Integrand size = 22, antiderivative size = 337 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {b^2}{16 c^4 d^3 (1+c x)^2}-\frac {21 b^2}{16 c^4 d^3 (1+c x)}+\frac {21 b^2 \text {arctanh}(c x)}{16 c^4 d^3}+\frac {b (a+b \text {arctanh}(c x))}{4 c^4 d^3 (1+c x)^2}-\frac {11 b (a+b \text {arctanh}(c x))}{4 c^4 d^3 (1+c x)}+\frac {19 (a+b \text {arctanh}(c x))^2}{8 c^4 d^3}+\frac {x (a+b \text {arctanh}(c x))^2}{c^3 d^3}+\frac {(a+b \text {arctanh}(c x))^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 (a+b \text {arctanh}(c x))^2}{c^4 d^3 (1+c x)}-\frac {2 b (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 (a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^4 d^3} \]
1/16*b^2/c^4/d^3/(c*x+1)^2-21/16*b^2/c^4/d^3/(c*x+1)+21/16*b^2*arctanh(c*x )/c^4/d^3+1/4*b*(a+b*arctanh(c*x))/c^4/d^3/(c*x+1)^2-11/4*b*(a+b*arctanh(c *x))/c^4/d^3/(c*x+1)+19/8*(a+b*arctanh(c*x))^2/c^4/d^3+x*(a+b*arctanh(c*x) )^2/c^3/d^3+1/2*(a+b*arctanh(c*x))^2/c^4/d^3/(c*x+1)^2-3*(a+b*arctanh(c*x) )^2/c^4/d^3/(c*x+1)-2*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^4/d^3+3*(a+b*a rctanh(c*x))^2*ln(2/(c*x+1))/c^4/d^3-b^2*polylog(2,1-2/(-c*x+1))/c^4/d^3-3 *b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^4/d^3-3/2*b^2*polylog(3,1-2 /(c*x+1))/c^4/d^3
Time = 0.99 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.24 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {64 a^2 c x+\frac {32 a^2}{(1+c x)^2}-\frac {192 a^2}{1+c x}-192 a^2 \log (1+c x)+4 a b \left (-20 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+16 \log \left (1-c^2 x^2\right )-48 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+20 \sinh (2 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (8 c x-10 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+24 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+10 \sinh (2 \text {arctanh}(c x))-\sinh (4 \text {arctanh}(c x))\right )-\sinh (4 \text {arctanh}(c x))\right )+b^2 \left (-64 \text {arctanh}(c x)^2+64 c x \text {arctanh}(c x)^2-40 \cosh (2 \text {arctanh}(c x))-80 \text {arctanh}(c x) \cosh (2 \text {arctanh}(c x))-80 \text {arctanh}(c x)^2 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+4 \text {arctanh}(c x) \cosh (4 \text {arctanh}(c x))+8 \text {arctanh}(c x)^2 \cosh (4 \text {arctanh}(c x))-128 \text {arctanh}(c x) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+192 \text {arctanh}(c x)^2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-64 (-1+3 \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )-96 \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )+40 \sinh (2 \text {arctanh}(c x))+80 \text {arctanh}(c x) \sinh (2 \text {arctanh}(c x))+80 \text {arctanh}(c x)^2 \sinh (2 \text {arctanh}(c x))-\sinh (4 \text {arctanh}(c x))-4 \text {arctanh}(c x) \sinh (4 \text {arctanh}(c x))-8 \text {arctanh}(c x)^2 \sinh (4 \text {arctanh}(c x))\right )}{64 c^4 d^3} \]
(64*a^2*c*x + (32*a^2)/(1 + c*x)^2 - (192*a^2)/(1 + c*x) - 192*a^2*Log[1 + c*x] + 4*a*b*(-20*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 16*Log[1 - c^2*x^2] - 48*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 20*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(8*c*x - 10*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 24*Log[1 + E^(-2*ArcTanh[c*x])] + 10*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTa nh[c*x]]) - Sinh[4*ArcTanh[c*x]]) + b^2*(-64*ArcTanh[c*x]^2 + 64*c*x*ArcTa nh[c*x]^2 - 40*Cosh[2*ArcTanh[c*x]] - 80*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] - 80*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 4*ArcTa nh[c*x]*Cosh[4*ArcTanh[c*x]] + 8*ArcTanh[c*x]^2*Cosh[4*ArcTanh[c*x]] - 128 *ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + 192*ArcTanh[c*x]^2*Log[1 + E^ (-2*ArcTanh[c*x])] - 64*(-1 + 3*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c* x])] - 96*PolyLog[3, -E^(-2*ArcTanh[c*x])] + 40*Sinh[2*ArcTanh[c*x]] + 80* ArcTanh[c*x]*Sinh[2*ArcTanh[c*x]] + 80*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x]] - 4*ArcTanh[c*x]*Sinh[4*ArcTanh[c*x]] - 8*ArcTanh[ c*x]^2*Sinh[4*ArcTanh[c*x]]))/(64*c^4*d^3)
Time = 0.91 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
| \(\displaystyle \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(c d x+d)^3} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (-\frac {3 (a+b \text {arctanh}(c x))^2}{c^3 d^3 (c x+1)}+\frac {3 (a+b \text {arctanh}(c x))^2}{c^3 d^3 (c x+1)^2}+\frac {(a+b \text {arctanh}(c x))^2}{c^3 d^3}-\frac {(a+b \text {arctanh}(c x))^2}{c^3 d^3 (c x+1)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^4 d^3}-\frac {11 b (a+b \text {arctanh}(c x))}{4 c^4 d^3 (c x+1)}+\frac {b (a+b \text {arctanh}(c x))}{4 c^4 d^3 (c x+1)^2}-\frac {3 (a+b \text {arctanh}(c x))^2}{c^4 d^3 (c x+1)}+\frac {(a+b \text {arctanh}(c x))^2}{2 c^4 d^3 (c x+1)^2}+\frac {19 (a+b \text {arctanh}(c x))^2}{8 c^4 d^3}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c^4 d^3}+\frac {3 \log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c^4 d^3}+\frac {x (a+b \text {arctanh}(c x))^2}{c^3 d^3}+\frac {21 b^2 \text {arctanh}(c x)}{16 c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {21 b^2}{16 c^4 d^3 (c x+1)}+\frac {b^2}{16 c^4 d^3 (c x+1)^2}\) |
