Integrand size = 22, antiderivative size = 408 \[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {a b x}{c^4 d^3}-\frac {b^2}{16 c^5 d^3 (1+c x)^2}+\frac {29 b^2}{16 c^5 d^3 (1+c x)}-\frac {29 b^2 \text {arctanh}(c x)}{16 c^5 d^3}+\frac {b^2 x \text {arctanh}(c x)}{c^4 d^3}-\frac {b (a+b \text {arctanh}(c x))}{4 c^5 d^3 (1+c x)^2}+\frac {15 b (a+b \text {arctanh}(c x))}{4 c^5 d^3 (1+c x)}-\frac {43 (a+b \text {arctanh}(c x))^2}{8 c^5 d^3}-\frac {3 x (a+b \text {arctanh}(c x))^2}{c^4 d^3}+\frac {x^2 (a+b \text {arctanh}(c x))^2}{2 c^3 d^3}-\frac {(a+b \text {arctanh}(c x))^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 (a+b \text {arctanh}(c x))^2}{c^5 d^3 (1+c x)}+\frac {6 b (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 (a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {6 b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{c^5 d^3} \]
a*b*x/c^4/d^3-1/16*b^2/c^5/d^3/(c*x+1)^2+29/16*b^2/c^5/d^3/(c*x+1)-29/16*b ^2*arctanh(c*x)/c^5/d^3+b^2*x*arctanh(c*x)/c^4/d^3-1/4*b*(a+b*arctanh(c*x) )/c^5/d^3/(c*x+1)^2+15/4*b*(a+b*arctanh(c*x))/c^5/d^3/(c*x+1)-43/8*(a+b*ar ctanh(c*x))^2/c^5/d^3-3*x*(a+b*arctanh(c*x))^2/c^4/d^3+1/2*x^2*(a+b*arctan h(c*x))^2/c^3/d^3-1/2*(a+b*arctanh(c*x))^2/c^5/d^3/(c*x+1)^2+4*(a+b*arctan h(c*x))^2/c^5/d^3/(c*x+1)+6*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^5/d^3-6* (a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^5/d^3+1/2*b^2*ln(-c^2*x^2+1)/c^5/d^3+ 3*b^2*polylog(2,1-2/(-c*x+1))/c^5/d^3+6*b*(a+b*arctanh(c*x))*polylog(2,1-2 /(c*x+1))/c^5/d^3+3*b^2*polylog(3,1-2/(c*x+1))/c^5/d^3
Time = 1.54 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {-48 a^2 c x+8 a^2 c^2 x^2-\frac {8 a^2}{(1+c x)^2}+\frac {64 a^2}{1+c x}+96 a^2 \log (1+c x)+a b \left (16 c x+28 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))-48 \log \left (1-c^2 x^2\right )+96 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )-28 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (-4-24 c x+4 c^2 x^2+14 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))-48 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-14 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))\right )\right )+16 b^2 \left ((-3+6 \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+\frac {1}{64} \left (56 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))+32 \log \left (1-c^2 x^2\right )+192 \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )-56 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (16 c x+28 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))+96 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-28 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))\right )+8 \text {arctanh}(c x)^2 \left (20-24 c x+4 c^2 x^2+14 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))-48 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-14 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))\right )\right )\right )}{16 c^5 d^3} \]
(-48*a^2*c*x + 8*a^2*c^2*x^2 - (8*a^2)/(1 + c*x)^2 + (64*a^2)/(1 + c*x) + 96*a^2*Log[1 + c*x] + a*b*(16*c*x + 28*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTa nh[c*x]] - 48*Log[1 - c^2*x^2] + 96*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 28* Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-4 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])) + 16*b^2*((-3 + 6*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + (56*Cosh[ 2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 32*Log[1 - c^2*x^2] + 192*PolyLog [3, -E^(-2*ArcTanh[c*x])] - 56*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(16*c*x + 28*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 96*Log[1 + E^(-2*ArcTanh[c*x])] - 28*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcT anh[c*x]]) + 8*ArcTanh[c*x]^2*(20 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2*ArcTanh [c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[ 2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]]))/64))/(16*c^5*d^3)
Time = 1.09 (sec) , antiderivative size = 408, 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.
