\(\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx\) [106]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6306, 100, 156, 160, 12, 95, 304, 209, 212} \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {55 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {55 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {287 \sqrt [4]{1-\frac {1}{a x}}}{24 a^3 \sqrt [4]{\frac {1}{a x}+1}}+\frac {61 x \sqrt [4]{1-\frac {1}{a x}}}{24 a^2 \sqrt [4]{\frac {1}{a x}+1}}+\frac {x^3 \sqrt [4]{1-\frac {1}{a x}}}{3 \sqrt [4]{\frac {1}{a x}+1}}-\frac {13 x^2 \sqrt [4]{1-\frac {1}{a x}}}{12 a \sqrt [4]{\frac {1}{a x}+1}} \]

[In]

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

[Out]

(287*(1 - 1/(a*x))^(1/4))/(24*a^3*(1 + 1/(a*x))^(1/4)) + (61*(1 - 1/(a*x))^(1/4)*x)/(24*a^2*(1 + 1/(a*x))^(1/4
)) - (13*(1 - 1/(a*x))^(1/4)*x^2)/(12*a*(1 + 1/(a*x))^(1/4)) + ((1 - 1/(a*x))^(1/4)*x^3)/(3*(1 + 1/(a*x))^(1/4
)) + (55*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8*a^3) - (55*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*
x))^(1/4)])/(8*a^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[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 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] + Dist[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 160

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] + Dist[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 + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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

Rule 6306

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]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{5/4}}{x^4 \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\sqrt [4]{1-\frac {1}{a x}} x^3}{3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {1}{3} \text {Subst}\left (\int \frac {\frac {13}{2 a}-\frac {6 x}{a^2}}{x^3 \left (1-\frac {x}{a}\right )^{3/4} \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\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 {1}{6} \text {Subst}\left (\int \frac {\frac {61}{4 a^2}-\frac {13 x}{a^3}}{x^2 \left (1-\frac {x}{a}\right )^{3/4} \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \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 {1}{6} \text {Subst}\left (\int \frac {\frac {165}{8 a^3}-\frac {61 x}{4 a^4}}{x \left (1-\frac {x}{a}\right )^{3/4} \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \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 {1}{3} a \text {Subst}\left (\int \frac {165}{16 a^4 x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \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 \text {Subst}\left (\int \frac {1}{x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{16 a^3} \\ & = \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 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^3} \\ & = \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 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {55 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.64 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {384 e^{-\frac {1}{2} \coth ^{-1}(a x)}+\frac {128 e^{\frac {11}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^3}-\frac {400 e^{\frac {7}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+\frac {548 e^{\frac {3}{2} \coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}-330 \arctan \left (e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )+165 \log \left (1-e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )-165 \log \left (1+e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )}{48 a^3} \]

[In]

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

[Out]

(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 + (548*E^((3*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x])) - 330*ArcTan[E^(-1/2
*ArcCoth[a*x])] + 165*Log[1 - E^(-1/2*ArcCoth[a*x])] - 165*Log[1 + E^(-1/2*ArcCoth[a*x])])/(48*a^3)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.48 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} - 26 \, a^{2} x^{2} + 61 \, a x + 287\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.97 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (137 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 174 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 69 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}} - \frac {384 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{4}}\right )} \]

[In]

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

[Out]

-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)

none

Time = 0.31 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.90 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {165 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} - \frac {384 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{4}} - \frac {4 \, {\left (\frac {174 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {137 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} - 69 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]

[In]

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

[Out]

-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 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {23\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{4}-\frac {29\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}+\frac {137\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{12}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a^3}-\frac {55\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}+\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,55{}\mathrm {i}}{8\,a^3} \]

[In]

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

[Out]

((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))/1
2)/(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)) + (at
an(((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)