\(\int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx\) [117]

Optimal result

Integrand size = 12, antiderivative size = 233 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=\sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}-\frac {1}{3} \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {1}{3} \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {2}{3} \arctan \left (\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {\log \left (1-\frac {\sqrt {3} \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{\frac {-1+x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}}-\frac {\log \left (1+\frac {\sqrt {3} \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{\frac {-1+x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}} \]

[Out]

(1+1/x)^(1/6)*((-1+x)/x)^(5/6)+2/3*arctan(((-1+x)/x)^(1/6)/(1+1/x)^(1/6))+1/3*arctan(2*((-1+x)/x)^(1/6)/(1+1/x
)^(1/6)-3^(1/2))+1/3*arctan(2*((-1+x)/x)^(1/6)/(1+1/x)^(1/6)+3^(1/2))+1/6*ln(1+((-1+x)/x)^(1/3)/(1+1/x)^(1/3)-
((-1+x)/x)^(1/6)*3^(1/2)/(1+1/x)^(1/6))*3^(1/2)-1/6*ln(1+((-1+x)/x)^(1/3)/(1+1/x)^(1/3)+((-1+x)/x)^(1/6)*3^(1/
2)/(1+1/x)^(1/6))*3^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6306, 52, 65, 338, 301, 648, 632, 210, 642, 209} \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=-\frac {1}{3} \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{\frac {x-1}{x}}}{\sqrt [6]{\frac {1}{x}+1}}\right )+\frac {1}{3} \arctan \left (\frac {2 \sqrt [6]{\frac {x-1}{x}}}{\sqrt [6]{\frac {1}{x}+1}}+\sqrt {3}\right )+\frac {2}{3} \arctan \left (\frac {\sqrt [6]{\frac {x-1}{x}}}{\sqrt [6]{\frac {1}{x}+1}}\right )+\sqrt [6]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{5/6}+\frac {\log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}-\frac {\sqrt {3} \sqrt [6]{\frac {x-1}{x}}}{\sqrt [6]{\frac {1}{x}+1}}+1\right )}{2 \sqrt {3}}-\frac {\log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+\frac {\sqrt {3} \sqrt [6]{\frac {x-1}{x}}}{\sqrt [6]{\frac {1}{x}+1}}+1\right )}{2 \sqrt {3}} \]

[In]

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

[Out]

(1 + x^(-1))^(1/6)*((-1 + x)/x)^(5/6) - ArcTan[Sqrt[3] - (2*((-1 + x)/x)^(1/6))/(1 + x^(-1))^(1/6)]/3 + ArcTan
[Sqrt[3] + (2*((-1 + x)/x)^(1/6))/(1 + x^(-1))^(1/6)]/3 + (2*ArcTan[((-1 + x)/x)^(1/6)/(1 + x^(-1))^(1/6)])/3
+ Log[1 - (Sqrt[3]*((-1 + x)/x)^(1/6))/(1 + x^(-1))^(1/6) + ((-1 + x)/x)^(1/3)/(1 + x^(-1))^(1/3)]/(2*Sqrt[3])
- Log[1 + (Sqrt[3]*((-1 + x)/x)^(1/6))/(1 + x^(-1))^(1/6) + ((-1 + x)/x)^(1/3)/(1 + x^(-1))^(1/3)]/(2*Sqrt[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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 301

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(
2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 +
2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))*Int[1/(r^2 + s^2*x^2), x] +
Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 [6]{1+x}}{\sqrt [6]{1-x}} \, dx,x,\frac {1}{x}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt [6]{1-x} (1+x)^{5/6}} \, dx,x,\frac {1}{x}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}+2 \text {Subst}\left (\int \frac {x^4}{\left (2-x^6\right )^{5/6}} \, dx,x,\sqrt [6]{\frac {-1+x}{x}}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}+2 \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}+\frac {2}{3} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {2}{3} \text {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {2}{3} \text {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}+\frac {2}{3} \arctan \left (\frac {\sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {\text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}} \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}+\frac {2}{3} \arctan \left (\frac {\sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {\log \left (1-\frac {\sqrt {3} \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}}-\frac {\log \left (1+\frac {\sqrt {3} \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+\frac {2 \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+\frac {2 \sqrt [6]{\frac {-1+x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right ) \\ & = \sqrt [6]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{5/6}-\frac {1}{3} \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {1}{3} \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {2}{3} \arctan \left (\frac {\sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}\right )+\frac {\log \left (1-\frac {\sqrt {3} \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}}-\frac {\log \left (1+\frac {\sqrt {3} \sqrt [6]{-\frac {1-x}{x}}}{\sqrt [6]{1+\frac {1}{x}}}+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )}{2 \sqrt {3}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.17 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=-2 e^{\frac {1}{3} \coth ^{-1}(x)} \left (-\frac {1}{1+e^{2 \coth ^{-1}(x)}}+\operatorname {Hypergeometric2F1}\left (\frac {1}{6},1,\frac {7}{6},-e^{2 \coth ^{-1}(x)}\right )\right ) \]

