\(\int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx\) [125]

Optimal result

Integrand size = 12, antiderivative size = 116 \[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=-\frac {(1-x)^{2/3} \sqrt [3]{1+x}}{3 x}-\frac {(1-x)^{2/3} (1+x)^{4/3}}{2 x^2}+\frac {2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{1+x}}\right )}{3 \sqrt {3}}-\frac {\log (x)}{9}+\frac {1}{3} \log \left (\sqrt [3]{1-x}-\sqrt [3]{1+x}\right ) \] Output:

-1/3*(1-x)^(2/3)*(1+x)^(1/3)/x-1/2*(1-x)^(2/3)*(1+x)^(4/3)/x^2+2/9*3^(1/2) 
*arctan(1/3*3^(1/2)+2/3*(1-x)^(1/3)*3^(1/2)/(1+x)^(1/3))-1/9*ln(x)+1/3*ln( 
(1-x)^(1/3)-(1+x)^(1/3))
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.49 \[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=-\frac {(1-x)^{2/3} \left (3+8 x+5 x^2+2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {1-x}{1+x}\right )\right )}{6 x^2 (1+x)^{2/3}} \] Input:

Integrate[E^((2*ArcTanh[x])/3)/x^3,x]
 

Output:

-1/6*((1 - x)^(2/3)*(3 + 8*x + 5*x^2 + 2*x^2*Hypergeometric2F1[2/3, 1, 5/3 
, (1 - x)/(1 + x)]))/(x^2*(1 + x)^(2/3))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6676, 107, 105, 102}

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.

\(\displaystyle \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx\)

\(\Big \downarrow \) 6676

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

\(\Big \downarrow \) 107

\(\displaystyle \frac {1}{3} \int \frac {\sqrt [3]{x+1}}{\sqrt [3]{1-x} x^2}dx-\frac {(1-x)^{2/3} (x+1)^{4/3}}{2 x^2}\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 102

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

Input:

Int[E^((2*ArcTanh[x])/3)/x^3,x]
 

Output:

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

Defintions of rubi rules used

rule 102

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.) 
*(x_))), x_] :> With[{q = Rt[(d*e - c*f)/(b*e - a*f), 3]}, Simp[(-Sqrt[3])* 
q*(ArcTan[1/Sqrt[3] + 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))]/(d*e 
 - c*f)), x] + (Simp[q*(Log[e + f*x]/(2*(d*e - c*f))), x] - Simp[3*q*(Log[q 
*(a + b*x)^(1/3) - (c + d*x)^(1/3)]/(2*(d*e - c*f))), x])] /; FreeQ[{a, b, 
c, d, e, f}, 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 6676

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x) 
^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, m, n}, x] &&  !Int 
egerQ[(n - 1)/2]
 

Maple [F]

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

Input:

int(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x)
 

Output:

int(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=-\frac {4 \, \sqrt {3} x^{2} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - 4 \, x^{2} \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - 1\right ) + 2 \, x^{2} \log \left (\frac {{\left (x - 1\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + x - \sqrt {-x^{2} + 1} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} - 1}{x - 1}\right ) - 3 \, {\left (5 \, x^{2} - 2 \, x - 3\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}}}{18 \, x^{2}} \] Input:

integrate(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x, algorithm="fricas")
 

Output:

-1/18*(4*sqrt(3)*x^2*arctan(2/3*sqrt(3)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) + 
1/3*sqrt(3)) - 4*x^2*log((-sqrt(-x^2 + 1)/(x - 1))^(2/3) - 1) + 2*x^2*log( 
((x - 1)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) + x - sqrt(-x^2 + 1)*(-sqrt(-x^2 
+ 1)/(x - 1))^(1/3) - 1)/(x - 1)) - 3*(5*x^2 - 2*x - 3)*(-sqrt(-x^2 + 1)/( 
x - 1))^(2/3))/x^2
 

Sympy [F]

\[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=\int \frac {\left (\frac {x + 1}{\sqrt {1 - x^{2}}}\right )^{\frac {2}{3}}}{x^{3}}\, dx \] Input:

integrate(((1+x)/(-x**2+1)**(1/2))**(2/3)/x**3,x)
 

Output:

Integral(((x + 1)/sqrt(1 - x**2))**(2/3)/x**3, x)
 

Maxima [F]

\[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=\int { \frac {\left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{3}} \,d x } \] Input:

integrate(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x, algorithm="maxima")
 

Output:

integrate(((x + 1)/sqrt(-x^2 + 1))^(2/3)/x^3, x)
 

Giac [F]

\[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=\int { \frac {\left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{3}} \,d x } \] Input:

integrate(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x, algorithm="giac")
 

Output:

integrate(((x + 1)/sqrt(-x^2 + 1))^(2/3)/x^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {2 \text {arctanh}(x)}{3}}}{x^3} \, dx=\int \frac {{\left (\frac {x+1}{\sqrt {1-x^2}}\right )}^{2/3}}{x^3} \,d x \] Input:

int(((x + 1)/(1 - x^2)^(1/2))^(2/3)/x^3,x)
 

Output:

int(((x + 1)/(1 - x^2)^(1/2))^(2/3)/x^3, x)
 

Reduce [F]

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

int(((1+x)/(-x^2+1)^(1/2))^(2/3)/x^3,x)
 

Output:

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