\(\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx\) [115]

Optimal result

Integrand size = 12, antiderivative size = 176 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {25 \sqrt [4]{1-\frac {1}{a x}}}{2 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {5 \left (1-\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {25 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \] Output:

-25/2*(1-1/a/x)^(1/4)/a^2/(1+1/a/x)^(1/4)-5/4*(1-1/a/x)^(5/4)*x/a/(1+1/a/x 
)^(1/4)+1/2*(1-1/a/x)^(9/4)*x^2/(1+1/a/x)^(1/4)-25/4*arctan((1+1/a/x)^(1/4 
)/(1-1/a/x)^(1/4))/a^2+25/4*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^2
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.69 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {e^{-\frac {1}{2} \coth ^{-1}(a x)} \left (32-90 e^{2 \coth ^{-1}(a x)}+50 e^{4 \coth ^{-1}(a x)}+25 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (-1+e^{2 \coth ^{-1}(a x)}\right )^2 \arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-25 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (-1+e^{2 \coth ^{-1}(a x)}\right )^2 \text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )\right )}{4 a^2 \left (-1+e^{2 \coth ^{-1}(a x)}\right )^2} \] Input:

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

Output:

-1/4*(32 - 90*E^(2*ArcCoth[a*x]) + 50*E^(4*ArcCoth[a*x]) + 25*E^(ArcCoth[a 
*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2*ArcTan[E^(ArcCoth[a*x]/2)] - 25*E^(ArcC 
oth[a*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2*ArcTanh[E^(ArcCoth[a*x]/2)])/(a^2* 
E^(ArcCoth[a*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6721, 107, 105, 105, 104, 25, 827, 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.

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

\(\Big \downarrow \) 6721

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

\(\Big \downarrow \) 107

\(\displaystyle \frac {5 \int \frac {\left (1-\frac {1}{a x}\right )^{5/4} x^2}{\left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {5 \left (-\frac {5 \int \frac {\sqrt [4]{1-\frac {1}{a x}} x}{\left (1+\frac {1}{a x}\right )^{5/4}}d\frac {1}{x}}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {5 \left (-\frac {5 \left (\int \frac {x}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}+\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {5 \left (-\frac {5 \left (4 \int -\frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}+\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5 \left (-\frac {5 \left (\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}-4 \int \frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {5 \left (-\frac {5 \left (4 \left (\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )+\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {5 \left (-\frac {5 \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )+\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (-\frac {5 \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )+\frac {4 \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{2 a}-\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 a}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}\)

Input:

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

Output:

((1 - 1/(a*x))^(9/4)*x^2)/(2*(1 + 1/(a*x))^(1/4)) + (5*(-(((1 - 1/(a*x))^( 
5/4)*x)/(1 + 1/(a*x))^(1/4)) - (5*((4*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^( 
1/4) + 4*(ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2 - ArcTanh[(1 + 
 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2)))/(2*a)))/(4*a)
 

Defintions of rubi rules used

rule 25

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

rule 104

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]
 

rule 105

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

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(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[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 216

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])
 

rule 219

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])
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.54 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {2 \, {\left (2 \, a^{2} x^{2} - 9 \, a x - 43\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{8 \, a^{2}} \] Input:

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

Output:

1/8*(2*(2*a^2*x^2 - 9*a*x - 43)*((a*x - 1)/(a*x + 1))^(1/4) + 50*arctan((( 
a*x - 1)/(a*x + 1))^(1/4)) + 25*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - 25* 
log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {4 \, {\left (13 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}} + \frac {64 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{3}}\right )} \] Input:

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

Output:

-1/8*a*(4*(13*((a*x - 1)/(a*x + 1))^(5/4) - 9*((a*x - 1)/(a*x + 1))^(1/4)) 
/(2*(a*x - 1)*a^3/(a*x + 1) - (a*x - 1)^2*a^3/(a*x + 1)^2 - a^3) - 50*arct 
an(((a*x - 1)/(a*x + 1))^(1/4))/a^3 - 25*log(((a*x - 1)/(a*x + 1))^(1/4) + 
 1)/a^3 + 25*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3 + 64*((a*x - 1)/(a*x 
 + 1))^(1/4)/a^3)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {25 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} - \frac {64 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{3}} + \frac {4 \, {\left (\frac {13 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \] Input:

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

Output:

1/8*a*(50*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 + 25*log(((a*x - 1)/(a*x 
 + 1))^(1/4) + 1)/a^3 - 25*log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3 - 
 64*((a*x - 1)/(a*x + 1))^(1/4)/a^3 + 4*(13*(a*x - 1)*((a*x - 1)/(a*x + 1) 
)^(1/4)/(a*x + 1) - 9*((a*x - 1)/(a*x + 1))^(1/4))/(a^3*((a*x - 1)/(a*x + 
1) - 1)^2))
 

Mupad [B] (verification not implemented)

Time = 24.03 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.82 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {25\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a^2}-\frac {\frac {9\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{2}-\frac {13\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}-\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{4\,a^2} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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