\(\int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx\) [87]

Optimal result

Integrand size = 14, antiderivative size = 211 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {287 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}+\frac {113 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}-\frac {8 \sqrt [4]{1+\frac {1}{a x}} x^3}{\sqrt [4]{1-\frac {1}{a x}}}+\frac {25}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3+\frac {55 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {55 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \] Output:

287/24*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x/a^2+113/12*(1-1/a/x)^(3/4)*(1+1/a 
/x)^(1/4)*x^2/a-8*(1+1/a/x)^(1/4)*x^3/(1-1/a/x)^(1/4)+25/3*(1-1/a/x)^(3/4) 
*(1+1/a/x)^(1/4)*x^3+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 = 5.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.65 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {-384 e^{\frac {1}{2} \coth ^{-1}(a x)}+\frac {128 e^{\frac {1}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^3}+\frac {400 e^{\frac {1}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+\frac {548 e^{\frac {1}{2} \coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}+330 \arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-165 \log \left (1-e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+165 \log \left (1+e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{48 a^3} \] Input:

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

Output:

(-384*E^(ArcCoth[a*x]/2) + (128*E^(ArcCoth[a*x]/2))/(-1 + E^(2*ArcCoth[a*x 
]))^3 + (400*E^(ArcCoth[a*x]/2))/(-1 + E^(2*ArcCoth[a*x]))^2 + (548*E^(Arc 
Coth[a*x]/2))/(-1 + E^(2*ArcCoth[a*x])) + 330*ArcTan[E^(ArcCoth[a*x]/2)] - 
 165*Log[1 - E^(ArcCoth[a*x]/2)] + 165*Log[1 + E^(ArcCoth[a*x]/2)])/(48*a^ 
3)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {6721, 109, 27, 168, 27, 168, 27, 172, 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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 172

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[E^((5*ArcCoth[a*x])/2)*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*( 
-1/2*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)] - 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 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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {330 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (8 \, a^{4} x^{4} + 34 \, a^{3} x^{3} + 87 \, a^{2} x^{2} - 226 \, a x - 287\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{48 \, {\left (a^{4} x - a^{3}\right )}} \] Input:

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

Output:

-1/48*(330*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 165*(a*x - 1)*l 
og(((a*x - 1)/(a*x + 1))^(1/4) + 1) + 165*(a*x - 1)*log(((a*x - 1)/(a*x + 
1))^(1/4) - 1) - 2*(8*a^4*x^4 + 34*a^3*x^3 + 87*a^2*x^2 - 226*a*x - 287)*( 
(a*x - 1)/(a*x + 1))^(3/4))/(a^4*x - a^3)
 

Sympy [F]

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

integrate(1/((a*x-1)/(a*x+1))**(5/4)*x**2,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 = 203, normalized size of antiderivative = 0.96 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (\frac {425 \, {\left (a x - 1\right )}}{a x + 1} - \frac {462 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {165 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 96\right )}}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {13}{4}} - 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} + 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \] Input:

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

Output:

-1/48*a*(4*(425*(a*x - 1)/(a*x + 1) - 462*(a*x - 1)^2/(a*x + 1)^2 + 165*(a 
*x - 1)^3/(a*x + 1)^3 - 96)/(a^4*((a*x - 1)/(a*x + 1))^(13/4) - 3*a^4*((a* 
x - 1)/(a*x + 1))^(9/4) + 3*a^4*((a*x - 1)/(a*x + 1))^(5/4) - a^4*((a*x - 
1)/(a*x + 1))^(1/4)) + 330*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 - 165*l 
og(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 + 165*log(((a*x - 1)/(a*x + 1))^(1 
/4) - 1)/a^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.91 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} + \frac {384}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} - \frac {4 \, {\left (\frac {174 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - \frac {69 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} - 137 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x^2,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^4*((a*x - 1)/(a*x + 1))^(1/4)) - 4*(174*(a*x - 1)*((a*x - 1)/ 
(a*x + 1))^(3/4)/(a*x + 1) - 69*(a*x - 1)^2*((a*x - 1)/(a*x + 1))^(3/4)/(a 
*x + 1)^2 - 137*((a*x - 1)/(a*x + 1))^(3/4))/(a^4*((a*x - 1)/(a*x + 1) - 1 
)^3))
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.83 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {55\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {55\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {\frac {77\,{\left (a\,x-1\right )}^2}{2\,{\left (a\,x+1\right )}^2}-\frac {55\,{\left (a\,x-1\right )}^3}{4\,{\left (a\,x+1\right )}^3}-\frac {425\,\left (a\,x-1\right )}{12\,\left (a\,x+1\right )}+8}{a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}+3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}-a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{13/4}} \] Input:

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

Output:

(55*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3) - (55*atan(((a*x - 1)/(a*x 
 + 1))^(1/4)))/(8*a^3) - ((77*(a*x - 1)^2)/(2*(a*x + 1)^2) - (55*(a*x - 1) 
^3)/(4*(a*x + 1)^3) - (425*(a*x - 1))/(12*(a*x + 1)) + 8)/(a^3*((a*x - 1)/ 
(a*x + 1))^(1/4) - 3*a^3*((a*x - 1)/(a*x + 1))^(5/4) + 3*a^3*((a*x - 1)/(a 
*x + 1))^(9/4) - a^3*((a*x - 1)/(a*x + 1))^(13/4))
 

Reduce [F]

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

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