\(\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx\) [114]

Optimal result

Integrand size = 14, antiderivative size = 213 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {287 \sqrt [4]{1-\frac {1}{a x}}}{24 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {61 \sqrt [4]{1-\frac {1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {13 \sqrt [4]{1-\frac {1}{a x}} x^2}{12 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^3}{3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {55 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {55 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \] Output:

287/24*(1-1/a/x)^(1/4)/a^3/(1+1/a/x)^(1/4)+61/24*(1-1/a/x)^(1/4)*x/a^2/(1+ 
1/a/x)^(1/4)-13/12*(1-1/a/x)^(1/4)*x^2/a/(1+1/a/x)^(1/4)+1/3*(1-1/a/x)^(1/ 
4)*x^3/(1+1/a/x)^(1/4)+55/8*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^3-55 
/8*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^3
 

Mathematica [A] (verified)

Time = 5.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.64 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {384 e^{-\frac {1}{2} \coth ^{-1}(a x)}+\frac {128 e^{\frac {11}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^3}-\frac {400 e^{\frac {7}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+\frac {548 e^{\frac {3}{2} \coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}-330 \arctan \left (e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )+165 \log \left (1-e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )-165 \log \left (1+e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )}{48 a^3} \] Input:

Integrate[x^2/E^((5*ArcCoth[a*x])/2),x]
 

Output:

(384/E^(ArcCoth[a*x]/2) + (128*E^((11*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth 
[a*x]))^3 - (400*E^((7*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x]))^2 + (54 
8*E^((3*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x])) - 330*ArcTan[E^(-1/2*A 
rcCoth[a*x])] + 165*Log[1 - E^(-1/2*ArcCoth[a*x])] - 165*Log[1 + E^(-1/2*A 
rcCoth[a*x])])/(48*a^3)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6721, 109, 27, 168, 27, 168, 27, 172, 27, 104, 25, 827, 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.

\(\displaystyle \int x^2 e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6721

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{3} \int \frac {\left (13 a-\frac {12}{x}\right ) x^3}{2 a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (13 a-\frac {12}{x}\right ) x^3}{\left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {1}{2} \int \frac {\left (61 a-\frac {52}{x}\right ) x^2}{2 a \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\left (61 a-\frac {52}{x}\right ) x^2}{\left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {-\int \frac {\left (165 a-\frac {122}{x}\right ) x}{2 a \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\left (165 a-\frac {122}{x}\right ) x}{\left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 172

\(\displaystyle \frac {-\frac {-\frac {2 a \int \frac {165 x}{2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {165 a \int \frac {x}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {-\frac {-\frac {660 a \int -\frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {-\frac {\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}-660 a \int \frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {-\frac {-\frac {660 a \left (\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {-\frac {660 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {660 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )+\frac {574 a \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{2 a}-\frac {61 a x \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}}{4 a}-\frac {13 a x^2 \sqrt [4]{1-\frac {1}{a x}}}{2 \sqrt [4]{\frac {1}{a x}+1}}}{6 a^2}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}\)

Input:

Int[x^2/E^((5*ArcCoth[a*x])/2),x]
 

Output:

((1 - 1/(a*x))^(1/4)*x^3)/(3*(1 + 1/(a*x))^(1/4)) + ((-13*a*(1 - 1/(a*x))^ 
(1/4)*x^2)/(2*(1 + 1/(a*x))^(1/4)) - ((-61*a*(1 - 1/(a*x))^(1/4)*x)/(1 + 1 
/(a*x))^(1/4) - ((574*a*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4) + 660*a*( 
ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2 - ArcTanh[(1 + 1/(a*x))^ 
(1/4)/(1 - 1/(a*x))^(1/4)]/2))/(2*a))/(4*a))/(6*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 172

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> With[{mnp = Simplify[m + n + p]}, 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] + Simp[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)*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1 
])))] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && NeQ[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 827

