\(\int \frac {(1-a^2 x^2)^2 \text {arctanh}(a x)^2}{x^4} \, dx\) [211]

Optimal result

Integrand size = 22, antiderivative size = 167 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=-\frac {a^2}{3 x}+\frac {1}{3} a^3 \text {arctanh}(a x)-\frac {a \text {arctanh}(a x)}{3 x^2}-\frac {2}{3} a^3 \text {arctanh}(a x)^2-\frac {\text {arctanh}(a x)^2}{3 x^3}+\frac {2 a^2 \text {arctanh}(a x)^2}{x}+a^4 x \text {arctanh}(a x)^2-2 a^3 \text {arctanh}(a x) \log \left (\frac {2}{1-a x}\right )-\frac {10}{3} a^3 \text {arctanh}(a x) \log \left (2-\frac {2}{1+a x}\right )-a^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-a x}\right )+\frac {5}{3} a^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+a x}\right ) \] Output:

-1/3*a^2/x+1/3*a^3*arctanh(a*x)-1/3*a*arctanh(a*x)/x^2-2/3*a^3*arctanh(a*x 
)^2-1/3*arctanh(a*x)^2/x^3+2*a^2*arctanh(a*x)^2/x+a^4*x*arctanh(a*x)^2-2*a 
^3*arctanh(a*x)*ln(2/(-a*x+1))-10/3*a^3*arctanh(a*x)*ln(2-2/(a*x+1))-a^3*p 
olylog(2,1-2/(-a*x+1))+5/3*a^3*polylog(2,-1+2/(a*x+1))
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\frac {1}{3} \left (-\frac {a^2}{x}+a^3 \text {arctanh}(a x)-\frac {a \text {arctanh}(a x)}{x^2}-8 a^3 \text {arctanh}(a x)^2-\frac {\text {arctanh}(a x)^2}{x^3}+\frac {6 a^2 \text {arctanh}(a x)^2}{x}+3 a^4 x \text {arctanh}(a x)^2-10 a^3 \text {arctanh}(a x) \log \left (1-e^{-2 \text {arctanh}(a x)}\right )-6 a^3 \text {arctanh}(a x) \log \left (1+e^{-2 \text {arctanh}(a x)}\right )+3 a^3 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(a x)}\right )+5 a^3 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(a x)}\right )\right ) \] Input:

Integrate[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^4,x]
 

Output:

(-(a^2/x) + a^3*ArcTanh[a*x] - (a*ArcTanh[a*x])/x^2 - 8*a^3*ArcTanh[a*x]^2 
 - ArcTanh[a*x]^2/x^3 + (6*a^2*ArcTanh[a*x]^2)/x + 3*a^4*x*ArcTanh[a*x]^2 
- 10*a^3*ArcTanh[a*x]*Log[1 - E^(-2*ArcTanh[a*x])] - 6*a^3*ArcTanh[a*x]*Lo 
g[1 + E^(-2*ArcTanh[a*x])] + 3*a^3*PolyLog[2, -E^(-2*ArcTanh[a*x])] + 5*a^ 
3*PolyLog[2, E^(-2*ArcTanh[a*x])])/3
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 167, 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.

\(\displaystyle \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx\)

\(\Big \downarrow \) 6574

\(\displaystyle \int \left (a^4 \text {arctanh}(a x)^2-\frac {2 a^2 \text {arctanh}(a x)^2}{x^2}+\frac {\text {arctanh}(a x)^2}{x^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle a^4 x \text {arctanh}(a x)^2-\frac {2}{3} a^3 \text {arctanh}(a x)^2+\frac {1}{3} a^3 \text {arctanh}(a x)-2 a^3 \text {arctanh}(a x) \log \left (\frac {2}{1-a x}\right )-\frac {10}{3} a^3 \text {arctanh}(a x) \log \left (2-\frac {2}{a x+1}\right )-a^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-a x}\right )+\frac {5}{3} a^3 \operatorname {PolyLog}\left (2,\frac {2}{a x+1}-1\right )+\frac {2 a^2 \text {arctanh}(a x)^2}{x}-\frac {a^2}{3 x}-\frac {\text {arctanh}(a x)^2}{3 x^3}-\frac {a \text {arctanh}(a x)}{3 x^2}\)

