\(\int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx\) [428]

Optimal result

Integrand size = 22, antiderivative size = 194 \[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^3}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {\arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{4 a^3}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3} \] Output:

1/8*(-a^2*x^2+1)^(1/2)/a^3-1/12*(-a^2*x^2+1)^(3/2)/a^3-1/8*x*(-a^2*x^2+1)^ 
(1/2)*arctanh(a*x)/a^2+1/4*x^3*(-a^2*x^2+1)^(1/2)*arctanh(a*x)-1/4*arctan( 
(-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^3-1/8*I*polylog(2,-I*(-a*x+1) 
^(1/2)/(a*x+1)^(1/2))/a^3+1/8*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/ 
a^3
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.82 \[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\frac {\sqrt {1-a^2 x^2} \left (1+2 a^2 x^2+3 a x \text {arctanh}(a x)+6 a x \left (-1+a^2 x^2\right ) \text {arctanh}(a x)-\frac {3 i \text {arctanh}(a x) \left (\log \left (1-i e^{-\text {arctanh}(a x)}\right )-\log \left (1+i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-\frac {3 i \left (\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{24 a^3} \] Input:

Integrate[x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]
 

Output:

(Sqrt[1 - a^2*x^2]*(1 + 2*a^2*x^2 + 3*a*x*ArcTanh[a*x] + 6*a*x*(-1 + a^2*x 
^2)*ArcTanh[a*x] - ((3*I)*ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 
+ I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2] - ((3*I)*(PolyLog[2, (-I)/E^ArcTan 
h[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(24*a^3)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6572, 243, 53, 2009, 6578, 241, 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.

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

\(\Big \downarrow \) 6572

\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{4} a \int \frac {x^3}{\sqrt {1-a^2 x^2}}dx+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{8} a \int \frac {x^2}{\sqrt {1-a^2 x^2}}dx^2+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{8} a \int \left (\frac {1}{a^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a^2}\right )dx^2+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\)

\(\Big \downarrow \) 6578

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}\right )+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\)

\(\Big \downarrow \) 6512

\(\displaystyle \frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )+\frac {1}{4} \left (\frac {-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )\)

Input:

Int[x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]
 

Output:

-1/8*(a*((-2*Sqrt[1 - a^2*x^2])/a^4 + (2*(1 - a^2*x^2)^(3/2))/(3*a^4))) + 
(x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/4 + (-1/2*Sqrt[1 - a^2*x^2]/a^3 - (x* 
Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*a^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])/Sqrt[1 + a*x]])/a)/(2*a^2))/4
 

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 241

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

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 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 6578

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) 
 + (e_.)*(x_)^2], x_Symbol] :> Simp[(-f)*(f*x)^(m - 1)*Sqrt[d + e*x^2]*((a 
+ b*ArcTanh[c*x])^p/(c^2*d*m)), x] + (Simp[b*f*(p/(c*m))   Int[(f*x)^(m - 1 
)*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Simp[f^2*((m - 1 
)/(c^2*m))   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] && GtQ[p, 0] && 
GtQ[m, 1]
 

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (6 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )+2 a^{2} x^{2}-3 a x \,\operatorname {arctanh}\left (a x \right )+1\right )}{24 a^{3}}-\frac {i \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}+\frac {i \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}\)	\(175\)

Input:

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

Output:

1/24/a^3*(-(a*x-1)*(a*x+1))^(1/2)*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2-3*a*x* 
arctanh(a*x)+1)-1/8*I/a^3*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+ 
1/8*I/a^3*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))-1/8*I*dilog(1+I* 
(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/8*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2) 
)/a^3
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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