Integrand size = 14, antiderivative size = 179 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {3 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \]
11/24*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)*x/a^2-5/12*(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+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
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.31 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.17 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {e^{-\frac {5}{2} \coth ^{-1}(a x)} \left (-22034705-26688365 e^{2 \coth ^{-1}(a x)}-3731255 e^{4 \coth ^{-1}(a x)}+3122405 e^{6 \coth ^{-1}(a x)}+22034705 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+17244920 e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-9077530 e^{4 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-7043960 e^{6 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+446985 e^{8 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+256 e^{6 \coth ^{-1}(a x)} \left (685+1090 e^{2 \coth ^{-1}(a x)}+437 e^{4 \coth ^{-1}(a x)}\right ) \, _4F_3\left (\frac {7}{4},2,2,2;1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+2048 e^{6 \coth ^{-1}(a x)} \left (21+38 e^{2 \coth ^{-1}(a x)}+17 e^{4 \coth ^{-1}(a x)}\right ) \, _5F_4\left (\frac {7}{4},2,2,2,2;1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{6 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+8192 e^{8 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{10 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )\right )}{221760 a^3} \]
-1/221760*(-22034705 - 26688365*E^(2*ArcCoth[a*x]) - 3731255*E^(4*ArcCoth[ a*x]) + 3122405*E^(6*ArcCoth[a*x]) + 22034705*Hypergeometric2F1[3/4, 1, 7/ 4, E^(2*ArcCoth[a*x])] + 17244920*E^(2*ArcCoth[a*x])*Hypergeometric2F1[3/4 , 1, 7/4, E^(2*ArcCoth[a*x])] - 9077530*E^(4*ArcCoth[a*x])*Hypergeometric2 F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] - 7043960*E^(6*ArcCoth[a*x])*Hypergeom etric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] + 446985*E^(8*ArcCoth[a*x])*Hype rgeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] + 256*E^(6*ArcCoth[a*x])*(6 85 + 1090*E^(2*ArcCoth[a*x]) + 437*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{ 7/4, 2, 2, 2}, {1, 1, 19/4}, E^(2*ArcCoth[a*x])] + 2048*E^(6*ArcCoth[a*x]) *(21 + 38*E^(2*ArcCoth[a*x]) + 17*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{7 /4, 2, 2, 2, 2}, {1, 1, 1, 19/4}, E^(2*ArcCoth[a*x])] + 4096*E^(6*ArcCoth[ a*x])*HypergeometricPFQ[{7/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 19/4}, E^(2*Arc Coth[a*x])] + 8192*E^(8*ArcCoth[a*x])*HypergeometricPFQ[{7/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 19/4}, E^(2*ArcCoth[a*x])] + 4096*E^(10*ArcCoth[a*x])*Hyp ergeometricPFQ[{7/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 19/4}, E^(2*ArcCoth[a*x] )])/(a^3*E^((5*ArcCoth[a*x])/2))
Time = 0.32 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6721, 110, 27, 168, 27, 168, 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.
| \(\displaystyle \int x^2 e^{-\frac {1}{2} \coth ^{-1}(a x)} \, dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle \frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {1}{3} \int -\frac {\left (5 a-\frac {4}{x}\right ) x^3}{2 a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\left (5 a-\frac {4}{x}\right ) x^3}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {-\frac {1}{2} \int \frac {\left (11 a-\frac {10}{x}\right ) x^2}{2 a \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\left (11 a-\frac {10}{x}\right ) x^2}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {-\frac {-\int \frac {9 x}{2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {9}{2} \int \frac {x}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {-\frac {-18 \int -\frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {18 \int \frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {-\frac {-18 \left (\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-18 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {-18 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )-11 a x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}-\frac {5}{2} a x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{6 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
((1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x^3)/3 + ((-5*a*(1 - 1/(a*x))^(1/ 4)*(1 + 1/(a*x))^(3/4)*x^2)/2 - (-11*a*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^( 3/4)*x - 18*(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))/(4*a))/(6*a^2)
3.1.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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]
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])
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])
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]
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]
\[\int x^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}}d x\]
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.57 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 11\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \]
1/48*(2*(8*a^3*x^3 - 2*a^2*x^2 + a*x + 11)*((a*x - 1)/(a*x + 1))^(1/4) - 1 8*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
\[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\int x^{2} \sqrt [4]{\frac {a x - 1}{a x + 1}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (29 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 6 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]
-1/48*a*(4*(29*((a*x - 1)/(a*x + 1))^(9/4) - 6*((a*x - 1)/(a*x + 1))^(5/4) + 9*((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) + 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)
Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {9 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} - \frac {4 \, {\left (\frac {6 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {29 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]
-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))^(1/4)/(a*x + 1) - 29*(a*x - 1)^2*((a *x - 1)/(a*x + 1))^(1/4)/(a*x + 1)^2 - 9*((a*x - 1)/(a*x + 1))^(1/4))/(a^4 *((a*x - 1)/(a*x + 1) - 1)^3))
Time = 4.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{4}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}+\frac {29\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{12}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}-\frac {3\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {3\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3} \]
((3*((a*x - 1)/(a*x + 1))^(1/4))/4 - ((a*x - 1)/(a*x + 1))^(5/4)/2 + (29*( (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)) - (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)