\(\int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx\) [111]

Optimal result

Integrand size = 14, antiderivative size = 351 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {25 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {25 a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}-\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \]

[Out]

-2*a^2*(1-1/a/x)^(9/4)/(1+1/a/x)^(1/4)-25/4*a^2*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)-5/2*a^2*(1-1/a/x)^(5/4)*(1+1/a
/x)^(3/4)+25/8*a^2*arctan(-1+(1-1/a/x)^(1/4)*2^(1/2)/(1+1/a/x)^(1/4))*2^(1/2)+25/8*a^2*arctan(1+(1-1/a/x)^(1/4
)*2^(1/2)/(1+1/a/x)^(1/4))*2^(1/2)-25/16*a^2*ln(1-(1-1/a/x)^(1/4)*2^(1/2)/(1+1/a/x)^(1/4)+(1-1/a/x)^(1/2)/(1+1
/a/x)^(1/2))*2^(1/2)+25/16*a^2*ln(1+(1-1/a/x)^(1/4)*2^(1/2)/(1+1/a/x)^(1/4)+(1-1/a/x)^(1/2)/(1+1/a/x)^(1/2))*2
^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6306, 79, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {25 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 \sqrt {2}}+\frac {25 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{4 \sqrt {2}}-\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{\frac {1}{a x}+1}}-\frac {5}{2} a^2 \left (\frac {1}{a x}+1\right )^{3/4} \left (1-\frac {1}{a x}\right )^{5/4}-\frac {25}{4} a^2 \left (\frac {1}{a x}+1\right )^{3/4} \sqrt [4]{1-\frac {1}{a x}}-\frac {25 a^2 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}}+\frac {25 a^2 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}} \]

[In]

Int[1/(E^((5*ArcCoth[a*x])/2)*x^3),x]

[Out]

(-2*a^2*(1 - 1/(a*x))^(9/4))/(1 + 1/(a*x))^(1/4) - (25*a^2*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4))/4 - (5*a^2
*(1 - 1/(a*x))^(5/4)*(1 + 1/(a*x))^(3/4))/2 - (25*a^2*ArcTan[1 - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(
1/4)])/(4*Sqrt[2]) + (25*a^2*ArcTan[1 + (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(4*Sqrt[2]) - (25*
a^2*Log[1 + Sqrt[1 - 1/(a*x)]/Sqrt[1 + 1/(a*x)] - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(8*Sqrt[
2]) + (25*a^2*Log[1 + Sqrt[1 - 1/(a*x)]/Sqrt[1 + 1/(a*x)] + (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)]
)/(8*Sqrt[2])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 {x \left (1-\frac {x}{a}\right )^{5/4}}{\left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-(5 a) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{5/4}}{\sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {1}{4} (25 a) \text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{\sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {1}{8} (25 a) \text {Subst}\left (\int \frac {1}{\left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {1}{2} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-\frac {1}{a x}}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {1}{2} \left (25 a^2\right ) \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 ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {1}{4} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )+\frac {1}{4} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right ) \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {1}{8} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )+\frac {1}{8} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )-\frac {\left (25 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}-\frac {\left (25 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {\left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}-\frac {\left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}} \\ & = -\frac {2 a^2 \left (1-\frac {1}{a x}\right )^{9/4}}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {25}{4} a^2 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {5}{2} a^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {25 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {25 a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}-\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {25 a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.29 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {8}{3} a^2 e^{-\frac {1}{2} \coth ^{-1}(a x)} \left (3+e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-e^{2 \coth ^{-1}(a x)}\right )+e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {7}{4},-e^{2 \coth ^{-1}(a x)}\right )+2 e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},3,\frac {7}{4},-e^{2 \coth ^{-1}(a x)}\right )\right ) \]

[In]

Integrate[1/(E^((5*ArcCoth[a*x])/2)*x^3),x]

[Out]

(-8*a^2*(3 + E^(2*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, -E^(2*ArcCoth[a*x])] + E^(2*ArcCoth[a*x])*Hyper
geometric2F1[3/4, 2, 7/4, -E^(2*ArcCoth[a*x])] + 2*E^(2*ArcCoth[a*x])*Hypergeometric2F1[3/4, 3, 7/4, -E^(2*Arc
Coth[a*x])]))/(3*E^(ArcCoth[a*x]/2))

Maple [F]

\[\int \frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}}{x^{3}}d x\]

[In]

int(((a*x-1)/(a*x+1))^(5/4)/x^3,x)

