Integrand size = 22, antiderivative size = 265 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=-\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {13 b^2 \text {arctanh}(c x)}{16 c^3 d^3}-\frac {b (a+b \text {arctanh}(c x))}{4 c^3 d^3 (1+c x)^2}+\frac {7 b (a+b \text {arctanh}(c x))}{4 c^3 d^3 (1+c x)}-\frac {7 (a+b \text {arctanh}(c x))^2}{8 c^3 d^3}-\frac {(a+b \text {arctanh}(c x))^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 (a+b \text {arctanh}(c x))^2}{c^3 d^3 (1+c x)}-\frac {(a+b \text {arctanh}(c x))^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^3 d^3} \]
-1/16*b^2/c^3/d^3/(c*x+1)^2+13/16*b^2/c^3/d^3/(c*x+1)-13/16*b^2*arctanh(c* x)/c^3/d^3-1/4*b*(a+b*arctanh(c*x))/c^3/d^3/(c*x+1)^2+7/4*b*(a+b*arctanh(c *x))/c^3/d^3/(c*x+1)-7/8*(a+b*arctanh(c*x))^2/c^3/d^3-1/2*(a+b*arctanh(c*x ))^2/c^3/d^3/(c*x+1)^2+2*(a+b*arctanh(c*x))^2/c^3/d^3/(c*x+1)-(a+b*arctanh (c*x))^2*ln(2/(c*x+1))/c^3/d^3+b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1)) /c^3/d^3+1/2*b^2*polylog(3,1-2/(c*x+1))/c^3/d^3
Time = 1.13 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.17 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {-\frac {8 a^2}{(1+c x)^2}+\frac {32 a^2}{1+c x}+16 a^2 \log (1+c x)+16 b^2 \left (\text {arctanh}(c x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )+\frac {1}{64} (-\cosh (2 \text {arctanh}(c x))+\sinh (2 \text {arctanh}(c x))) \left (-24+\cosh (2 \text {arctanh}(c x))+4 \text {arctanh}(c x) (-12+\cosh (2 \text {arctanh}(c x))-\sinh (2 \text {arctanh}(c x)))-\sinh (2 \text {arctanh}(c x))+8 \text {arctanh}(c x)^2 \left (-6+\cosh (2 \text {arctanh}(c x)) \left (1+8 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )+\left (-1+8 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right ) \sinh (2 \text {arctanh}(c x))\right )\right )\right )+a b \left (12 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))+16 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )-12 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (6 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))-8 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-6 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))\right )\right )}{16 c^3 d^3} \]
((-8*a^2)/(1 + c*x)^2 + (32*a^2)/(1 + c*x) + 16*a^2*Log[1 + c*x] + 16*b^2* (ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + PolyLog[3, -E^(-2*ArcTanh [c*x])]/2 + ((-Cosh[2*ArcTanh[c*x]] + Sinh[2*ArcTanh[c*x]])*(-24 + Cosh[2* ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-12 + Cosh[2*ArcTanh[c*x]] - Sinh[2*ArcTan h[c*x]]) - Sinh[2*ArcTanh[c*x]] + 8*ArcTanh[c*x]^2*(-6 + Cosh[2*ArcTanh[c* x]]*(1 + 8*Log[1 + E^(-2*ArcTanh[c*x])]) + (-1 + 8*Log[1 + E^(-2*ArcTanh[c *x])])*Sinh[2*ArcTanh[c*x]])))/64) + a*b*(12*Cosh[2*ArcTanh[c*x]] - Cosh[4 *ArcTanh[c*x]] + 16*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 12*Sinh[2*ArcTanh[c *x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(6*Cosh[2*ArcTanh[c*x]] - Cos h[4*ArcTanh[c*x]] - 8*Log[1 + E^(-2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x] ] + Sinh[4*ArcTanh[c*x]])))/(16*c^3*d^3)
Time = 0.77 (sec) , antiderivative size = 265, 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.
