\(\int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx\) [110]

Optimal result

Integrand size = 14, antiderivative size = 299 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {5 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}-\frac {5 a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}+\frac {5 a \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{2 \sqrt {2}}-\frac {5 a \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{2 \sqrt {2}} \]

[Out]

4*a*(1-1/a/x)^(5/4)/(1+1/a/x)^(1/4)+5*a*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)-5/2*a*arctan(-1+(1-1/a/x)^(1/4)*2^(1/2
)/(1+1/a/x)^(1/4))*2^(1/2)-5/2*a*arctan(1+(1-1/a/x)^(1/4)*2^(1/2)/(1+1/a/x)^(1/4))*2^(1/2)+5/4*a*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)-5/4*a*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 = 0.19 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6306, 49, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=\frac {5 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{\sqrt {2}}-\frac {5 a \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{\sqrt {2}}+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}+5 a \left (\frac {1}{a x}+1\right )^{3/4} \sqrt [4]{1-\frac {1}{a x}}+\frac {5 a \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{2 \sqrt {2}}-\frac {5 a \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{2 \sqrt {2}} \]

[In]

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

[Out]

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

Rule 49

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

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 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 {\left (1-\frac {x}{a}\right )^{5/4}}{\left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 \text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{\sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {5}{2} \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 {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-(10 a) \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-\frac {1}{a x}}\right ) \\ & = \frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-(10 a) \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 {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-(5 a) \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 )-(5 a) \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 {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}-\frac {1}{2} (5 a) \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}{2} (5 a) \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 {(5 a) \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 )}{2 \sqrt {2}}+\frac {(5 a) \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 )}{2 \sqrt {2}} \\ & = \frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {5 a \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 )}{2 \sqrt {2}}-\frac {5 a \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 )}{2 \sqrt {2}}-\frac {(5 a) \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 )}{\sqrt {2}}+\frac {(5 a) \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 )}{\sqrt {2}} \\ & = \frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {5 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}-\frac {5 a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}+\frac {5 a \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 )}{2 \sqrt {2}}-\frac {5 a \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 )}{2 \sqrt {2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.10 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=8 a e^{-\frac {1}{2} \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},-e^{2 \coth ^{-1}(a x)}\right ) \]

[In]

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

[Out]

(8*a*Hypergeometric2F1[-1/4, 2, 3/4, -E^(2*ArcCoth[a*x])])/E^(ArcCoth[a*x]/2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {5 \, \left (-a^{4}\right )^{\frac {1}{4}} x \log \left (5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 5 \, \left (-a^{4}\right )^{\frac {1}{4}}\right ) + 5 i \, \left (-a^{4}\right )^{\frac {1}{4}} x \log \left (5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 5 i \, \left (-a^{4}\right )^{\frac {1}{4}}\right ) - 5 i \, \left (-a^{4}\right )^{\frac {1}{4}} x \log \left (5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 5 i \, \left (-a^{4}\right )^{\frac {1}{4}}\right ) - 5 \, \left (-a^{4}\right )^{\frac {1}{4}} x \log \left (5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 5 \, \left (-a^{4}\right )^{\frac {1}{4}}\right ) - 2 \, {\left (9 \, a x + 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{2 \, x} \]

[In]

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

[Out]

-1/2*(5*(-a^4)^(1/4)*x*log(5*a*((a*x - 1)/(a*x + 1))^(1/4) + 5*(-a^4)^(1/4)) + 5*I*(-a^4)^(1/4)*x*log(5*a*((a*
x - 1)/(a*x + 1))^(1/4) + 5*I*(-a^4)^(1/4)) - 5*I*(-a^4)^(1/4)*x*log(5*a*((a*x - 1)/(a*x + 1))^(1/4) - 5*I*(-a
^4)^(1/4)) - 5*(-a^4)^(1/4)*x*log(5*a*((a*x - 1)/(a*x + 1))^(1/4) - 5*(-a^4)^(1/4)) - 2*(9*a*x + 1)*((a*x - 1)
/(a*x + 1))^(1/4))/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{4} \, {\left (10 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 10 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 5 \, \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 32 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {a x - 1}{a x + 1} + 1}\right )} a \]

[In]

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

[Out]

-1/4*(10*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 10*sqrt(2)*arctan(-1/2*sqrt(2
)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) + 5*sqrt(2)*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x -
1)/(a*x + 1)) + 1) - 5*sqrt(2)*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)) + 1) - 32
*((a*x - 1)/(a*x + 1))^(1/4) - 8*((a*x - 1)/(a*x + 1))^(1/4)/((a*x - 1)/(a*x + 1) + 1))*a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{4} \, {\left (10 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 10 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 5 \, \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 32 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {a x - 1}{a x + 1} + 1}\right )} a \]

[In]

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

[Out]

-1/4*(10*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 10*sqrt(2)*arctan(-1/2*sqrt(2
)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) + 5*sqrt(2)*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x -
1)/(a*x + 1)) + 1) - 5*sqrt(2)*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)) + 1) - 32
*((a*x - 1)/(a*x + 1))^(1/4) - 8*((a*x - 1)/(a*x + 1))^(1/4)/((a*x - 1)/(a*x + 1) + 1))*a

Mupad [B] (verification not implemented)

Time = 4.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.35 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=8\,a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}+5\,{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )+\frac {2\,a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{\frac {a\,x-1}{a\,x+1}+1}+{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )\,5{}\mathrm {i} \]

[In]

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

[Out]

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