Integrand size = 22, antiderivative size = 214 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=-\frac {a^2}{12 x^2}-\frac {a \text {arctanh}(a x)}{6 x^3}+\frac {3 a^3 \text {arctanh}(a x)}{2 x}-\frac {3}{4} a^4 \text {arctanh}(a x)^2-\frac {\text {arctanh}(a x)^2}{4 x^4}+\frac {a^2 \text {arctanh}(a x)^2}{x^2}+2 a^4 \text {arctanh}(a x)^2 \text {arctanh}\left (1-\frac {2}{1-a x}\right )-\frac {4}{3} a^4 \log (x)+\frac {2}{3} a^4 \log \left (1-a^2 x^2\right )-a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,1-\frac {2}{1-a x}\right )+a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-a x}\right )+\frac {1}{2} a^4 \operatorname {PolyLog}\left (3,1-\frac {2}{1-a x}\right )-\frac {1}{2} a^4 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-a x}\right ) \]
-1/12*a^2/x^2-1/6*a*arctanh(a*x)/x^3+3/2*a^3*arctanh(a*x)/x-3/4*a^4*arctan h(a*x)^2-1/4*arctanh(a*x)^2/x^4+a^2*arctanh(a*x)^2/x^2-2*a^4*arctanh(a*x)^ 2*arctanh(-1+2/(-a*x+1))-4/3*a^4*ln(x)+2/3*a^4*ln(-a^2*x^2+1)-a^4*arctanh( a*x)*polylog(2,1-2/(-a*x+1))+a^4*arctanh(a*x)*polylog(2,-1+2/(-a*x+1))+1/2 *a^4*polylog(3,1-2/(-a*x+1))-1/2*a^4*polylog(3,-1+2/(-a*x+1))
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.11 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\frac {1}{24} \left (2 a^4+i a^4 \pi ^3-\frac {2 a^2}{x^2}-\frac {4 a \text {arctanh}(a x)}{x^3}+\frac {36 a^3 \text {arctanh}(a x)}{x}-18 a^4 \text {arctanh}(a x)^2-\frac {6 \text {arctanh}(a x)^2}{x^4}+\frac {24 a^2 \text {arctanh}(a x)^2}{x^2}-16 a^4 \text {arctanh}(a x)^3-24 a^4 \text {arctanh}(a x)^2 \log \left (1+e^{-2 \text {arctanh}(a x)}\right )+24 a^4 \text {arctanh}(a x)^2 \log \left (1-e^{2 \text {arctanh}(a x)}\right )-32 a^4 \log \left (\frac {a x}{\sqrt {1-a^2 x^2}}\right )+24 a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(a x)}\right )+24 a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(a x)}\right )+12 a^4 \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(a x)}\right )-12 a^4 \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}(a x)}\right )\right ) \]
(2*a^4 + I*a^4*Pi^3 - (2*a^2)/x^2 - (4*a*ArcTanh[a*x])/x^3 + (36*a^3*ArcTa nh[a*x])/x - 18*a^4*ArcTanh[a*x]^2 - (6*ArcTanh[a*x]^2)/x^4 + (24*a^2*ArcT anh[a*x]^2)/x^2 - 16*a^4*ArcTanh[a*x]^3 - 24*a^4*ArcTanh[a*x]^2*Log[1 + E^ (-2*ArcTanh[a*x])] + 24*a^4*ArcTanh[a*x]^2*Log[1 - E^(2*ArcTanh[a*x])] - 3 2*a^4*Log[(a*x)/Sqrt[1 - a^2*x^2]] + 24*a^4*ArcTanh[a*x]*PolyLog[2, -E^(-2 *ArcTanh[a*x])] + 24*a^4*ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] + 12* a^4*PolyLog[3, -E^(-2*ArcTanh[a*x])] - 12*a^4*PolyLog[3, E^(2*ArcTanh[a*x] )])/24
Time = 0.81 (sec) , antiderivative size = 214, 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 = {6574, 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.