b^2/(16*c^4*d^3*(1 + c*x)^2) - (21*b^2)/(16*c^4*d^3*(1 + c*x)) + (21*b^2*A rcTanh[c*x])/(16*c^4*d^3) + (b*(a + b*ArcTanh[c*x]))/(4*c^4*d^3*(1 + c*x)^ 2) - (11*b*(a + b*ArcTanh[c*x]))/(4*c^4*d^3*(1 + c*x)) + (19*(a + b*ArcTan h[c*x])^2)/(8*c^4*d^3) + (x*(a + b*ArcTanh[c*x])^2)/(c^3*d^3) + (a + b*Arc Tanh[c*x])^2/(2*c^4*d^3*(1 + c*x)^2) - (3*(a + b*ArcTanh[c*x])^2)/(c^4*d^3 *(1 + c*x)) - (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(c^4*d^3) + (3*( a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^4*d^3) - (b^2*PolyLog[2, 1 - 2/ (1 - c*x)])/(c^4*d^3) - (3*b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c* x)])/(c^4*d^3) - (3*b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^4*d^3)
3.2.12.3.1 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 = 5.31 (sec) , antiderivative size = 2772, normalized size of antiderivative = 8.23
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2772\) |
| default | \(\text {Expression too large to display}\) | \(2772\) |
| parts | \(\text {Expression too large to display}\) | \(2785\) |
1/c^4*(a^2/d^3*(c*x-3*ln(c*x+1)-3/(c*x+1)+1/2/(c*x+1)^2)+b^2/d^3*(c*x*arct anh(c*x)^2-19/8*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))+3/8*dilog(1+I*(c *x+1)/(-c^2*x^2+1)^(1/2))+3/8*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3*arct anh(c*x)^2*ln(c*x+1)+3*arctanh(c*x)^2*ln(2)+6*arctanh(c*x)^2*ln((c*x+1)/(- c^2*x^2+1)^(1/2))+19/8*arctanh(c*x)^2-19/16*polylog(2,-(c*x+1)^2/(-c^2*x^2 +1))-2*arctanh(c*x)^3+5/8*(c*x-1)/(c*x+1)+3/8*arctanh(c*x)*ln(1+I*(c*x+1)/ (-c^2*x^2+1)^(1/2))+3/8*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3/ 2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+3*arctanh(c*x)*polylog(2,-(c*x+1)^2/( -c^2*x^2+1))+1/64/(c*x+1)^2*(c*x-1)^2+3*ln(2)*arctanh(c*x)*ln(1+I*(c*x+1)/ (-c^2*x^2+1)^(1/2))+3*ln(2)*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2) )-3*ln(2)*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))+3*ln(2)*dilog(1+I*(c*x +1)/(-c^2*x^2+1)^(1/2))+3*ln(2)*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3/2* ln(2)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-3/(c*x+1)*arctanh(c*x)^2+5/4*arct anh(c*x)*(c*x-1)/(c*x+1)+1/2/(c*x+1)^2*arctanh(c*x)^2+1/16*arctanh(c*x)*(c *x-1)^2/(c*x+1)^2-3/2*I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(c*x+1 )^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))*( arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c*x)*ln(1-I*(c*x+1 )/(-c^2*x^2+1)^(1/2))+dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+dilog(1-I*(c*x +1)/(-c^2*x^2+1)^(1/2)))+3/2*I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*(arctanh (c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+arctanh(c*x)*ln(1-I*(c*x+1)/(-...
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{3}}{{\left (c d x + d\right )}^{3}} \,d x } \]
integral((b^2*x^3*arctanh(c*x)^2 + 2*a*b*x^3*arctanh(c*x) + a^2*x^3)/(c^3* d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {\int \frac {a^{2} x^{3}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{3} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{3} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]
(Integral(a**2*x**3/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b **2*x**3*atanh(c*x)**2/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integra l(2*a*b*x**3*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{3}}{{\left (c d x + d\right )}^{3}} \,d x } \]
-1/2*a^2*((6*c*x + 5)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 2*x/(c^3*d^3 ) + 6*log(c*x + 1)/(c^4*d^3)) + 1/8*(2*b^2*c^3*x^3 + 4*b^2*c^2*x^2 - 4*b^2 *c*x - 5*b^2 - 6*(b^2*c^2*x^2 + 2*b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1)^2/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - integrate(-1/4*((b^2*c^4*x^4 - b^2*c^3*x^3)*log(c*x + 1)^2 + 4*(a*b*c^4*x^4 - a*b*c^3*x^3)*log(c*x + 1) - (2*(2*a*b*c^4 + b^2*c^4)*x^4 - 9*b^2*c*x - 2*(2*a*b*c^3 - 3*b^2*c^3)*x^ 3 - 5*b^2 + 2*(b^2*c^4*x^4 - 4*b^2*c^3*x^3 - 9*b^2*c^2*x^2 - 9*b^2*c*x - 3 *b^2)*log(c*x + 1))*log(-c*x + 1))/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^ 3*x - c^3*d^3), x)
\[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{3}}{{\left (c d x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^3} \,d x \]