\(\int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx\) [124]

Optimal result

Integrand size = 12, antiderivative size = 99 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {-1+x}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )}{\sqrt {3}}-\log \left (1+\frac {\sqrt [3]{\frac {-1+x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )-\frac {1}{3} \log \left (1+\frac {1}{x}\right ) \]

[Out]

(1+1/x)^(1/3)*((-1+x)/x)^(2/3)-ln(1+((-1+x)/x)^(1/3)/(1+1/x)^(1/3))-1/3*ln(1+1/x)+2/3*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 = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6306, 52, 62} \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )}{\sqrt {3}}+\sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3}-\log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{3} \log \left (\frac {1}{x}+1\right ) \]

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2 e^{\frac {2}{3} \coth ^{-1}(x)}}{1+e^{2 \coth ^{-1}(x)}}-\frac {2 \arctan \left (\frac {-1+2 e^{\frac {2}{3} \coth ^{-1}(x)}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log \left (1+e^{\frac {2}{3} \coth ^{-1}(x)}\right )+\frac {1}{3} \log \left (1-e^{\frac {2}{3} \coth ^{-1}(x)}+e^{\frac {4}{3} \coth ^{-1}(x)}\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 0.76 (sec) , antiderivative size = 517, normalized size of antiderivative = 5.22

method	result	size
trager	\(\frac {\left (1+x \right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}}}{x}-\frac {2 \ln \left (\frac {9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}} x +9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}}-3 \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}} x +9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}} x -3 \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}}+9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}}-36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+6 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +12 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-x -1}{x}\right )}{3}+\frac {2 \ln \left (-\frac {9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}} x +9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}}+3 \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}} x -18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +3 \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-x +1}{x}\right )}{3}-2 \ln \left (-\frac {9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}} x +9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (-\frac {1-x}{1+x}\right )^{\frac {2}{3}}+3 \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}} x -18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +3 \left (-\frac {1-x}{1+x}\right )^{\frac {1}{3}}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-x +1}{x}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )\)	\(517\)
risch	\(\text {Expression too large to display}\)	\(733\)

[In]

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

[Out]

(1+x)/x*(-(1-x)/(1+x))^(2/3)-2/3*ln((9*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x+9*RootOf(9*_Z^2-3*_Z+1)*(-
(1-x)/(1+x))^(2/3)-3*(-(1-x)/(1+x))^(2/3)*x+9*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)*x-3*(-(1-x)/(1+x))^(2
/3)+9*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(1/3)-36*RootOf(9*_Z^2-3*_Z+1)^2+6*RootOf(9*_Z^2-3*_Z+1)*x+12*RootO
f(9*_Z^2-3*_Z+1)-x-1)/x)+2/3*ln(-(9*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x+9*RootOf(9*_Z^2-3*_Z+1)*(-(1-
x)/(1+x))^(2/3)+3*(-(1-x)/(1+x))^(1/3)*x-18*RootOf(9*_Z^2-3*_Z+1)^2-3*RootOf(9*_Z^2-3*_Z+1)*x+3*(-(1-x)/(1+x))
^(1/3)-3*RootOf(9*_Z^2-3*_Z+1)-x+1)/x)-2*ln(-(9*RootOf(9*_Z^2-3*_Z+1)*(-(1-x)/(1+x))^(2/3)*x+9*RootOf(9*_Z^2-3
*_Z+1)*(-(1-x)/(1+x))^(2/3)+3*(-(1-x)/(1+x))^(1/3)*x-18*RootOf(9*_Z^2-3*_Z+1)^2-3*RootOf(9*_Z^2-3*_Z+1)*x+3*(-
(1-x)/(1+x))^(1/3)-3*RootOf(9*_Z^2-3*_Z+1)-x+1)/x)*RootOf(9*_Z^2-3*_Z+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2 \, \sqrt {3} x \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + x \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - 2 \, x \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + 3 \, {\left (x + 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{3 \, x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{3} \, \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2\,{\left (\frac {x-1}{x+1}\right )}^{2/3}}{\frac {x-1}{x+1}+1}-\ln \left (9\,{\left (-\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )}^2+4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )+\ln \left (9\,{\left (\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )}^2+4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )-\frac {2\,\ln \left (4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}+4\right )}{3} \]

[In]

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

[Out]

log(9*((3^(1/2)*1i)/3 + 1/3)^2 + 4*((x - 1)/(x + 1))^(1/3))*((3^(1/2)*1i)/3 + 1/3) - log(9*((3^(1/2)*1i)/3 - 1
/3)^2 + 4*((x - 1)/(x + 1))^(1/3))*((3^(1/2)*1i)/3 - 1/3) - (2*log(4*((x - 1)/(x + 1))^(1/3) + 4))/3 + (2*((x
- 1)/(x + 1))^(2/3))/((x - 1)/(x + 1) + 1)