\(\displaystyle \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx\) |
\(\Big \downarrow \) 6574 |
\(\displaystyle \int \left (\frac {a^4 \text {arctanh}(a x)^2}{x}-\frac {2 a^2 \text {arctanh}(a x)^2}{x^3}+\frac {\text {arctanh}(a x)^2}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,1-\frac {2}{1-a x}\right )+a^4 \text {arctanh}(a x) \operatorname {PolyLog}\left (2,\frac {2}{1-a x}-1\right )-\frac {3}{4} a^4 \text {arctanh}(a x)^2+2 a^4 \text {arctanh}(a x)^2 \text {arctanh}\left (1-\frac {2}{1-a x}\right )+\frac {1}{2} a^4 \operatorname {PolyLog}\left (3,1-\frac {2}{1-a x}\right )-\frac {1}{2} a^4 \operatorname {PolyLog}\left (3,\frac {2}{1-a x}-1\right )-\frac {4}{3} a^4 \log (x)+\frac {3 a^3 \text {arctanh}(a x)}{2 x}+\frac {a^2 \text {arctanh}(a x)^2}{x^2}-\frac {a^2}{12 x^2}+\frac {2}{3} a^4 \log \left (1-a^2 x^2\right )-\frac {\text {arctanh}(a x)^2}{4 x^4}-\frac {a \text {arctanh}(a x)}{6 x^3}\) |
-1/12*a^2/x^2 - (a*ArcTanh[a*x])/(6*x^3) + (3*a^3*ArcTanh[a*x])/(2*x) - (3 *a^4*ArcTanh[a*x]^2)/4 - ArcTanh[a*x]^2/(4*x^4) + (a^2*ArcTanh[a*x]^2)/x^2 + 2*a^4*ArcTanh[a*x]^2*ArcTanh[1 - 2/(1 - a*x)] - (4*a^4*Log[x])/3 + (2*a ^4*Log[1 - a^2*x^2])/3 - a^4*ArcTanh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)] + a^ 4*ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 - a*x)] + (a^4*PolyLog[3, 1 - 2/(1 - a *x)])/2 - (a^4*PolyLog[3, -1 + 2/(1 - a*x)])/2
3.3.12.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_ .)*(x_)^2)^(q_), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0] && IGtQ[q, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 5.43 (sec) , antiderivative size = 1124, normalized size of antiderivative = 5.25
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1124\) |
default | \(\text {Expression too large to display}\) | \(1124\) |
parts | \(\text {Expression too large to display}\) | \(1829\) |
a^4*(arctanh(a*x)^2/a^2/x^2+1/2*I*Pi*csgn(I*(-(a*x+1)^2/(a^2*x^2-1)-1))*cs gn(I*(-(a*x+1)^2/(a^2*x^2-1)-1)/(1-(a*x+1)^2/(a^2*x^2-1)))*csgn(I/(1-(a*x+ 1)^2/(a^2*x^2-1)))*arctanh(a*x)^2+1/8*(-a*x*(-a^2*x^2+1)^(1/2)+2*a^2*x^2+( -a^2*x^2+1)^(1/2)+a*x-1)*arctanh(a*x)/a^2/x^2+1/8*(a*x*(-a^2*x^2+1)^(1/2)+ 2*a^2*x^2-(-a^2*x^2+1)^(1/2)+a*x-1)*arctanh(a*x)/a^2/x^2-1/24*((-a^2*x^2+1 )^(1/2)*a^2*x^2+5*a^3*x^3-3*a*x*(-a^2*x^2+1)^(1/2)+2*(-a^2*x^2+1)^(1/2)-3* a*x+2)*arctanh(a*x)/a^3/x^3-1/24*(-(-a^2*x^2+1)^(1/2)*a^2*x^2+5*a^3*x^3+3* a*x*(-a^2*x^2+1)^(1/2)-2*(-a^2*x^2+1)^(1/2)-3*a*x+2)*arctanh(a*x)/a^3/x^3- 1/4*arctanh(a*x)^2/a^4/x^4+2*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^ (1/2))+2*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-2*polylog(3,-( a*x+1)/(-a^2*x^2+1)^(1/2))-2*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))-3/4*arc tanh(a*x)^2-1/12/(a*x-(-a^2*x^2+1)^(1/2)+1)*(-a^2*x^2+1)^(1/2)-1/24*(a*x-1 )/((-a^2*x^2+1)^(1/2)-1)+1/24*(a*x-1)/((-a^2*x^2+1)^(1/2)+1)+1/12/(a*x+(-a ^2*x^2+1)^(1/2)+1)*(-a^2*x^2+1)^(1/2)+1/2*I*Pi*csgn(I*(-(a*x+1)^2/(a^2*x^2 -1)-1)/(1-(a*x+1)^2/(a^2*x^2-1)))^3*arctanh(a*x)^2-1/2*I*Pi*csgn(I*(-(a*x+ 1)^2/(a^2*x^2-1)-1)/(1-(a*x+1)^2/(a^2*x^2-1)))^2*csgn(I/(1-(a*x+1)^2/(a^2* x^2-1)))*arctanh(a*x)^2-1/2*I*Pi*csgn(I*(-(a*x+1)^2/(a^2*x^2-1)-1))*csgn(I *(-(a*x+1)^2/(a^2*x^2-1)-1)/(1-(a*x+1)^2/(a^2*x^2-1)))^2*arctanh(a*x)^2-4/ 3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-arctanh(a*x)*polylog(2,-(a*x+1)^2/(-a^2 *x^2+1))+arctanh(a*x)^2*ln(a*x)-4/3*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-1)-ar...
\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{5}} \,d x } \]
\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\int \frac {\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{x^{5}}\, dx \]
\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{5}} \,d x } \]
1/16*(4*a^2*x^2 - 1)*log(-a*x + 1)^2/x^4 - 1/4*integrate(-1/2*(2*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x + 1)^2 - (4*a^3*x^3 - a*x + 4*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x + 1 ))*log(-a*x + 1))/(a*x^6 - x^5), x)
\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^5} \, dx=\int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2}{x^5} \,d x \]