\(\int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx\) [372]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=-\frac {a \sqrt {1-a^2 x^2}}{2 x}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-a^2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {1}{2} a^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{2} a^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \] Output:

-1/2*a*(-a^2*x^2+1)^(1/2)/x-1/2*(-a^2*x^2+1)^(1/2)*arctanh(a*x)/x^2-a^2*ar 
ctanh(a*x)*arctanh((-a*x+1)^(1/2)/(a*x+1)^(1/2))+1/2*a^2*polylog(2,-(-a*x+ 
1)^(1/2)/(a*x+1)^(1/2))-1/2*a^2*polylog(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.92 \[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\frac {1}{8} a^2 \left (-2 \coth \left (\frac {1}{2} \text {arctanh}(a x)\right )-\text {arctanh}(a x) \text {csch}^2\left (\frac {1}{2} \text {arctanh}(a x)\right )+4 \text {arctanh}(a x) \log \left (1-e^{-\text {arctanh}(a x)}\right )-4 \text {arctanh}(a x) \log \left (1+e^{-\text {arctanh}(a x)}\right )+4 \operatorname {PolyLog}\left (2,-e^{-\text {arctanh}(a x)}\right )-4 \operatorname {PolyLog}\left (2,e^{-\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \text {sech}^2\left (\frac {1}{2} \text {arctanh}(a x)\right )+2 \tanh \left (\frac {1}{2} \text {arctanh}(a x)\right )\right ) \] Input:

Integrate[ArcTanh[a*x]/(x^3*Sqrt[1 - a^2*x^2]),x]
 

Output:

(a^2*(-2*Coth[ArcTanh[a*x]/2] - ArcTanh[a*x]*Csch[ArcTanh[a*x]/2]^2 + 4*Ar 
cTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] - 4*ArcTanh[a*x]*Log[1 + E^(-ArcTanh 
[a*x])] + 4*PolyLog[2, -E^(-ArcTanh[a*x])] - 4*PolyLog[2, E^(-ArcTanh[a*x] 
)] - ArcTanh[a*x]*Sech[ArcTanh[a*x]/2]^2 + 2*Tanh[ArcTanh[a*x]/2]))/8
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6588, 242, 6580}

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.

\(\displaystyle \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx\)

\(\Big \downarrow \) 6588

\(\displaystyle \frac {1}{2} a^2 \int \frac {\text {arctanh}(a x)}{x \sqrt {1-a^2 x^2}}dx+\frac {1}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}\)

\(\Big \downarrow \) 242

\(\displaystyle \frac {1}{2} a^2 \int \frac {\text {arctanh}(a x)}{x \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-\frac {a \sqrt {1-a^2 x^2}}{2 x}\)

\(\Big \downarrow \) 6580

\(\displaystyle \frac {1}{2} a^2 \left (-2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )\right )-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-\frac {a \sqrt {1-a^2 x^2}}{2 x}\)

Input:

Int[ArcTanh[a*x]/(x^3*Sqrt[1 - a^2*x^2]),x]
 

Output:

-1/2*(a*Sqrt[1 - a^2*x^2])/x - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*x^2) + 
(a^2*(-2*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] + PolyLog[2, -( 
Sqrt[1 - a*x]/Sqrt[1 + a*x])] - PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]]))/ 
2
 

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 6580

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x 
_Symbol] :> Simp[(-2/Sqrt[d])*(a + b*ArcTanh[c*x])*ArcTanh[Sqrt[1 - c*x]/Sq 
rt[1 + c*x]], x] + (Simp[(b/Sqrt[d])*PolyLog[2, -Sqrt[1 - c*x]/Sqrt[1 + c*x 
]], x] - Simp[(b/Sqrt[d])*PolyLog[2, Sqrt[1 - c*x]/Sqrt[1 + c*x]], x]) /; F 
reeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]
 

rule 6588

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) 
 + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*A 
rcTanh[c*x])^p/(d*f*(m + 1))), x] + (-Simp[b*c*(p/(f*(m + 1)))   Int[(f*x)^ 
(m + 1)*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Simp[c^2*( 
(m + 2)/(f^2*(m + 1)))   Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/Sqrt[d + 
 e*x^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && G 
tQ[p, 0] && LtQ[m, -1] && NeQ[m, -2]
 

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {\left (a x +\operatorname {arctanh}\left (a x \right )\right ) \sqrt {-a^{2} x^{2}+1}}{2 x^{2}}+\frac {a^{2} \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {a^{2} \operatorname {polylog}\left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}-\frac {a^{2} \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}-\frac {a^{2} \operatorname {polylog}\left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\)	\(139\)

Input:

int(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(a*x+arctanh(a*x))*(-a^2*x^2+1)^(1/2)/x^2+1/2*a^2*arctanh(a*x)*ln(1-( 
a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*a^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-1 
/2*a^2*arctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*a^2*polylog(2,-(a 
*x+1)/(-a^2*x^2+1)^(1/2))
 

Fricas [F]

\[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\int { \frac {\operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \] Input:

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(-a^2*x^2 + 1)*arctanh(a*x)/(a^2*x^5 - x^3), x)
 

Sympy [F]

\[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\int \frac {\operatorname {atanh}{\left (a x \right )}}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:

integrate(atanh(a*x)/x**3/(-a**2*x**2+1)**(1/2),x)
 

Output:

Integral(atanh(a*x)/(x**3*sqrt(-(a*x - 1)*(a*x + 1))), x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\int { \frac {\operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \] Input:

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")
 

Output:

integrate(arctanh(a*x)/(sqrt(-a^2*x^2 + 1)*x^3), x)
 

Giac [F]

\[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\int { \frac {\operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \] Input:

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="giac")
 

Output:

integrate(arctanh(a*x)/(sqrt(-a^2*x^2 + 1)*x^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx=\int \frac {\mathrm {atanh}\left (a\,x\right )}{x^3\,\sqrt {1-a^2\,x^2}} \,d x \] Input:

int(atanh(a*x)/(x^3*(1 - a^2*x^2)^(1/2)),x)
 

Output:

int(atanh(a*x)/(x^3*(1 - a^2*x^2)^(1/2)), x)
 

Reduce [F]

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

int(atanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x)
 

Output:

int(atanh(a*x)/(sqrt( - a**2*x**2 + 1)*x**3),x)