\(\int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx\) [131]

Optimal result

Integrand size = 12, antiderivative size = 155 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-\frac {1}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )-\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-\frac {1}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )-\frac {3}{2} \log \left (1+\frac {\sqrt [3]{1-\frac {1}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )-\frac {3}{2} \log \left (\sqrt [3]{1-\frac {1}{x}}-\sqrt [3]{1+\frac {1}{x}}\right )-\frac {1}{2} \log \left (1+\frac {1}{x}\right )-\frac {\log (x)}{2} \] Output:

3^(1/2)*arctan(-1/3*3^(1/2)+2/3*(1-1/x)^(1/3)*3^(1/2)/(1+1/x)^(1/3))-arcta 
n(1/3*3^(1/2)+2/3*(1-1/x)^(1/3)*3^(1/2)/(1+1/x)^(1/3))*3^(1/2)-3/2*ln(1+(1 
-1/x)^(1/3)/(1+1/x)^(1/3))-3/2*ln((1-1/x)^(1/3)-(1+1/x)^(1/3))-1/2*ln(1+1/ 
x)-1/2*ln(x)
 

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.17 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\frac {3}{2} e^{\frac {8}{3} \coth ^{-1}(x)} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},e^{4 \coth ^{-1}(x)}\right ) \] Input:

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

Output:

(3*E^((8*ArcCoth[x])/3)*Hypergeometric2F1[2/3, 1, 5/3, E^(4*ArcCoth[x])])/ 
2
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6721, 140, 72, 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.

\(\displaystyle \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx\)

\(\Big \downarrow \) 6721

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

\(\Big \downarrow \) 140

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

\(\Big \downarrow \) 72

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

\(\Big \downarrow \) 102

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 72

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> 
 With[{q = Rt[-d/b, 3]}, Simp[Sqrt[3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a + b* 
x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 

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 140

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

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 [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.58 (sec) , antiderivative size = 1039, normalized size of antiderivative = 6.70

method	result	size
trager	\(\text {Expression too large to display}\)	\(1039\)

Input:

int(1/((x-1)/(1+x))^(1/3)/x,x,method=_RETURNVERBOSE)
 

Output:

3*RootOf(9*_Z^2-3*_Z+1)*ln(-(-945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/ 
3)*x^2+1224*RootOf(9*_Z^2-3*_Z+1)^2*x^2-1890*RootOf(9*_Z^2-3*_Z+1)*(-(1-x) 
/(1+x))^(2/3)*x-945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)*x^2+168*(-( 
1-x)/(1+x))^(2/3)*x^2-3060*RootOf(9*_Z^2-3*_Z+1)^2*x-945*RootOf(9*_Z^2-3*_ 
Z+1)*(-(1-x)/(1+x))^(2/3)-1353*RootOf(9*_Z^2-3*_Z+1)*x^2+336*(-(1-x)/(1+x) 
)^(2/3)*x+168*(-(1-x)/(1+x))^(1/3)*x^2+1224*RootOf(9*_Z^2-3*_Z+1)^2+945*Ro 
otOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)+1062*RootOf(9*_Z^2-3*_Z+1)*x+168* 
(-(1-x)/(1+x))^(2/3)+304*x^2-1353*RootOf(9*_Z^2-3*_Z+1)-168*(-(1-x)/(1+x)) 
^(1/3)-32*x+304)/x)-3*ln(-(945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)* 
x^2+1224*RootOf(9*_Z^2-3*_Z+1)^2*x^2+1890*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1 
+x))^(2/3)*x+945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)*x^2-147*(-(1-x 
)/(1+x))^(2/3)*x^2-3060*RootOf(9*_Z^2-3*_Z+1)^2*x+945*RootOf(9*_Z^2-3*_Z+1 
)*(-(1-x)/(1+x))^(2/3)+537*RootOf(9*_Z^2-3*_Z+1)*x^2-294*(-(1-x)/(1+x))^(2 
/3)*x-147*(-(1-x)/(1+x))^(1/3)*x^2+1224*RootOf(9*_Z^2-3*_Z+1)^2-945*RootOf 
(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)+978*RootOf(9*_Z^2-3*_Z+1)*x-147*(-(1- 
x)/(1+x))^(2/3)-11*x^2+537*RootOf(9*_Z^2-3*_Z+1)+147*(-(1-x)/(1+x))^(1/3)- 
18*x-11)/x)*RootOf(9*_Z^2-3*_Z+1)+ln(-(945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/( 
1+x))^(2/3)*x^2+1224*RootOf(9*_Z^2-3*_Z+1)^2*x^2+1890*RootOf(9*_Z^2-3*_Z+1 
)*(-(1-x)/(1+x))^(2/3)*x+945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)*x^ 
2-147*(-(1-x)/(1+x))^(2/3)*x^2-3060*RootOf(9*_Z^2-3*_Z+1)^2*x+945*RootO...
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - 1\right ) + \frac {1}{2} \, \log \left (\frac {{\left (x + 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + {\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + x + 1}{x + 1}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right )}\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right ) \] Input:

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

Output:

-sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) + 1)) + sqrt(3)*arc 
tan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) - 1)) + 1/2*log(((x - 1)/(x + 1 
))^(2/3) + ((x - 1)/(x + 1))^(1/3) + 1) + 1/2*log(((x - 1)/(x + 1))^(2/3) 
- ((x - 1)/(x + 1))^(1/3) + 1) - log(((x - 1)/(x + 1))^(1/3) + 1) - log((( 
x - 1)/(x + 1))^(1/3) - 1)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.51 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \frac {{\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}}}{x + 1} + 1\right ) - \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - 1 \right |}\right ) \] Input:

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

Output:

sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(2/3) + 1)) + 1/2*log(((x 
- 1)/(x + 1))^(2/3) + (x - 1)*((x - 1)/(x + 1))^(1/3)/(x + 1) + 1) - log(a 
bs(((x - 1)/(x + 1))^(2/3) - 1))
 

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}-1296\right )-\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}+648-\sqrt {3}\,648{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}+648+\sqrt {3}\,648{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right ) \] Input:

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

Output:

log(3^(1/2)*648i + 1296*((x - 1)/(x + 1))^(2/3) + 648)*((3^(1/2)*1i)/2 + 1 
/2) - log(1296*((x - 1)/(x + 1))^(2/3) - 3^(1/2)*648i + 648)*((3^(1/2)*1i) 
/2 - 1/2) - log(1296*((x - 1)/(x + 1))^(2/3) - 1296)
 

Reduce [F]

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

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

Output:

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