\(\int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx\) [88]

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 = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {-\frac {2 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (25-45 e^{2 \coth ^{-1}(a x)}+16 e^{4 \coth ^{-1}(a x)}\right )}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+25 \arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+25 \text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \] Input:

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

Output:

((-2*E^(ArcCoth[a*x]/2)*(25 - 45*E^(2*ArcCoth[a*x]) + 16*E^(4*ArcCoth[a*x] 
)))/(-1 + E^(2*ArcCoth[a*x]))^2 + 25*ArcTan[E^(ArcCoth[a*x]/2)] + 25*ArcTa 
nh[E^(ArcCoth[a*x]/2)])/(4*a^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6721, 107, 105, 105, 104, 756, 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

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[E^((5*ArcCoth[a*x])/2)*x,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*(-1/2*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)] - ArcTanh[( 
1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2)))/(2*a)))/(4*a)
 

Defintions of rubi rules used

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 756

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.73 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {50 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 25 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 25 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (2 \, a^{3} x^{3} + 11 \, a^{2} x^{2} - 34 \, a x - 43\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{8 \, {\left (a^{3} x - a^{2}\right )}} \] Input:

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

Output:

-1/8*(50*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 25*(a*x - 1)*log( 
((a*x - 1)/(a*x + 1))^(1/4) + 1) + 25*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^ 
(1/4) - 1) - 2*(2*a^3*x^3 + 11*a^2*x^2 - 34*a*x - 43)*((a*x - 1)/(a*x + 1) 
)^(3/4))/(a^3*x - a^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.94 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {4 \, {\left (\frac {45 \, {\left (a x - 1\right )}}{a x + 1} - \frac {25 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 16\right )}}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 2 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} - \frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}}\right )} \] Input:

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

Output:

1/8*a*(4*(45*(a*x - 1)/(a*x + 1) - 25*(a*x - 1)^2/(a*x + 1)^2 - 16)/(a^3*( 
(a*x - 1)/(a*x + 1))^(9/4) - 2*a^3*((a*x - 1)/(a*x + 1))^(5/4) + a^3*((a*x 
 - 1)/(a*x + 1))^(1/4)) - 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(((a*x - 1)/(a*x + 1))^(1 
/4) - 1)/a^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {25 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} + \frac {64}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {4 \, {\left (\frac {9 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - 13 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x,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^3*((a*x - 1)/(a*x + 1))^(1/4)) + 4*(9*(a*x - 1)*((a*x - 1)/(a*x + 
1))^(3/4)/(a*x + 1) - 13*((a*x - 1)/(a*x + 1))^(3/4))/(a^3*((a*x - 1)/(a*x 
 + 1) - 1)^2))
 

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {25\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {25\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {\frac {25\,{\left (a\,x-1\right )}^2}{2\,{\left (a\,x+1\right )}^2}-\frac {45\,\left (a\,x-1\right )}{2\,\left (a\,x+1\right )}+8}{a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-2\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}+a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}} \] Input:

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

Output:

(25*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/(4*a^2) - (25*atan(((a*x - 1)/(a*x 
 + 1))^(1/4)))/(4*a^2) - ((25*(a*x - 1)^2)/(2*(a*x + 1)^2) - (45*(a*x - 1) 
)/(2*(a*x + 1)) + 8)/(a^2*((a*x - 1)/(a*x + 1))^(1/4) - 2*a^2*((a*x - 1)/( 
a*x + 1))^(5/4) + a^2*((a*x - 1)/(a*x + 1))^(9/4))
 

Reduce [F]

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

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