\(\int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx\) [123]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6306, 132, 62, 93} \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}+\frac {1}{\sqrt {3}}\right )-\frac {3}{2} \log \left (\sqrt [3]{\frac {1}{x}+1}-\sqrt [3]{\frac {x-1}{x}}\right )-\frac {3}{2} \log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{2} \log \left (\frac {1}{x}+1\right )-\frac {\log (x)}{2} \]

[In]

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

[Out]

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

Rule 62

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])] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && NegQ[d/b]

Rule 93

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> 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 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[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)*ExpandTo
Sum[(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 6306

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]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x} x} \, dx,x,\frac {1}{x}\right ) \\ & = -\text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )-\text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right ) \\ & = -\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {-1+x}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )-\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{\frac {-1+x}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )-\frac {3}{2} \log \left (\sqrt [3]{1+\frac {1}{x}}-\sqrt [3]{\frac {-1+x}{x}}\right )-\frac {3}{2} \log \left (1+\frac {\sqrt [3]{\frac {-1+x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )-\frac {1}{2} \log \left (1+\frac {1}{x}\right )-\frac {\log (x)}{2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.17 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\frac {3}{2} e^{\frac {8}{3} \coth ^{-1}(x)} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},e^{4 \coth ^{-1}(x)}\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 0.66 (sec) , antiderivative size = 1026, normalized size of antiderivative = 6.62

method	result	size
trager	\(\text {Expression too large to display}\)	\(1026\)

[In]

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

[Out]

-3*ln((945*(-(1-x)/(1+x))^(2/3)*RootOf(9*_Z^2-3*_Z+1)*x^2+1890*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x-16
8*(-(1-x)/(1+x))^(2/3)*x^2-504*(-(1-x)/(1+x))^(1/3)*RootOf(9*_Z^2-3*_Z+1)*x^2+72*RootOf(9*_Z^2-3*_Z+1)^2*x^2+9
45*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)-336*(-(1-x)/(1+x))^(2/3)*x-147*(-(1-x)/(1+x))^(1/3)*x^2-180*Root
Of(9*_Z^2-3*_Z+1)^2*x-465*RootOf(9*_Z^2-3*_Z+1)*x^2-168*(-(1-x)/(1+x))^(2/3)+504*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)
/(1+x))^(1/3)+72*RootOf(9*_Z^2-3*_Z+1)^2+1026*RootOf(9*_Z^2-3*_Z+1)*x+323*x^2+147*(-(1-x)/(1+x))^(1/3)-465*Roo
tOf(9*_Z^2-3*_Z+1)-34*x+323)/x)*RootOf(9*_Z^2-3*_Z+1)-ln(-(945*(-(1-x)/(1+x))^(2/3)*RootOf(9*_Z^2-3*_Z+1)*x^2+
1890*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x-147*(-(1-x)/(1+x))^(2/3)*x^2-441*(-(1-x)/(1+x))^(1/3)*RootOf
(9*_Z^2-3*_Z+1)*x^2-1152*RootOf(9*_Z^2-3*_Z+1)^2*x^2+945*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)-294*(-(1-x
)/(1+x))^(2/3)*x-168*(-(1-x)/(1+x))^(1/3)*x^2+2880*RootOf(9*_Z^2-3*_Z+1)^2*x-120*RootOf(9*_Z^2-3*_Z+1)*x^2-147
*(-(1-x)/(1+x))^(2/3)+441*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)-1152*RootOf(9*_Z^2-3*_Z+1)^2-1884*RootOf(
9*_Z^2-3*_Z+1)*x+187*x^2+168*(-(1-x)/(1+x))^(1/3)-120*RootOf(9*_Z^2-3*_Z+1)+306*x+187)/x)+ln((945*(-(1-x)/(1+x
))^(2/3)*RootOf(9*_Z^2-3*_Z+1)*x^2+1890*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x-168*(-(1-x)/(1+x))^(2/3)*
x^2-504*(-(1-x)/(1+x))^(1/3)*RootOf(9*_Z^2-3*_Z+1)*x^2+72*RootOf(9*_Z^2-3*_Z+1)^2*x^2+945*RootOf(9*_Z^2-3*_Z+1
)*(-(1-x)/(1+x))^(2/3)-336*(-(1-x)/(1+x))^(2/3)*x-147*(-(1-x)/(1+x))^(1/3)*x^2-180*RootOf(9*_Z^2-3*_Z+1)^2*x-4
65*RootOf(9*_Z^2-3*_Z+1)*x^2-168*(-(1-x)/(1+x))^(2/3)+504*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)+72*RootOf
(9*_Z^2-3*_Z+1)^2+1026*RootOf(9*_Z^2-3*_Z+1)*x+323*x^2+147*(-(1-x)/(1+x))^(1/3)-465*RootOf(9*_Z^2-3*_Z+1)-34*x
+323)/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - 1\right ) + \frac {1}{2} \, \log \left (\frac {{\left (x + 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + {\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + x + 1}{x + 1}\right ) \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right )}\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.51 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \frac {{\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}}}{x + 1} + 1\right ) - \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

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(abs(((x - 1)/(x + 1))^(2/3) - 1))

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x} \, dx=-\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}-1296\right )-\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}+648-\sqrt {3}\,648{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+\ln \left (1296\,{\left (\frac {x-1}{x+1}\right )}^{2/3}+648+\sqrt {3}\,648{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right ) \]

[In]

int(1/(x*((x - 1)/(x + 1))^(1/3)),x)

[Out]

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)