[In]

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

[Out]

-2*E^(ArcCoth[x]/3)*(-(1 + E^(2*ArcCoth[x]))^(-1) + Hypergeometric2F1[1/6, 1, 7/6, -E^(2*ArcCoth[x])])

Maple [C] (verified)

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

Time = 16.72 (sec) , antiderivative size = 1487, normalized size of antiderivative = 6.38

method	result	size
trager	\(\text {Expression too large to display}\)	\(1487\)
risch	\(\text {Expression too large to display}\)	\(2991\)

[In]

int(1/((x-1)/(1+x))^(1/6)/x^2,x,method=_RETURNVERBOSE)

[Out]

(1+x)*(-(1-x)/(1+x))^(5/6)/x-9*RootOf(81*_Z^4-9*_Z^2+1)^3*ln((-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(2
/3)*x-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(2/3)+(-(1-x)/(1+x))^(5/6)*x-3*RootOf(81*_Z^4-9*_Z^2+1)*(-(
1-x)/(1+x))^(2/3)*x+18*x*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/2)-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)
/(1+x))^(1/3)*x+(-(1-x)/(1+x))^(5/6)-3*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(2/3)+18*RootOf(81*_Z^4-9*_Z^2+
1)^2*(-(1-x)/(1+x))^(1/2)-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(1/3)-2*(-(1-x)/(1+x))^(1/2)*x+6*RootOf
(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3)*x-9*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/6)*x+18*RootOf(81*_Z^
4-9*_Z^2+1)^3*x-2*(-(1-x)/(1+x))^(1/2)+6*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3)-9*RootOf(81*_Z^4-9*_Z^2
+1)^2*(-(1-x)/(1+x))^(1/6)-RootOf(81*_Z^4-9*_Z^2+1)*x)/x)+RootOf(81*_Z^4-9*_Z^2+1)*ln((-27*RootOf(81*_Z^4-9*_Z
^2+1)^3*(-(1-x)/(1+x))^(2/3)*x-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(2/3)+(-(1-x)/(1+x))^(5/6)*x-3*Roo
tOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(2/3)*x+18*x*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/2)-27*RootOf(81
*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(1/3)*x+(-(1-x)/(1+x))^(5/6)-3*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(2/3)+
18*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/2)-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(1/3)-2*(-(1-x
)/(1+x))^(1/2)*x+6*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3)*x-9*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))
^(1/6)*x+18*RootOf(81*_Z^4-9*_Z^2+1)^3*x-2*(-(1-x)/(1+x))^(1/2)+6*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3
)-9*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/6)-RootOf(81*_Z^4-9*_Z^2+1)*x)/x)+RootOf(81*_Z^4-9*_Z^2+1)*ln
((54*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(2/3)*x+54*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(2/3)+(-(1
-x)/(1+x))^(5/6)*x-3*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(2/3)*x-18*x*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(
1+x))^(1/2)-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-x)/(1+x))^(1/3)*x+(-(1-x)/(1+x))^(5/6)-3*RootOf(81*_Z^4-9*_Z^2+
1)*(-(1-x)/(1+x))^(2/3)-18*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^(1/2)-27*RootOf(81*_Z^4-9*_Z^2+1)^3*(-(1-
x)/(1+x))^(1/3)+6*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3)*x+9*RootOf(81*_Z^4-9*_Z^2+1)^2*(-(1-x)/(1+x))^
(1/6)*x-9*RootOf(81*_Z^4-9*_Z^2+1)^3*x+6*RootOf(81*_Z^4-9*_Z^2+1)*(-(1-x)/(1+x))^(1/3)+9*RootOf(81*_Z^4-9*_Z^2
+1)^2*(-(1-x)/(1+x))^(1/6)-(-(1-x)/(1+x))^(1/6)*x-RootOf(81*_Z^4-9*_Z^2+1)*x-(-(1-x)/(1+x))^(1/6))/x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {x \sqrt {2 i \, \sqrt {3} + 2} \log \left (\sqrt {2 i \, \sqrt {3} + 2} {\left (i \, \sqrt {3} - 1\right )} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) - x \sqrt {2 i \, \sqrt {3} + 2} \log \left (\sqrt {2 i \, \sqrt {3} + 2} {\left (-i \, \sqrt {3} + 1\right )} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) - x \sqrt {-2 i \, \sqrt {3} + 2} \log \left ({\left (i \, \sqrt {3} + 1\right )} \sqrt {-2 i \, \sqrt {3} + 2} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + x \sqrt {-2 i \, \sqrt {3} + 2} \log \left ({\left (-i \, \sqrt {3} - 1\right )} \sqrt {-2 i \, \sqrt {3} + 2} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + 4 \, x \arctan \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + 6 \, {\left (x + 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{6}}}{6 \, x} \]

[In]

integrate(1/((-1+x)/(1+x))^(1/6)/x^2,x, algorithm="fricas")

[Out]

