\(\int \frac {x^2 \text {arctanh}(a x)}{(1-a^2 x^2)^{3/2}} \, dx\) [389]

Optimal result

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

-1/a^3/(-a^2*x^2+1)^(1/2)+x*arctanh(a*x)/a^2/(-a^2*x^2+1)^(1/2)+2*arctan(( 
-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^3+I*polylog(2,-I*(-a*x+1)^(1/2 
)/(a*x+1)^(1/2))/a^3-I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {i \left (\frac {i}{\sqrt {1-a^2 x^2}}-\frac {i a x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}+\text {arctanh}(a x) \log \left (1-i e^{-\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \log \left (1+i e^{-\text {arctanh}(a x)}\right )+\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{a^3} \] Input:

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

Output:

(I*(I/Sqrt[1 - a^2*x^2] - (I*a*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] + ArcTanh 
[a*x]*Log[1 - I/E^ArcTanh[a*x]] - ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] + 
 PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/a^3
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6560, 6512}

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.

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

\(\Big \downarrow \) 6560

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

\(\Big \downarrow \) 6512

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 6512

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol 
] :> Simp[-2*(a + b*ArcTanh[c*x])*(ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]]/(c*S 
qrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/( 
c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/(c* 
Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 
0]
 

rule 6560

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), 
x_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (-Si 
mp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])/(2*c^2*d*(q + 1))), x] + Sim 
p[1/(2*c^2*d*(q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x 
]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q 
, -5/2]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.39

method	result	size
default	\(-\frac {\left (\operatorname {arctanh}\left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x -1\right )}-\frac {\left (\operatorname {arctanh}\left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x +1\right )}+\frac {i \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}-\frac {i \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}+\frac {i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}-\frac {i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}\)	\(190\)

Input:

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

Output:

-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(a*x-1)-1/2*(arctanh(a* 
x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(a*x+1)+I/a^3*arctanh(a*x)*ln(1+I*(a*x+ 
1)/(-a^2*x^2+1)^(1/2))-I/a^3*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2 
))+I/a^3*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))-I/a^3*dilog(1-I*(a*x+1)/(-a 
^2*x^2+1)^(1/2))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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