\(\int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx\) [79]

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 = 0.09 (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, 218, 212, 209} \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {55 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {55 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {287 \sqrt [4]{\frac {1}{a x}+1}}{24 a^3 \sqrt [4]{1-\frac {1}{a x}}}+\frac {61 x \sqrt [4]{\frac {1}{a x}+1}}{24 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {x^3 \sqrt [4]{\frac {1}{a x}+1}}{3 \sqrt [4]{1-\frac {1}{a x}}}+\frac {13 x^2 \sqrt [4]{\frac {1}{a x}+1}}{12 a \sqrt [4]{1-\frac {1}{a x}}} \]

[In]

Int[E^((5*ArcCoth[a*x])/2)*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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 )^{5/4} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 )^{5/4} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 )^{5/4} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 {55 \text {Subst}\left (\int \frac {1}{x \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 {1}{-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 [C] (warning: unable to verify)

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 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {8 e^{\frac {9}{2} \coth ^{-1}(a x)} \left (-\frac {27653}{195}-\frac {899079}{512} e^{-8 \coth ^{-1}(a x)}-\frac {3309759 e^{-6 \coth ^{-1}(a x)}}{2560}+\frac {8521937 e^{-4 \coth ^{-1}(a x)}}{7680}+\frac {69571361 e^{-2 \coth ^{-1}(a x)}}{99840}-\frac {653}{390} e^{2 \coth ^{-1}(a x)}+\frac {133407}{512} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {899079}{512} e^{-8 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {60267}{64} e^{-6 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac {382227}{256} e^{-4 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac {40827}{64} e^{-2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac {e^{2 \coth ^{-1}(a x)} \left (1117+1906 e^{2 \coth ^{-1}(a x)}+821 e^{4 \coth ^{-1}(a x)}\right ) \, _4F_3\left (2,2,2,\frac {13}{4};1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{3094}+\frac {4 e^{2 \coth ^{-1}(a x)} \left (27+50 e^{2 \coth ^{-1}(a x)}+23 e^{4 \coth ^{-1}(a x)}\right ) \, _5F_4\left (2,2,2,2,\frac {13}{4};1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {8 e^{2 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {16 e^{4 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac {8 e^{6 \coth ^{-1}(a x)} \, _6F_5\left (2,2,2,2,2,\frac {13}{4};1,1,1,1,\frac {25}{4};e^{2 \coth ^{-1}(a x)}\right )}{1547}\right )}{9 a^3} \]

[In]

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

[Out]

(-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])) + 69571361/(99840*E^(2*ArcCoth[a*x])) - (653*E^(2*ArcCoth[a*x]))/390 + (133
407*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/512 + (899079*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcC
oth[a*x])])/(512*E^(8*ArcCoth[a*x])) + (60267*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(64*E^(6*Arc
Coth[a*x])) - (382227*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(256*E^(4*ArcCoth[a*x])) - (40827*Hy
pergeometric2F1[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*ArcCoth[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, 13/4}, {1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (8*E^(2*ArcCoth[a*x])*HypergeometricPFQ[{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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {330 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 165 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (8 \, a^{4} x^{4} + 34 \, a^{3} x^{3} + 87 \, a^{2} x^{2} - 226 \, a x - 287\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{48 \, {\left (a^{4} x - a^{3}\right )}} \]

[In]

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

[Out]

-1/48*(330*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 165*(a*x - 1)*log(((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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.95 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (\frac {425 \, {\left (a x - 1\right )}}{a x + 1} - \frac {462 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {165 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 96\right )}}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {13}{4}} - 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} + 3 \, a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]

[In]

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

[Out]

-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*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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.90 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {330 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {165 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {165 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} + \frac {384}{a^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} - \frac {4 \, {\left (\frac {174 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - \frac {69 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} - 137 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x^2,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^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 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {55\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {55\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {\frac {77\,{\left (a\,x-1\right )}^2}{2\,{\left (a\,x+1\right )}^2}-\frac {55\,{\left (a\,x-1\right )}^3}{4\,{\left (a\,x+1\right )}^3}-\frac {425\,\left (a\,x-1\right )}{12\,\left (a\,x+1\right )}+8}{a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}+3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}-a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{13/4}} \]

[In]

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

[Out]

(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))