| \(\displaystyle \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(c d x+d)^3} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (\frac {6 (a+b \text {arctanh}(c x))^2}{c^4 d^3 (c x+1)}-\frac {4 (a+b \text {arctanh}(c x))^2}{c^4 d^3 (c x+1)^2}-\frac {3 (a+b \text {arctanh}(c x))^2}{c^4 d^3}+\frac {(a+b \text {arctanh}(c x))^2}{c^4 d^3 (c x+1)^3}+\frac {x (a+b \text {arctanh}(c x))^2}{c^3 d^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {6 b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^5 d^3}+\frac {15 b (a+b \text {arctanh}(c x))}{4 c^5 d^3 (c x+1)}-\frac {b (a+b \text {arctanh}(c x))}{4 c^5 d^3 (c x+1)^2}+\frac {4 (a+b \text {arctanh}(c x))^2}{c^5 d^3 (c x+1)}-\frac {(a+b \text {arctanh}(c x))^2}{2 c^5 d^3 (c x+1)^2}-\frac {43 (a+b \text {arctanh}(c x))^2}{8 c^5 d^3}+\frac {6 b \log \left (\frac {2}{1-c x}\right ) (a+b \text {arctanh}(c x))}{c^5 d^3}-\frac {6 \log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c^5 d^3}-\frac {3 x (a+b \text {arctanh}(c x))^2}{c^4 d^3}+\frac {x^2 (a+b \text {arctanh}(c x))^2}{2 c^3 d^3}+\frac {a b x}{c^4 d^3}-\frac {29 b^2 \text {arctanh}(c x)}{16 c^5 d^3}+\frac {b^2 x \text {arctanh}(c x)}{c^4 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{c^5 d^3}+\frac {29 b^2}{16 c^5 d^3 (c x+1)}-\frac {b^2}{16 c^5 d^3 (c x+1)^2}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}\) |
(a*b*x)/(c^4*d^3) - b^2/(16*c^5*d^3*(1 + c*x)^2) + (29*b^2)/(16*c^5*d^3*(1 + c*x)) - (29*b^2*ArcTanh[c*x])/(16*c^5*d^3) + (b^2*x*ArcTanh[c*x])/(c^4* d^3) - (b*(a + b*ArcTanh[c*x]))/(4*c^5*d^3*(1 + c*x)^2) + (15*b*(a + b*Arc Tanh[c*x]))/(4*c^5*d^3*(1 + c*x)) - (43*(a + b*ArcTanh[c*x])^2)/(8*c^5*d^3 ) - (3*x*(a + b*ArcTanh[c*x])^2)/(c^4*d^3) + (x^2*(a + b*ArcTanh[c*x])^2)/ (2*c^3*d^3) - (a + b*ArcTanh[c*x])^2/(2*c^5*d^3*(1 + c*x)^2) + (4*(a + b*A rcTanh[c*x])^2)/(c^5*d^3*(1 + c*x)) + (6*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(c^5*d^3) - (6*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^5*d^3) + (b^2*Log[1 - c^2*x^2])/(2*c^5*d^3) + (3*b^2*PolyLog[2, 1 - 2/(1 - c*x)]) /(c^5*d^3) + (6*b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^5*d ^3) + (3*b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(c^5*d^3)
3.2.11.3.1 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 = 8.82 (sec) , antiderivative size = 1069, normalized size of antiderivative = 2.62
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1069\) |
| default | \(\text {Expression too large to display}\) | \(1069\) |
| parts | \(\text {Expression too large to display}\) | \(1082\) |
1/c^5*(a^2/d^3*(1/2*c^2*x^2-3*c*x+6*ln(c*x+1)+4/(c*x+1)-1/2/(c*x+1)^2)+b^2 /d^3*(-3*c*x*arctanh(c*x)^2+6*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+6*dilo g(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*c^2*x^2*arctanh(c*x)^2+6*arctanh(c*x )^2*ln(c*x+1)-6*arctanh(c*x)^2*ln(2)-12*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^ 2+1)^(1/2))-43/8*arctanh(c*x)^2-ln(1+(c*x+1)^2/(-c^2*x^2+1))+4*arctanh(c*x )^3-7/8*(c*x-1)/(c*x+1)+6*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+ 6*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3*polylog(3,-(c*x+1)^2/( -c^2*x^2+1))-6*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-1/64/(c*x+1 )^2*(c*x-1)^2+(c*x+1)*arctanh(c*x)+4/(c*x+1)*arctanh(c*x)^2-7/4*arctanh(c* x)*(c*x-1)/(c*x+1)+3*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)))*ar ctanh(c*x)^2-1/2/(c*x+1)^2*arctanh(c*x)^2-1/16*arctanh(c*x)*(c*x-1)^2/(c*x +1)^2-3*I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3*arc tanh(c*x)^2-3*I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-3*I*Pi*c sgn(I/(1-(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)))^2*arctanh(c*x)^2-3*I*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^ 2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2-6*I*Pi*csgn(I*(c*x+1)/(-c^2 *x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2+3*I*Pi*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)))^2*arctanh(c*x)^2)+2*a*b/d^3*(1/2*c^2*x^2*arctanh(c*x)-3*c*x*arct...
\[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{4}}{{\left (c d x + d\right )}^{3}} \,d x } \]
integral((b^2*x^4*arctanh(c*x)^2 + 2*a*b*x^4*arctanh(c*x) + a^2*x^4)/(c^3* d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
\[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {\int \frac {a^{2} x^{4}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]
(Integral(a**2*x**4/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b **2*x**4*atanh(c*x)**2/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integra l(2*a*b*x**4*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3
\[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{4}}{{\left (c d x + d\right )}^{3}} \,d x } \]
1/2*a^2*((8*c*x + 7)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + (c*x^2 - 6*x) /(c^4*d^3) + 12*log(c*x + 1)/(c^5*d^3)) + 1/8*(b^2*c^4*x^4 - 4*b^2*c^3*x^3 - 11*b^2*c^2*x^2 + 2*b^2*c*x + 7*b^2 + 12*(b^2*c^2*x^2 + 2*b^2*c*x + b^2) *log(c*x + 1))*log(-c*x + 1)^2/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - int egrate(-1/4*((b^2*c^5*x^5 - b^2*c^4*x^4)*log(c*x + 1)^2 + 4*(a*b*c^5*x^5 - a*b*c^4*x^4)*log(c*x + 1) + (15*b^2*c^3*x^3 + 9*b^2*c^2*x^2 - (4*a*b*c^5 + b^2*c^5)*x^5 + (4*a*b*c^4 + 3*b^2*c^4)*x^4 - 9*b^2*c*x - 7*b^2 - 2*(b^2* c^5*x^5 - b^2*c^4*x^4 + 6*b^2*c^3*x^3 + 18*b^2*c^2*x^2 + 18*b^2*c*x + 6*b^ 2)*log(c*x + 1))*log(-c*x + 1))/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5*d^3*x - c^4*d^3), x)
\[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{4}}{{\left (c d x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^4 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int \frac {x^4\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^3} \,d x \]