Input:

Int[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^4,x]
 

Output:

-1/3*a^2/x + (a^3*ArcTanh[a*x])/3 - (a*ArcTanh[a*x])/(3*x^2) - (2*a^3*ArcT 
anh[a*x]^2)/3 - ArcTanh[a*x]^2/(3*x^3) + (2*a^2*ArcTanh[a*x]^2)/x + a^4*x* 
ArcTanh[a*x]^2 - 2*a^3*ArcTanh[a*x]*Log[2/(1 - a*x)] - (10*a^3*ArcTanh[a*x 
]*Log[2 - 2/(1 + a*x)])/3 - a^3*PolyLog[2, 1 - 2/(1 - a*x)] + (5*a^3*PolyL 
og[2, -1 + 2/(1 + a*x)])/3
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6574

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]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(a^{3} \left (\operatorname {arctanh}\left (a x \right )^{2} a x +\frac {2 \operatorname {arctanh}\left (a x \right )^{2}}{a x}-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {\operatorname {arctanh}\left (a x \right )}{3 a^{2} x^{2}}-\frac {10 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x \right )}{3}+\frac {8 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {8 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {1}{3 a x}-\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}+\frac {5 \operatorname {dilog}\left (a x \right )}{3}+\frac {5 \operatorname {dilog}\left (a x +1\right )}{3}+\frac {5 \ln \left (a x \right ) \ln \left (a x +1\right )}{3}+\frac {2 \ln \left (a x -1\right )^{2}}{3}-\frac {8 \operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {2 \ln \left (a x +1\right )^{2}}{3}\right )\)	\(208\)
default	\(a^{3} \left (\operatorname {arctanh}\left (a x \right )^{2} a x +\frac {2 \operatorname {arctanh}\left (a x \right )^{2}}{a x}-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {\operatorname {arctanh}\left (a x \right )}{3 a^{2} x^{2}}-\frac {10 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x \right )}{3}+\frac {8 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {8 \,\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {1}{3 a x}-\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}+\frac {5 \operatorname {dilog}\left (a x \right )}{3}+\frac {5 \operatorname {dilog}\left (a x +1\right )}{3}+\frac {5 \ln \left (a x \right ) \ln \left (a x +1\right )}{3}+\frac {2 \ln \left (a x -1\right )^{2}}{3}-\frac {8 \operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {2 \ln \left (a x +1\right )^{2}}{3}\right )\)	\(208\)
parts	\(a^{4} x \operatorname {arctanh}\left (a x \right )^{2}-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{3 x^{3}}+\frac {2 a^{2} \operatorname {arctanh}\left (a x \right )^{2}}{x}-\frac {a \,\operatorname {arctanh}\left (a x \right )}{3 x^{2}}-\frac {10 a^{3} \operatorname {arctanh}\left (a x \right ) \ln \left (a x \right )}{3}+\frac {8 a^{3} \operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {8 a^{3} \operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {a^{3} \left (\frac {1}{a x}+\frac {\ln \left (a x -1\right )}{2}-\frac {\ln \left (a x +1\right )}{2}-5 \operatorname {dilog}\left (a x \right )-5 \operatorname {dilog}\left (a x +1\right )-5 \ln \left (a x \right ) \ln \left (a x +1\right )-2 \ln \left (a x -1\right )^{2}+8 \operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )+4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )-4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )+2 \ln \left (a x +1\right )^{2}\right )}{3}\)	\(215\)

Input:

int((-a^2*x^2+1)^2*arctanh(a*x)^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

a^3*(arctanh(a*x)^2*a*x+2*arctanh(a*x)^2/a/x-1/3*arctanh(a*x)^2/a^3/x^3-1/ 
3*arctanh(a*x)/a^2/x^2-10/3*arctanh(a*x)*ln(a*x)+8/3*arctanh(a*x)*ln(a*x-1 
)+8/3*arctanh(a*x)*ln(a*x+1)-1/3/a/x-1/6*ln(a*x-1)+1/6*ln(a*x+1)+5/3*dilog 
(a*x)+5/3*dilog(a*x+1)+5/3*ln(a*x)*ln(a*x+1)+2/3*ln(a*x-1)^2-8/3*dilog(1/2 
*a*x+1/2)-4/3*ln(a*x-1)*ln(1/2*a*x+1/2)+4/3*(ln(a*x+1)-ln(1/2*a*x+1/2))*ln 
(-1/2*a*x+1/2)-2/3*ln(a*x+1)^2)
 

