Integrand size = 14, antiderivative size = 213 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {287 \sqrt [4]{1+\frac {1}{a x}}}{24 a^3 \sqrt [4]{1-\frac {1}{a x}}}+\frac {61 \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {13 \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\sqrt [4]{1+\frac {1}{a x}} x^3}{3 \sqrt [4]{1-\frac {1}{a x}}}+\frac {55 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {55 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \]
-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+5 5/8*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 = 6.98 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.07 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {8 e^{\frac {9}{2} \coth ^{-1}(a x)} \left (-\frac {27653}{195}-\frac {899079}{512} e^{-8 \coth ^{-1}(a x)}-\frac {3309759 e^{-6 \coth ^{-1}(a x)}}{2560}+\frac {8521937 e^{-4 \coth ^{-1}(a x)}}{7680}+\frac {69571361 e^{-2 \coth ^{-1}(a x)}}{99840}-\frac {653}{390} e^{2 \coth ^{-1}(a x)}+\frac {133407}{512} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {899079}{512} e^{-8 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {60267}{64} e^{-6 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac {382227}{256} e^{-4 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac {40827}{64} e^{-2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {e^{2 \coth ^{-1}(a x)} \left (1117+1906 e^{2 \coth ^{-1}(a x)}+821 e^{4 \coth ^{-1}(a x)}\right ) \, _4F_3\left (2,2,2,\frac {13}{4};1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{3094}+\frac {4 e^{2 \coth ^{-1}(a x)} \left (27+50 e^{2 \coth ^{-1}(a x)}+23 e^{4 \coth ^{-1}(a x)}\right ) \, _5F_4\left (2,2,2,2,\frac {13}{4};1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {8 e^{2 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {16 e^{4 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {8 e^{6 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}\right )}{9 a^3} \]
(-8*E^((9*ArcCoth[a*x])/2)*(-27653/195 - 899079/(512*E^(8*ArcCoth[a*x])) - 3309759/(2560*E^(6*ArcCoth[a*x])) + 8521937/(7680*E^(4*ArcCoth[a*x])) + 6 9571361/(99840*E^(2*ArcCoth[a*x])) - (653*E^(2*ArcCoth[a*x]))/390 + (13340 7*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/512 + (899079*Hyperg eometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(512*E^(8*ArcCoth[a*x])) + ( 60267*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(64*E^(6*ArcCoth [a*x])) - (382227*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(256 *E^(4*ArcCoth[a*x])) - (40827*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[ a*x])])/(64*E^(2*ArcCoth[a*x])) + (E^(2*ArcCoth[a*x])*(1117 + 1906*E^(2*Ar cCoth[a*x]) + 821*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{2, 2, 2, 13/4}, { 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/3094 + (4*E^(2*ArcCoth[a*x])*(27 + 50*E^ (2*ArcCoth[a*x]) + 23*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{2, 2, 2, 2, 1 3/4}, {1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (8*E^(2*ArcCoth[a*x])*H ypergeometricPFQ[{2, 2, 2, 2, 2, 13/4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[a *x])])/1547 + (16*E^(4*ArcCoth[a*x])*HypergeometricPFQ[{2, 2, 2, 2, 2, 13/ 4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (8*E^(6*ArcCoth[a*x])* HypergeometricPFQ[{2, 2, 2, 2, 2, 13/4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[ a*x])])/1547))/(9*a^3)
Time = 0.35 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.06, 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 |
| \(\displaystyle \frac {x^3 \sqrt [4]{\frac {1}{a x}+1}}{3 \sqrt [4]{1-\frac {1}{a x}}}-\frac {\frac {\frac {660 a \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )+\frac {574 a \sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}}{2 a}-\frac {61 a x \sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}}{4 a}-\frac {13 a x^2 \sqrt [4]{\frac {1}{a x}+1}}{2 \sqrt [4]{1-\frac {1}{a x}}}}{6 a^2}\) |
((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)
3.1.79.3.1 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[(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])
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_))^(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]
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[((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]
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]
\[\int \frac {x^{2}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}}d x\]
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {330 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (8 \, a^{4} x^{4} + 34 \, a^{3} x^{3} + 87 \, a^{2} x^{2} - 226 \, a x - 287\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{48 \, {\left (a^{4} x - a^{3}\right )}} \]
-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)
\[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\int \frac {x^{2}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.95 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (\frac {425 \, {\left (a x - 1\right )}}{a x + 1} - \frac {462 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {165 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 96\right )}}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {13}{4}} - 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} + 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]
-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)
Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.90 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} + \frac {384}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} - \frac {4 \, {\left (\frac {174 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - \frac {69 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} - 137 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]
-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))
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.83 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {55\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {55\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {\frac {77\,{\left (a\,x-1\right )}^2}{2\,{\left (a\,x+1\right )}^2}-\frac {55\,{\left (a\,x-1\right )}^3}{4\,{\left (a\,x+1\right )}^3}-\frac {425\,\left (a\,x-1\right )}{12\,\left (a\,x+1\right )}+8}{a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}+3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}-a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{13/4}} \]
(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))