| \(\displaystyle \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(c d x+d)^3} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (\frac {(a+b \text {arctanh}(c x))^2}{c^2 d^3 (c x+1)}-\frac {2 (a+b \text {arctanh}(c x))^2}{c^2 d^3 (c x+1)^2}+\frac {(a+b \text {arctanh}(c x))^2}{c^2 d^3 (c x+1)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^3 d^3}+\frac {7 b (a+b \text {arctanh}(c x))}{4 c^3 d^3 (c x+1)}-\frac {b (a+b \text {arctanh}(c x))}{4 c^3 d^3 (c x+1)^2}+\frac {2 (a+b \text {arctanh}(c x))^2}{c^3 d^3 (c x+1)}-\frac {(a+b \text {arctanh}(c x))^2}{2 c^3 d^3 (c x+1)^2}-\frac {7 (a+b \text {arctanh}(c x))^2}{8 c^3 d^3}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2}{c^3 d^3}-\frac {13 b^2 \text {arctanh}(c x)}{16 c^3 d^3}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (c x+1)}-\frac {b^2}{16 c^3 d^3 (c x+1)^2}\) |
-1/16*b^2/(c^3*d^3*(1 + c*x)^2) + (13*b^2)/(16*c^3*d^3*(1 + c*x)) - (13*b^ 2*ArcTanh[c*x])/(16*c^3*d^3) - (b*(a + b*ArcTanh[c*x]))/(4*c^3*d^3*(1 + c* x)^2) + (7*b*(a + b*ArcTanh[c*x]))/(4*c^3*d^3*(1 + c*x)) - (7*(a + b*ArcTa nh[c*x])^2)/(8*c^3*d^3) - (a + b*ArcTanh[c*x])^2/(2*c^3*d^3*(1 + c*x)^2) + (2*(a + b*ArcTanh[c*x])^2)/(c^3*d^3*(1 + c*x)) - ((a + b*ArcTanh[c*x])^2* Log[2/(1 + c*x)])/(c^3*d^3) + (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^3*d^3) + (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^3*d^3)
3.2.13.3.1 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 = 4.45 (sec) , antiderivative size = 815, normalized size of antiderivative = 3.08
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(815\) |
| default | \(\text {Expression too large to display}\) | \(815\) |
| parts | \(\text {Expression too large to display}\) | \(827\) |
1/c^3*(a^2/d^3*(ln(c*x+1)+2/(c*x+1)-1/2/(c*x+1)^2)+b^2/d^3*(arctanh(c*x)^2 *ln(c*x+1)+2/(c*x+1)*arctanh(c*x)^2-1/2/(c*x+1)^2*arctanh(c*x)^2-2*arctanh (c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+2/3*arctanh(c*x)^3-1/8*(-4*I*Pi*csg n(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)))+4*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)/(1-(c*x+1)^2/(c^2*x^2-1)))^2+4*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))+8*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+4*I*Pi *csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-4*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+7+4*I*Pi*csgn(I*(c*x +1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3+8*ln(2))*arctanh(c*x)^2-arc tanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/2*polylog(3,-(c*x+1)^2/(-c^ 2*x^2+1))-1/16*arctanh(c*x)*(c*x-1)^2/(c*x+1)^2-1/64/(c*x+1)^2*(c*x-1)^2-3 /4*arctanh(c*x)*(c*x-1)/(c*x+1)-3/8*(c*x-1)/(c*x+1))+2*a*b/d^3*(arctanh(c* x)*ln(c*x+1)+2/(c*x+1)*arctanh(c*x)-1/2/(c*x+1)^2*arctanh(c*x)+7/16*ln(c*x -1)-1/8/(c*x+1)^2+7/8/(c*x+1)-7/16*ln(c*x+1)+1/2*(ln(c*x+1)-ln(1/2*c*x+1/2 ))*ln(-1/2*c*x+1/2)-1/2*dilog(1/2*c*x+1/2)-1/4*ln(c*x+1)^2))
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \]
integral((b^2*x^2*arctanh(c*x)^2 + 2*a*b*x^2*arctanh(c*x) + a^2*x^2)/(c^3* d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\frac {\int \frac {a^{2} x^{2}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]
(Integral(a**2*x**2/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b **2*x**2*atanh(c*x)**2/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integra l(2*a*b*x**2*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \]
1/2*a^2*((4*c*x + 3)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) + 2*log(c*x + 1 )/(c^3*d^3)) + 1/8*(4*b^2*c*x + 3*b^2 + 2*(b^2*c^2*x^2 + 2*b^2*c*x + b^2)* log(c*x + 1))*log(-c*x + 1)^2/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - inte grate(-1/4*((b^2*c^3*x^3 - b^2*c^2*x^2)*log(c*x + 1)^2 + 4*(a*b*c^3*x^3 - a*b*c^2*x^2)*log(c*x + 1) - (4*a*b*c^3*x^3 + 7*b^2*c*x - 4*(a*b*c^2 - b^2* c^2)*x^2 + 3*b^2 + 2*(2*b^2*c^3*x^3 + 2*b^2*c^2*x^2 + 3*b^2*c*x + b^2)*log (c*x + 1))*log(-c*x + 1))/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2 *d^3), x)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \text {arctanh}(c x))^2}{(d+c d x)^3} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^3} \,d x \]