\(\int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx\) [70]

Optimal result

Integrand size = 14, antiderivative size = 179 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {17 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {17 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \]

[Out]

23/24*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)*x/a^2+7/12*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)*x^2/a+1/3*(1-1/a/x)^(1/4)*(1+
1/a/x)^(3/4)*x^3-17/8*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^3+17/8*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4)
)/a^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6306, 101, 156, 12, 95, 304, 209, 212} \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {17 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {17 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {23 x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{24 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}+\frac {7 x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{12 a} \]

[In]

Int[E^((3*ArcCoth[a*x])/2)*x^2,x]

[Out]

(23*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x)/(24*a^2) + (7*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x^2)/(12*
a) + ((1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x^3)/3 - (17*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8
*a^3) + (17*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 101

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[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] && GtQ[n, 0] && (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 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 )^{3/4}}{x^4 \left (1-\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {1}{3} \text {Subst}\left (\int \frac {\frac {7}{2 a}+\frac {2 x}{a^2}}{x^3 \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {1}{6} \text {Subst}\left (\int \frac {-\frac {23}{4 a^2}-\frac {7 x}{2 a^3}}{x^2 \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {1}{6} \text {Subst}\left (\int \frac {51}{8 a^3 x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {17 \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 {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {17 \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 {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {17 \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 {17 \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 {23 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}+\frac {7 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {17 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {17 \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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.43 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.23 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {e^{-\frac {5}{2} \coth ^{-1}(a x)} \left (-1357846875-1400453615 e^{2 \coth ^{-1}(a x)}[phone] e^{4 \coth ^{-1}(a x)}[phone] e^{6 \coth ^{-1}(a x)}-12962560 e^{8 \coth ^{-1}(a x)}[phone] \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )[phone] e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-997722110 e^{4 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-501106760 e^{6 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )[phone] e^{8 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+1792 e^{8 \coth ^{-1}(a x)} \left (965+1618 e^{2 \coth ^{-1}(a x)}+685 e^{4 \coth ^{-1}(a x)}\right ) \, _4F_3\left (2,2,2,\frac {11}{4};1,1,\frac {23}{4};e^{2 \coth ^{-1}(a x)}\right )+14336 e^{8 \coth ^{-1}(a x)} \left (25+46 e^{2 \coth ^{-1}(a x)}+21 e^{4 \coth ^{-1}(a x)}\right ) \, _5F_4\left (2,2,2,2,\frac {11}{4};1,1,1,\frac {23}{4};e^{2 \coth ^{-1}(a x)}\right )+28672 e^{8 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {11}{4};1,1,1,1,\frac {23}{4};e^{2 \coth ^{-1}(a x)}\right )+57344 e^{10 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {11}{4};1,1,1,1,\frac {23}{4};e^{2 \coth ^{-1}(a x)}\right )+28672 e^{12 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {11}{4};1,1,1,1,\frac {23}{4};e^{2 \coth ^{-1}(a x)}\right )\right )}{4213440 a^3} \]

[In]

Integrate[E^((3*ArcCoth[a*x])/2)*x^2,x]

[Out]

-1/4213440*(-1357846875 - 1400453615*E^(2*ArcCoth[a*x]) + 276606275*E^(4*ArcCoth[a*x]) + 438715415*E^(6*ArcCot
h[a*x]) - 12962560*E^(8*ArcCoth[a*x]) + 1357846875*Hypergeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] + 818519
240*E^(2*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] - 997722110*E^(4*ArcCoth[a*x])*Hyper
geometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] - 501106760*E^(6*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^
(2*ArcCoth[a*x])] + 137997475*E^(8*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] + 1792*E^(
8*ArcCoth[a*x])*(965 + 1618*E^(2*ArcCoth[a*x]) + 685*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{2, 2, 2, 11/4}, {1
, 1, 23/4}, E^(2*ArcCoth[a*x])] + 14336*E^(8*ArcCoth[a*x])*(25 + 46*E^(2*ArcCoth[a*x]) + 21*E^(4*ArcCoth[a*x])
)*HypergeometricPFQ[{2, 2, 2, 2, 11/4}, {1, 1, 1, 23/4}, E^(2*ArcCoth[a*x])] + 28672*E^(8*ArcCoth[a*x])*Hyperg
eometricPFQ[{2, 2, 2, 2, 2, 11/4}, {1, 1, 1, 1, 23/4}, E^(2*ArcCoth[a*x])] + 57344*E^(10*ArcCoth[a*x])*Hyperge
ometricPFQ[{2, 2, 2, 2, 2, 11/4}, {1, 1, 1, 1, 23/4}, E^(2*ArcCoth[a*x])] + 28672*E^(12*ArcCoth[a*x])*Hypergeo
metricPFQ[{2, 2, 2, 2, 2, 11/4}, {1, 1, 1, 1, 23/4}, E^(2*ArcCoth[a*x])])/(a^3*E^((5*ArcCoth[a*x])/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.58 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} + 22 \, a^{2} x^{2} + 37 \, a x + 23\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x^2,x, algorithm="fricas")

[Out]

1/48*(2*(8*a^3*x^3 + 22*a^2*x^2 + 37*a*x + 23)*((a*x - 1)/(a*x + 1))^(1/4) + 102*arctan(((a*x - 1)/(a*x + 1))^
(1/4)) + 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - 51*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (17 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 30 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 45 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} - \frac {102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x^2,x, algorithm="maxima")

[Out]

-1/48*a*(4*(17*((a*x - 1)/(a*x + 1))^(9/4) - 30*((a*x - 1)/(a*x + 1))^(5/4) + 45*((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) - 102*arctan((
(a*x - 1)/(a*x + 1))^(1/4))/a^4 - 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 + 51*log(((a*x - 1)/(a*x + 1))^(
1/4) - 1)/a^4)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {1}{48} \, a {\left (\frac {102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {51 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} + \frac {4 \, {\left (\frac {30 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {17 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} - 45 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x^2,x, algorithm="giac")

[Out]

1/48*a*(102*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 + 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 - 51*log(abs
(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^4 + 4*(30*(a*x - 1)*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1) - 17*(a*x - 1)^
2*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1)^2 - 45*((a*x - 1)/(a*x + 1))^(1/4))/(a^4*((a*x - 1)/(a*x + 1) - 1)^3))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {15\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{4}-\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}+\frac {17\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{12}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {17\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}+\frac {17\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3} \]

[In]

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

[Out]

((15*((a*x - 1)/(a*x + 1))^(1/4))/4 - (5*((a*x - 1)/(a*x + 1))^(5/4))/2 + (17*((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)) + (17*a
tan(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3) + (17*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3)