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

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

Input:

int(x^2*((a*x-1)/(a*x+1))^(5/4),x)
 

Output:

int(x^2*((a*x-1)/(a*x+1))^(5/4),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.48 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} - 26 \, a^{2} x^{2} + 61 \, a x + 287\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \] Input:

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="fricas")
 

Output:

1/48*(2*(8*a^3*x^3 - 26*a^2*x^2 + 61*a*x + 287)*((a*x - 1)/(a*x + 1))^(1/4 
) - 330*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 165*log(((a*x - 1)/(a*x + 1) 
)^(1/4) + 1) + 165*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3
 

Sympy [F]

\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\int x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}\, dx \] Input:

integrate(x**2*((a*x-1)/(a*x+1))**(5/4),x)
 

Output:

Integral(x**2*((a*x - 1)/(a*x + 1))**(5/4), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.97 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (137 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 174 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 69 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}} - \frac {384 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{4}}\right )} \] Input:

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="maxima")
 

Output:

-1/48*a*(4*(137*((a*x - 1)/(a*x + 1))^(9/4) - 174*((a*x - 1)/(a*x + 1))^(5 
/4) + 69*((a*x - 1)/(a*x + 1))^(1/4))/(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) + 330*arctan(( 
(a*x - 1)/(a*x + 1))^(1/4))/a^4 + 165*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) 
/a^4 - 165*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^4 - 384*((a*x - 1)/(a*x 
+ 1))^(1/4)/a^4)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.90 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {165 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} - \frac {384 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{4}} - \frac {4 \, {\left (\frac {174 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {137 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} - 69 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \] Input:

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="giac")
 

Output:

-1/48*a*(330*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 + 165*log(((a*x - 1)/ 
(a*x + 1))^(1/4) + 1)/a^4 - 165*log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/ 
a^4 - 384*((a*x - 1)/(a*x + 1))^(1/4)/a^4 - 4*(174*(a*x - 1)*((a*x - 1)/(a 
*x + 1))^(1/4)/(a*x + 1) - 137*(a*x - 1)^2*((a*x - 1)/(a*x + 1))^(1/4)/(a* 
x + 1)^2 - 69*((a*x - 1)/(a*x + 1))^(1/4))/(a^4*((a*x - 1)/(a*x + 1) - 1)^ 
3))
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.85 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {23\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{4}-\frac {29\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}+\frac {137\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{12}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a^3}-\frac {55\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}+\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,55{}\mathrm {i}}{8\,a^3} \] Input:

int(x^2*((a*x - 1)/(a*x + 1))^(5/4),x)
 

Output:

((23*((a*x - 1)/(a*x + 1))^(1/4))/4 - (29*((a*x - 1)/(a*x + 1))^(5/4))/2 + 
 (137*((a*x - 1)/(a*x + 1))^(9/4))/12)/(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)) + (atan 
(((a*x - 1)/(a*x + 1))^(1/4)*1i)*55i)/(8*a^3) + (8*((a*x - 1)/(a*x + 1))^( 
1/4))/a^3 - (55*atan(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3)
 

Reduce [F]

\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\left (\int \frac {\left (a x -1\right )^{\frac {1}{4}} x^{3}}{\left (a x +1\right )^{\frac {1}{4}} a x +\left (a x +1\right )^{\frac {1}{4}}}d x \right ) a -\left (\int \frac {\left (a x -1\right )^{\frac {1}{4}} x^{2}}{\left (a x +1\right )^{\frac {1}{4}} a x +\left (a x +1\right )^{\frac {1}{4}}}d x \right ) \] Input:

int(x^2*((a*x-1)/(a*x+1))^(5/4),x)
 

Output:

int(((a*x - 1)**(1/4)*x**3)/((a*x + 1)**(1/4)*a*x + (a*x + 1)**(1/4)),x)*a 
 - int(((a*x - 1)**(1/4)*x**2)/((a*x + 1)**(1/4)*a*x + (a*x + 1)**(1/4)),x 
)