\(\int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^3} \, dx\) [433]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^3} \, dx=-\frac {a \sqrt {1-a^2 x^2}}{2 x}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}+a^2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{2} a^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {1}{2} a^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \] 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 = 0.93 \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^3} \, dx=\frac {1}{8} a^2 \left (-2 \coth \left (\frac {1}{2} \text {arctanh}(a x)\right )-\text {arctanh}(a x) \text {csch}^2\left (\frac {1}{2} \text {arctanh}(a x)\right )-4 \text {arctanh}(a x) \log \left (1-e^{-\text {arctanh}(a x)}\right )+4 \text {arctanh}(a x) \log \left (1+e^{-\text {arctanh}(a x)}\right )-4 \operatorname {PolyLog}\left (2,-e^{-\text {arctanh}(a x)}\right )+4 \operatorname {PolyLog}\left (2,e^{-\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \text {sech}^2\left (\frac {1}{2} \text {arctanh}(a x)\right )+2 \tanh \left (\frac {1}{2} \text {arctanh}(a x)\right )\right ) \] Input:

Integrate[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^3,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 = 0.54 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6572, 242, 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.

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

\(\Big \downarrow \) 6572

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

\(\Big \downarrow \) 242

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

\(\Big \downarrow \) 6588

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

\(\Big \downarrow \) 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[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x^3,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 6572

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

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 = 0.83 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.02

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

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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