\(\int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx\) [70]

Optimal result

Integrand size = 12, antiderivative size = 142 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4} x^2+\frac {\arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \] Output:

1/4*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x/a+1/2*(1-1/a/x)^(3/4)*(1+1/a/x)^(5/4 
)*x^2+1/4*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^2+1/4*arctanh((1+1/a/x 
)^(1/4)/(1-1/a/x)^(1/4))/a^2
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {\frac {2 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (-1+5 e^{2 \coth ^{-1}(a x)}\right )}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+\arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+\text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6721, 107, 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.

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

\(\Big \downarrow \) 6721

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {2 \, {\left (2 \, a^{2} x^{2} + 5 \, a x + 3\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}} - 2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{8 \, a^{2}} \] Input:

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

Output:

1/8*(2*(2*a^2*x^2 + 5*a*x + 3)*((a*x - 1)/(a*x + 1))^(3/4) - 2*arctan(((a* 
x - 1)/(a*x + 1))^(1/4)) + log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - log(((a* 
x - 1)/(a*x + 1))^(1/4) - 1))/a^2
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.05 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {4 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{4}} - 5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}}\right )} \] Input:

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

Output:

1/8*a*(4*(((a*x - 1)/(a*x + 1))^(7/4) - 5*((a*x - 1)/(a*x + 1))^(3/4))/(2* 
(a*x - 1)*a^3/(a*x + 1) - (a*x - 1)^2*a^3/(a*x + 1)^2 - a^3) - 2*arctan((( 
a*x - 1)/(a*x + 1))^(1/4))/a^3 + log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^3 
- log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.98 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {\log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (\frac {{\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - 5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \] Input:

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

Output:

-1/8*a*(2*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 - log(((a*x - 1)/(a*x + 
1))^(1/4) + 1)/a^3 + log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3 + 4*((a 
*x - 1)*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) - 5*((a*x - 1)/(a*x + 1))^(3 
/4))/(a^3*((a*x - 1)/(a*x + 1) - 1)^2))
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85 \[ \int e^{\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{2}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/4}}{2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}-\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}+\frac {\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2} \] Input:

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

Output:

((5*((a*x - 1)/(a*x + 1))^(3/4))/2 - ((a*x - 1)/(a*x + 1))^(7/4)/2)/(a^2 + 
 (a^2*(a*x - 1)^2)/(a*x + 1)^2 - (2*a^2*(a*x - 1))/(a*x + 1)) - atan(((a*x 
 - 1)/(a*x + 1))^(1/4))/(4*a^2) + atanh(((a*x - 1)/(a*x + 1))^(1/4))/(4*a^ 
2)
 

Reduce [F]

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

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

Output:

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