1/6*(x*sqrt(2*I*sqrt(3) + 2)*log(sqrt(2*I*sqrt(3) + 2)*(I*sqrt(3) - 1) + 4*((x - 1)/(x + 1))^(1/6)) - x*sqrt(2
*I*sqrt(3) + 2)*log(sqrt(2*I*sqrt(3) + 2)*(-I*sqrt(3) + 1) + 4*((x - 1)/(x + 1))^(1/6)) - x*sqrt(-2*I*sqrt(3)
+ 2)*log((I*sqrt(3) + 1)*sqrt(-2*I*sqrt(3) + 2) + 4*((x - 1)/(x + 1))^(1/6)) + x*sqrt(-2*I*sqrt(3) + 2)*log((-
I*sqrt(3) - 1)*sqrt(-2*I*sqrt(3) + 2) + 4*((x - 1)/(x + 1))^(1/6)) + 4*x*arctan(((x - 1)/(x + 1))^(1/6)) + 6*(
x + 1)*((x - 1)/(x + 1))^(5/6))/x

Sympy [F]

\[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=\int \frac {1}{x^{2} \sqrt [6]{\frac {x - 1}{x + 1}}}\, dx \]

[In]

integrate(1/((-1+x)/(1+x))**(1/6)/x**2,x)

[Out]

Integral(1/(x**2*((x - 1)/(x + 1))**(1/6)), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=-\frac {1}{6} \, \sqrt {3} \log \left (\sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{6}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \arctan \left (\sqrt {3} + 2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + \frac {1}{3} \, \arctan \left (-\sqrt {3} + 2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + \frac {2}{3} \, \arctan \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) \]

[In]

integrate(1/((-1+x)/(1+x))^(1/6)/x^2,x, algorithm="maxima")

[Out]

-1/6*sqrt(3)*log(sqrt(3)*((x - 1)/(x + 1))^(1/6) + ((x - 1)/(x + 1))^(1/3) + 1) + 1/6*sqrt(3)*log(-sqrt(3)*((x
- 1)/(x + 1))^(1/6) + ((x - 1)/(x + 1))^(1/3) + 1) + 2*((x - 1)/(x + 1))^(5/6)/((x - 1)/(x + 1) + 1) + 1/3*ar
ctan(sqrt(3) + 2*((x - 1)/(x + 1))^(1/6)) + 1/3*arctan(-sqrt(3) + 2*((x - 1)/(x + 1))^(1/6)) + 2/3*arctan(((x
- 1)/(x + 1))^(1/6))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=-\frac {1}{6} \, \sqrt {3} \log \left (\sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{6}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \arctan \left (\sqrt {3} + 2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + \frac {1}{3} \, \arctan \left (-\sqrt {3} + 2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) + \frac {2}{3} \, \arctan \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{6}}\right ) \]

[In]

integrate(1/((-1+x)/(1+x))^(1/6)/x^2,x, algorithm="giac")

[Out]

-1/6*sqrt(3)*log(sqrt(3)*((x - 1)/(x + 1))^(1/6) + ((x - 1)/(x + 1))^(1/3) + 1) + 1/6*sqrt(3)*log(-sqrt(3)*((x
- 1)/(x + 1))^(1/6) + ((x - 1)/(x + 1))^(1/3) + 1) + 2*((x - 1)/(x + 1))^(5/6)/((x - 1)/(x + 1) + 1) + 1/3*ar
ctan(sqrt(3) + 2*((x - 1)/(x + 1))^(1/6)) + 1/3*arctan(-sqrt(3) + 2*((x - 1)/(x + 1))^(1/6)) + 2/3*arctan(((x
- 1)/(x + 1))^(1/6))

Mupad [B] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.47 \[ \int \frac {e^{\frac {1}{3} \coth ^{-1}(x)}}{x^2} \, dx=\frac {2\,\mathrm {atan}\left ({\left (\frac {x-1}{x+1}\right )}^{1/6}\right )}{3}+\frac {2\,{\left (\frac {x-1}{x+1}\right )}^{5/6}}{\frac {x-1}{x+1}+1}-\mathrm {atan}\left (\frac {64\,{\left (\frac {x-1}{x+1}\right )}^{1/6}}{-32+\sqrt {3}\,32{}\mathrm {i}}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )-\mathrm {atan}\left (\frac {64\,{\left (\frac {x-1}{x+1}\right )}^{1/6}}{32+\sqrt {3}\,32{}\mathrm {i}}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right ) \]

[In]

int(1/(x^2*((x - 1)/(x + 1))^(1/6)),x)

[Out]

(2*atan(((x - 1)/(x + 1))^(1/6)))/3 + (2*((x - 1)/(x + 1))^(5/6))/((x - 1)/(x + 1) + 1) - atan((64*((x - 1)/(x
+ 1))^(1/6))/(3^(1/2)*32i - 32))*((3^(1/2)*1i)/3 + 1/3) - atan((64*((x - 1)/(x + 1))^(1/6))/(3^(1/2)*32i + 32
))*((3^(1/2)*1i)/3 - 1/3)