\(\int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx\) [69]

Optimal result

Integrand size = 14, antiderivative size = 179 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {11 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}+\frac {5 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3+\frac {3 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \] Output:

11/24*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x/a^2+5/12*(1-1/a/x)^(3/4)*(1+1/a/x) 
^(1/4)*x^2/a+1/3*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x^3+3/8*arctan((1+1/a/x)^ 
(1/4)/(1-1/a/x)^(1/4))/a^3+3/8*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^ 
3
 

Mathematica [A] (verified)

Time = 5.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.70 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {128 e^{\frac {1}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^3}+\frac {208 e^{\frac {1}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+\frac {116 e^{\frac {1}{2} \coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}+18 \arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-9 \log \left (1-e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+9 \log \left (1+e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{48 a^3} \] Input:

Integrate[E^(ArcCoth[a*x]/2)*x^2,x]
 

Output:

((128*E^(ArcCoth[a*x]/2))/(-1 + E^(2*ArcCoth[a*x]))^3 + (208*E^(ArcCoth[a* 
x]/2))/(-1 + E^(2*ArcCoth[a*x]))^2 + (116*E^(ArcCoth[a*x]/2))/(-1 + E^(2*A 
rcCoth[a*x])) + 18*ArcTan[E^(ArcCoth[a*x]/2)] - 9*Log[1 - E^(ArcCoth[a*x]/ 
2)] + 9*Log[1 + E^(ArcCoth[a*x]/2)])/(48*a^3)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6721, 110, 27, 168, 27, 168, 27, 104, 756, 216, 219}

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.

\(\displaystyle \int x^2 e^{\frac {1}{2} \coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6721

\(\displaystyle -\int \frac {\sqrt [4]{1+\frac {1}{a x}} x^4}{\sqrt [4]{1-\frac {1}{a x}}}d\frac {1}{x}\)

\(\Big \downarrow \) 110

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {1}{3} \int \frac {\left (5 a+\frac {4}{x}\right ) x^3}{2 a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\int \frac {\left (5 a+\frac {4}{x}\right ) x^3}{\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}}{6 a^2}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {-\frac {1}{2} \int -\frac {\left (11 a+\frac {10}{x}\right ) x^2}{2 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {\int \frac {\left (11 a+\frac {10}{x}\right ) x^2}{\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {-\int -\frac {9 x}{2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {\frac {9}{2} \int \frac {x}{\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {18 \int \frac {1}{\frac {1}{x^4}-1}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {18 \left (-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {18 \left (-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {\frac {18 \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )-11 a x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{4 a}-\frac {5}{2} a x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{6 a^2}\)

Input:

Int[E^(ArcCoth[a*x]/2)*x^2,x]
 

Output:

((1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x^3)/3 - ((-5*a*(1 - 1/(a*x))^(3/ 
4)*(1 + 1/(a*x))^(1/4)*x^2)/2 + (-11*a*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^( 
1/4)*x + 18*(-1/2*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)] - ArcTan 
h[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2))/(4*a))/(6*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 110

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 216

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

rule 219

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

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 6721

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x 
/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x] /; FreeQ[{a, n}, x] && 
!IntegerQ[n] && IntegerQ[m]
 

Maple [F]

\[\int \frac {x^{2}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}}}d x\]

Input:

int(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x)
 

Output:

int(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.58 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} + 18 \, a^{2} x^{2} + 21 \, a x + 11\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}} - 18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/48*(2*(8*a^3*x^3 + 18*a^2*x^2 + 21*a*x + 11)*((a*x - 1)/(a*x + 1))^(3/4) 
 - 18*arctan(((a*x - 1)/(a*x + 1))^(1/4)) + 9*log(((a*x - 1)/(a*x + 1))^(1 
/4) + 1) - 9*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3
 

Sympy [F]

\[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\int \frac {x^{2}}{\sqrt [4]{\frac {a x - 1}{a x + 1}}}\, dx \] Input:

integrate(1/((a*x-1)/(a*x+1))**(1/4)*x**2,x)
 

Output:

Integral(x**2/((a*x - 1)/(a*x + 1))**(1/4), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {11}{4}} - 6 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{4}} + 29 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x, algorithm="maxima")
 

Output:

-1/48*a*(4*(9*((a*x - 1)/(a*x + 1))^(11/4) - 6*((a*x - 1)/(a*x + 1))^(7/4) 
 + 29*((a*x - 1)/(a*x + 1))^(3/4))/(3*(a*x - 1)*a^4/(a*x + 1) - 3*(a*x - 1 
)^2*a^4/(a*x + 1)^2 + (a*x - 1)^3*a^4/(a*x + 1)^3 - a^4) + 18*arctan(((a*x 
 - 1)/(a*x + 1))^(1/4))/a^4 - 9*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 + 
 9*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^4)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {9 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} - \frac {4 \, {\left (\frac {6 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - \frac {9 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} - 29 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x, algorithm="giac")
 

Output:

-1/48*a*(18*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 - 9*log(((a*x - 1)/(a* 
x + 1))^(1/4) + 1)/a^4 + 9*log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^4 - 
 4*(6*(a*x - 1)*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) - 9*(a*x - 1)^2*((a* 
x - 1)/(a*x + 1))^(3/4)/(a*x + 1)^2 - 29*((a*x - 1)/(a*x + 1))^(3/4))/(a^4 
*((a*x - 1)/(a*x + 1) - 1)^3))
 

Mupad [B] (verification not implemented)

Time = 23.95 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {29\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{12}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/4}}{2}+\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{11/4}}{4}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}-\frac {3\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}+\frac {3\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3} \] Input:

int(x^2/((a*x - 1)/(a*x + 1))^(1/4),x)
 

Output:

((29*((a*x - 1)/(a*x + 1))^(3/4))/12 - ((a*x - 1)/(a*x + 1))^(7/4)/2 + (3* 
((a*x - 1)/(a*x + 1))^(11/4))/4)/(a^3 + (3*a^3*(a*x - 1)^2)/(a*x + 1)^2 - 
(a^3*(a*x - 1)^3)/(a*x + 1)^3 - (3*a^3*(a*x - 1))/(a*x + 1)) - (3*atan(((a 
*x - 1)/(a*x + 1))^(1/4)))/(8*a^3) + (3*atanh(((a*x - 1)/(a*x + 1))^(1/4)) 
)/(8*a^3)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\int \frac {\left (a x +1\right )^{\frac {1}{4}} x^{2}}{\left (a x -1\right )^{\frac {1}{4}}}d x \] Input:

int(1/((a*x-1)/(a*x+1))^(1/4)*x^2,x)
 

Output:

int(((a*x + 1)**(1/4)*x**2)/(a*x - 1)**(1/4),x)