\(\int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx\) [118]

Optimal result

Integrand size = 14, antiderivative size = 224 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{1+\frac {1}{a x}}}+5 a \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4}+\frac {5 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}-\frac {5 a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}}-\frac {5 a \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}\right ) \sqrt [4]{1+\frac {1}{a x}}}\right )}{\sqrt {2}} \] Output:

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+2^(1/2)*(1-1/a/x)^(1/4)/(1+1/a/x)^(1/4))*2^(1/2)-5/2*a*arcta 
n(1+2^(1/2)*(1-1/a/x)^(1/4)/(1+1/a/x)^(1/4))*2^(1/2)-5/2*a*arctanh(2^(1/2) 
*(1-1/a/x)^(1/4)/(1+(1-1/a/x)^(1/2)/(1+1/a/x)^(1/2))/(1+1/a/x)^(1/4))*2^(1 
/2)
 

Mathematica [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.14 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=8 a e^{-\frac {1}{2} \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},-e^{2 \coth ^{-1}(a x)}\right ) \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {6721, 57, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 6721

\(\displaystyle -\int \frac {\left (1-\frac {1}{a x}\right )^{5/4}}{\left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

\(\displaystyle 5 \left (\frac {1}{2} \int \frac {1}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}+a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 73

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \int \frac {1}{\sqrt [4]{2-\frac {1}{x^4}}}d\sqrt [4]{1-\frac {1}{a x}}\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 770

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \int \frac {1}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 755

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \int \frac {1-\frac {1}{x^2}}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{2} \int \frac {1+\frac {1}{x^2}}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 1476

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \int \frac {1-\frac {1}{x^2}}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 1082

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-\frac {1}{x^2}}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-\frac {1}{x^2}}d\left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\frac {1}{x^2}}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 217

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \int \frac {1-\frac {1}{x^2}}{1+\frac {1}{x^4}}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 1479

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1}{\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1}d\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 1103

\(\displaystyle 5 \left (a \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-2 a \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{2-\frac {1}{x^4}}}+\frac {1}{x^2}+1\right )}{2 \sqrt {2}}\right )\right )\right )+\frac {4 a \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 57

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] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& 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 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

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 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[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 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

rule 6721

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {10 \, \sqrt {2} a x \arctan \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 10 \, \sqrt {2} a x \arctan \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) + 5 \, \sqrt {2} a x \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, \sqrt {2} a x \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 4 \, {\left (9 \, a x + 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{4 \, x} \] Input:

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

Output:

-1/4*(10*sqrt(2)*a*x*arctan(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + 1) + 10* 
sqrt(2)*a*x*arctan(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) - 1) + 5*sqrt(2)*a* 
x*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)) + 1) 
 - 5*sqrt(2)*a*x*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1) 
/(a*x + 1)) + 1) - 4*(9*a*x + 1)*((a*x - 1)/(a*x + 1))^(1/4))/x
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{4} \, {\left (10 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 10 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 5 \, \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 32 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {a x - 1}{a x + 1} + 1}\right )} a \] Input:

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

Output:

-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) + s 
qrt((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)

Time = 0.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{4} \, {\left (10 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 10 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 5 \, \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 32 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \frac {8 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {a x - 1}{a x + 1} + 1}\right )} a \] Input:

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

Output:

-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) + s 
qrt((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 = 23.66 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.47 \[ \int \frac {e^{-\frac {5}{2} \coth ^{-1}(a x)}}{x^2} \, dx=8\,a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}+5\,{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )+\frac {2\,a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{\frac {a\,x-1}{a\,x+1}+1}+{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )\,5{}\mathrm {i} \] Input:

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

Output:

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)
 

Reduce [F]

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

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

Output:

 - int((a*x - 1)**(1/4)/((a*x + 1)**(1/4)*a*x**3 + (a*x + 1)**(1/4)*x**2), 
x) + int((a*x - 1)**(1/4)/((a*x + 1)**(1/4)*a*x**2 + (a*x + 1)**(1/4)*x),x 
)*a