\(\int \frac {(1-a^2 x^2)^{3/2} \text {arctanh}(a x)}{x} \, dx\) [452]

Optimal result

Integrand size = 22, antiderivative size = 144 \[ \int \frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}{x} \, dx=-\frac {1}{6} a x \sqrt {1-a^2 x^2}-\frac {7}{6} \arcsin (a x)+\sqrt {1-a^2 x^2} \text {arctanh}(a x)+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)-2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \] Output:

-1/6*a*x*(-a^2*x^2+1)^(1/2)-7/6*arcsin(a*x)+(-a^2*x^2+1)^(1/2)*arctanh(a*x 
)+1/3*(-a^2*x^2+1)^(3/2)*arctanh(a*x)-2*arctanh(a*x)*arctanh((-a*x+1)^(1/2 
)/(a*x+1)^(1/2))+polylog(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))-polylog(2,(-a*x+ 
1)^(1/2)/(a*x+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}{x} \, dx=\frac {1}{6} \left (-a x \sqrt {1-a^2 x^2}-14 \arctan \left (\tanh \left (\frac {1}{2} \text {arctanh}(a x)\right )\right )+8 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-2 a^2 x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x)+6 \text {arctanh}(a x) \log \left (1-e^{-\text {arctanh}(a x)}\right )-6 \text {arctanh}(a x) \log \left (1+e^{-\text {arctanh}(a x)}\right )+6 \operatorname {PolyLog}\left (2,-e^{-\text {arctanh}(a x)}\right )-6 \operatorname {PolyLog}\left (2,e^{-\text {arctanh}(a x)}\right )\right ) \] Input:

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

Output:

(-(a*x*Sqrt[1 - a^2*x^2]) - 14*ArcTan[Tanh[ArcTanh[a*x]/2]] + 8*Sqrt[1 - a 
^2*x^2]*ArcTanh[a*x] - 2*a^2*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + 6*ArcTan 
h[a*x]*Log[1 - E^(-ArcTanh[a*x])] - 6*ArcTanh[a*x]*Log[1 + E^(-ArcTanh[a*x 
])] + 6*PolyLog[2, -E^(-ArcTanh[a*x])] - 6*PolyLog[2, E^(-ArcTanh[a*x])])/ 
6
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6576, 6556, 211, 223, 6572, 223, 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.

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

\(\Big \downarrow \) 6576

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

\(\Big \downarrow \) 6556

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 6572

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 6580

\(\displaystyle -\left (a^2 \left (\frac {\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}}{3 a}-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}{3 a^2}\right )\right )+\sqrt {1-a^2 x^2} \text {arctanh}(a x)-\arcsin (a x)-2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )\)

Input:

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

Output:

-ArcSin[a*x] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - a^2*(((x*Sqrt[1 - a^2*x^2] 
)/2 + ArcSin[a*x]/(2*a))/(3*a) - ((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/(3*a^2 
)) - 2*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] + PolyLog[2, -(Sq 
rt[1 - a*x]/Sqrt[1 + a*x])] - PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]]
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 6556

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q 
_.), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p/(2*e*(q 
+ 1))), x] + Simp[b*(p/(2*c*(q + 1)))   Int[(d + e*x^2)^q*(a + b*ArcTanh[c* 
x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && 
 GtQ[p, 0] && NeQ[q, -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 6576

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[d   Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a 
+ b*ArcTanh[c*x])^p, x], x] - Simp[c^2*(d/f^2)   Int[(f*x)^(m + 2)*(d + e*x 
^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, 
x] && EqQ[c^2*d + e, 0] && GtQ[q, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ 
[p, 1] && IntegerQ[q]))
 

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]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {\left (2 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )+a x -8 \,\operatorname {arctanh}\left (a x \right )\right ) \sqrt {-a^{2} x^{2}+1}}{6}-\frac {7 \arctan \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-\operatorname {dilog}\left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\operatorname {dilog}\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(130\)

Input:

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

Output:

-1/6*(2*a^2*x^2*arctanh(a*x)+a*x-8*arctanh(a*x))*(-a^2*x^2+1)^(1/2)-7/3*ar 
ctan((a*x+1)/(-a^2*x^2+1)^(1/2))-dilog(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-arcta 
nh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-dilog((a*x+1)/(-a^2*x^2+1)^(1/2))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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