Fricas [F]

\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}} \,d x } \] Input:

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^4,x, algorithm="fricas")
 

Output:

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^2/x^4, x)
 

Sympy [F]

\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\int \frac {\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{x^{4}}\, dx \] Input:

integrate((-a**2*x**2+1)**2*atanh(a*x)**2/x**4,x)
 

Output:

Integral((a*x - 1)**2*(a*x + 1)**2*atanh(a*x)**2/x**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.22 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=-\frac {1}{6} \, {\left (16 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )} a - 10 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )} a + 10 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (a x\right )\right )} a - a \log \left (a x + 1\right ) + a \log \left (a x - 1\right ) + \frac {2 \, {\left (2 \, a x \log \left (a x + 1\right )^{2} - 4 \, a x \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 2 \, a x \log \left (a x - 1\right )^{2} + 1\right )}}{x}\right )} a^{2} + \frac {1}{3} \, {\left (8 \, a^{2} \log \left (a x + 1\right ) + 8 \, a^{2} \log \left (a x - 1\right ) - 10 \, a^{2} \log \left (x\right ) - \frac {1}{x^{2}}\right )} a \operatorname {artanh}\left (a x\right ) + \frac {1}{3} \, {\left (3 \, a^{4} x + \frac {6 \, a^{2} x^{2} - 1}{x^{3}}\right )} \operatorname {artanh}\left (a x\right )^{2} \] Input:

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^4,x, algorithm="maxima")
 

Output:

-1/6*(16*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))*a - 10* 
(log(a*x + 1)*log(x) + dilog(-a*x))*a + 10*(log(-a*x + 1)*log(x) + dilog(a 
*x))*a - a*log(a*x + 1) + a*log(a*x - 1) + 2*(2*a*x*log(a*x + 1)^2 - 4*a*x 
*log(a*x + 1)*log(a*x - 1) - 2*a*x*log(a*x - 1)^2 + 1)/x)*a^2 + 1/3*(8*a^2 
*log(a*x + 1) + 8*a^2*log(a*x - 1) - 10*a^2*log(x) - 1/x^2)*a*arctanh(a*x) 
 + 1/3*(3*a^4*x + (6*a^2*x^2 - 1)/x^3)*arctanh(a*x)^2
 

Giac [F]

\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}} \,d x } \] Input:

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^4,x, algorithm="giac")
 

Output:

integrate((a^2*x^2 - 1)^2*arctanh(a*x)^2/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2}{x^4} \,d x \] Input:

int((atanh(a*x)^2*(a^2*x^2 - 1)^2)/x^4,x)
 

Output:

int((atanh(a*x)^2*(a^2*x^2 - 1)^2)/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^2}{x^4} \, dx=\frac {6 \mathit {atanh} \left (a x \right )^{2} a^{2} x^{2}-\mathit {atanh} \left (a x \right )^{2}+\mathit {atanh} \left (a x \right ) a^{3} x^{3}-\mathit {atanh} \left (a x \right ) a x +3 \left (\int \mathit {atanh} \left (a x \right )^{2}d x \right ) a^{4} x^{3}+10 \left (\int \frac {\mathit {atanh} \left (a x \right )}{a^{2} x^{3}-x}d x \right ) a^{3} x^{3}-a^{2} x^{2}}{3 x^{3}} \] Input:

int((-a^2*x^2+1)^2*atanh(a*x)^2/x^4,x)
 

Output:

(6*atanh(a*x)**2*a**2*x**2 - atanh(a*x)**2 + atanh(a*x)*a**3*x**3 - atanh( 
a*x)*a*x + 3*int(atanh(a*x)**2,x)*a**4*x**3 + 10*int(atanh(a*x)/(a**2*x**3 
 - x),x)*a**3*x**3 - a**2*x**2)/(3*x**3)