog}\left (3,\frac {2}{-c-d x+1}-1\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{4} b \operatorname {PolyLog}\left (4,\frac {2}{-c-d x+1}-1\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right ) (a+b \text {arctanh}(c+d x))^2\right )\right )}{d e}\) |
(2*(a + b*ArcTanh[c + d*x])^3*ArcTanh[1 - 2/(1 - c - d*x)] - 6*b*((((a + b *ArcTanh[c + d*x])^2*PolyLog[2, 1 - 2/(1 - c - d*x)])/2 - b*(((a + b*ArcTa nh[c + d*x])*PolyLog[3, 1 - 2/(1 - c - d*x)])/2 - (b*PolyLog[4, 1 - 2/(1 - c - d*x)])/4))/2 + (-1/2*((a + b*ArcTanh[c + d*x])^2*PolyLog[2, -1 + 2/(1 - c - d*x)]) + b*(((a + b*ArcTanh[c + d*x])*PolyLog[3, -1 + 2/(1 - c - d* x)])/2 - (b*PolyLog[4, -1 + 2/(1 - c - d*x)])/4))/2))/(d*e)
3.1.25.3.1 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_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1 - c*x)], x] - Simp[2*b*c*p Int[(a + b *ArcTanh[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 - c*x)]/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]
Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*( x_)^2), x_Symbol] :> Simp[1/2 Int[Log[1 + u]*((a + b*ArcTanh[c*x])^p/(d + e*x^2)), x], x] - Simp[1/2 Int[Log[1 - u]*((a + b*ArcTanh[c*x])^p/(d + e *x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x))^2, 0]
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 2), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(PolyLog[2, 1 - u]/(2*c*d)) , x] + Simp[b*(p/2) Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/( d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2/(1 - c*x))^2, 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_ .)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[k + 1, u]/(2* c*d)), x] - Simp[b*(p/2) Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[k + 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && E qQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x))^2, 0]
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x ], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.48 (sec) , antiderivative size = 1442, normalized size of antiderivative = 5.61
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1442\) |
default | \(\text {Expression too large to display}\) | \(1442\) |
parts | \(\text {Expression too large to display}\) | \(1450\) |
1/d*(a^3/e*ln(d*x+c)+b^3/e*(ln(d*x+c)*arctanh(d*x+c)^3-arctanh(d*x+c)^3*ln ((d*x+c+1)^2/(1-(d*x+c)^2)-1)+arctanh(d*x+c)^3*ln(1-(d*x+c+1)/(1-(d*x+c)^2 )^(1/2))+3*arctanh(d*x+c)^2*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6*arc tanh(d*x+c)*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6*polylog(4,(d*x+c+1) /(1-(d*x+c)^2)^(1/2))+arctanh(d*x+c)^3*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2)) +3*arctanh(d*x+c)^2*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6*arctanh(d* x+c)*polylog(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6*polylog(4,-(d*x+c+1)/(1-( d*x+c)^2)^(1/2))+1/2*I*Pi*csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+ 1)^2/((d*x+c)^2-1)))*(csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1))*csgn(I/(1-(d* x+c+1)^2/((d*x+c)^2-1)))-csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1))*csgn(I*(-( d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))-csgn(I*(-(d*x+c +1)^2/((d*x+c)^2-1)-1)/(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*(-(d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d *x+c)^2-1)))^2)*arctanh(d*x+c)^3-3/2*arctanh(d*x+c)^2*polylog(2,-(d*x+c+1) ^2/(1-(d*x+c)^2))+3/2*arctanh(d*x+c)*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2)) -3/4*polylog(4,-(d*x+c+1)^2/(1-(d*x+c)^2)))+3*a*b^2/e*(ln(d*x+c)*arctanh(d *x+c)^2-arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))+1/2*polylog(3 ,-(d*x+c+1)^2/(1-(d*x+c)^2))-arctanh(d*x+c)^2*ln((d*x+c+1)^2/(1-(d*x+c)^2) -1)+arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+2*arctanh(d*x+c)* polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-2*polylog(3,(d*x+c+1)/(1-(d*x+...
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{d e x + c e} \,d x } \]
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)/(d*e*x + c*e), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\frac {\int \frac {a^{3}}{c + d x}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]
(Integral(a**3/(c + d*x), x) + Integral(b**3*atanh(c + d*x)**3/(c + d*x), x) + Integral(3*a*b**2*atanh(c + d*x)**2/(c + d*x), x) + Integral(3*a**2*b *atanh(c + d*x)/(c + d*x), x))/e
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{d e x + c e} \,d x } \]
a^3*log(d*e*x + c*e)/(d*e) + integrate(1/8*b^3*(log(d*x + c + 1) - log(-d* x - c + 1))^3/(d*e*x + c*e) + 3/4*a*b^2*(log(d*x + c + 1) - log(-d*x - c + 1))^2/(d*e*x + c*e) + 3/2*a^2*b*(log(d*x + c + 1) - log(-d*x - c + 1))/(d *e*x + c*e), x)
\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{d e x + c e} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{c e+d e x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{c\,e+d\,e\,x} \,d x \]