[Out]

int(((a*x-1)/(a*x+1))^(5/4)/x^3,x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {25 \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (25 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 25 \, \left (-a^{8}\right )^{\frac {1}{4}}\right ) + 25 i \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (25 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 25 i \, \left (-a^{8}\right )^{\frac {1}{4}}\right ) - 25 i \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (25 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 25 i \, \left (-a^{8}\right )^{\frac {1}{4}}\right ) - 25 \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (25 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 25 \, \left (-a^{8}\right )^{\frac {1}{4}}\right ) - 2 \, {\left (43 \, a^{2} x^{2} + 9 \, a x - 2\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{8 \, x^{2}} \]

[In]

integrate(((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="fricas")

[Out]

1/8*(25*(-a^8)^(1/4)*x^2*log(25*a^2*((a*x - 1)/(a*x + 1))^(1/4) + 25*(-a^8)^(1/4)) + 25*I*(-a^8)^(1/4)*x^2*log
(25*a^2*((a*x - 1)/(a*x + 1))^(1/4) + 25*I*(-a^8)^(1/4)) - 25*I*(-a^8)^(1/4)*x^2*log(25*a^2*((a*x - 1)/(a*x +
1))^(1/4) - 25*I*(-a^8)^(1/4)) - 25*(-a^8)^(1/4)*x^2*log(25*a^2*((a*x - 1)/(a*x + 1))^(1/4) - 25*(-a^8)^(1/4))
- 2*(43*a^2*x^2 + 9*a*x - 2)*((a*x - 1)/(a*x + 1))^(1/4))/x^2

Sympy [F]

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

[In]

integrate(((a*x-1)/(a*x+1))**(5/4)/x**3,x)

[Out]

Integral(((a*x - 1)/(a*x + 1))**(5/4)/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{16} \, {\left (50 \, \sqrt {2} a \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 50 \, \sqrt {2} a \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 25 \, \sqrt {2} a \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 25 \, \sqrt {2} a \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 128 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, {\left (13 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 9 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )}}{a x + 1} + \frac {{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + 1}\right )} a \]

[In]

integrate(((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="maxima")

[Out]

1/16*(50*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 50*sqrt(2)*a*arctan(-1/2*sq
rt(2)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) + 25*sqrt(2)*a*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt
((a*x - 1)/(a*x + 1)) + 1) - 25*sqrt(2)*a*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1))
+ 1) - 128*a*((a*x - 1)/(a*x + 1))^(1/4) - 8*(13*a*((a*x - 1)/(a*x + 1))^(5/4) + 9*a*((a*x - 1)/(a*x + 1))^(1
/4))/(2*(a*x - 1)/(a*x + 1) + (a*x - 1)^2/(a*x + 1)^2 + 1))*a

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{16} \, {\left (50 \, \sqrt {2} a \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 50 \, \sqrt {2} a \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 25 \, \sqrt {2} a \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 25 \, \sqrt {2} a \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 128 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, {\left (\frac {13 \, {\left (a x - 1\right )} a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} + 9 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{{\left (\frac {a x - 1}{a x + 1} + 1\right )}^{2}}\right )} a \]

[In]

integrate(((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="giac")

[Out]

1/16*(50*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 50*sqrt(2)*a*arctan(-1/2*sq
rt(2)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) + 25*sqrt(2)*a*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt
((a*x - 1)/(a*x + 1)) + 1) - 25*sqrt(2)*a*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1))
+ 1) - 128*a*((a*x - 1)/(a*x + 1))^(1/4) - 8*(13*(a*x - 1)*a*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1) + 9*a*((a*
x - 1)/(a*x + 1))^(1/4))/((a*x - 1)/(a*x + 1) + 1)^2)*a

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^3} \, dx=-8\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-\frac {\frac {9\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{2}+\frac {13\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}}{\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+\frac {2\,\left (a\,x-1\right )}{a\,x+1}+1}-\frac {{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )\,25{}\mathrm {i}}{4}-\frac {25\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )}{4} \]

[In]

int(((a*x - 1)/(a*x + 1))^(5/4)/x^3,x)

[Out]

- 8*a^2*((a*x - 1)/(a*x + 1))^(1/4) - ((9*a^2*((a*x - 1)/(a*x + 1))^(1/4))/2 + (13*a^2*((a*x - 1)/(a*x + 1))^(
5/4))/2)/((a*x - 1)^2/(a*x + 1)^2 + (2*(a*x - 1))/(a*x + 1) + 1) - ((-1)^(1/4)*a^2*atan((-1)^(1/4)*((a*x - 1)/
(a*x + 1))^(1/4))*25i)/4 - (25*(-1)^(1/4)*a^2*atan((-1)^(1/4)*((a*x - 1)/(a*x + 1))^(1/4)*1i))/4