3.126 \(\int \frac {e^{\frac {1}{3} \tanh ^{-1}(x)}}{x^3} \, dx\)

Optimal. Leaf size=224 \[ -\frac {(1-x)^{5/6} (x+1)^{7/6}}{2 x^2}-\frac {(1-x)^{5/6} \sqrt [6]{x+1}}{6 x}+\frac {1}{36} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{1-x}}-\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1\right )-\frac {1}{36} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{1-x}}+\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1\right )+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [6]{x+1}}{\sqrt [6]{1-x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {1}{9} \tanh ^{-1}\left (\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}\right ) \]

[Out]

-1/6*(1-x)^(5/6)*(1+x)^(1/6)/x-1/2*(1-x)^(5/6)*(1+x)^(7/6)/x^2-1/9*arctanh((1+x)^(1/6)/(1-x)^(1/6))+1/36*ln(1-
(1+x)^(1/6)/(1-x)^(1/6)+(1+x)^(1/3)/(1-x)^(1/3))-1/36*ln(1+(1+x)^(1/6)/(1-x)^(1/6)+(1+x)^(1/3)/(1-x)^(1/3))+1/
18*arctan(1/3*(1-2*(1+x)^(1/6)/(1-x)^(1/6))*3^(1/2))*3^(1/2)-1/18*arctan(1/3*(1+2*(1+x)^(1/6)/(1-x)^(1/6))*3^(
1/2))*3^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6126, 96, 94, 93, 210, 634, 618, 204, 628, 206} \[ -\frac {(1-x)^{5/6} (x+1)^{7/6}}{2 x^2}-\frac {(1-x)^{5/6} \sqrt [6]{x+1}}{6 x}+\frac {1}{36} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{1-x}}-\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1\right )-\frac {1}{36} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{1-x}}+\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1\right )+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [6]{x+1}}{\sqrt [6]{1-x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [6]{x+1}}{\sqrt [6]{1-x}}+1}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {1}{9} \tanh ^{-1}\left (\frac {\sqrt [6]{x+1}}{\sqrt [6]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - x)^(5/6)*(1 + x)^(1/6))/(6*x) - ((1 - x)^(5/6)*(1 + x)^(7/6))/(2*x^2) + ArcTan[(1 - (2*(1 + x)^(1/6))/(
1 - x)^(1/6))/Sqrt[3]]/(6*Sqrt[3]) - ArcTan[(1 + (2*(1 + x)^(1/6))/(1 - x)^(1/6))/Sqrt[3]]/(6*Sqrt[3]) - ArcTa
nh[(1 + x)^(1/6)/(1 - x)^(1/6)]/9 + Log[1 - (1 + x)^(1/6)/(1 - x)^(1/6) + (1 + x)^(1/3)/(1 - x)^(1/3)]/36 - Lo
g[1 + (1 + x)^(1/6)/(1 - x)^(1/6) + (1 + x)^(1/3)/(1 - x)^(1/3)]/36

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[(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] || Sum
SimplerQ[m, 1])

Rule 204

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

Rule 206

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

Rule 210

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

Rule 618

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 628

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 634

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 6126

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

Rubi steps

\begin {align*} \int \frac {e^{\frac {1}{3} \tanh ^{-1}(x)}}{x^3} \, dx &=\int \frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x} x^3} \, dx\\ &=-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}+\frac {1}{6} \int \frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x} x^2} \, dx\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}+\frac {1}{18} \int \frac {1}{\sqrt [6]{1-x} x (1+x)^{5/6}} \, dx\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1-\frac {x}{2}}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1+\frac {x}{2}}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}-\frac {1}{9} \tanh ^{-1}\left (\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )+\frac {1}{36} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )-\frac {1}{36} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )-\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )-\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}-\frac {1}{9} \tanh ^{-1}\left (\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )+\frac {1}{36} \log \left (1-\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}+\frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}}\right )-\frac {1}{36} \log \left (1+\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}+\frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 \sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )\\ &=-\frac {(1-x)^{5/6} \sqrt [6]{1+x}}{6 x}-\frac {(1-x)^{5/6} (1+x)^{7/6}}{2 x^2}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [6]{1+x}}{\sqrt [6]{1-x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [6]{1+x}}{\sqrt [6]{1-x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {1}{9} \tanh ^{-1}\left (\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}\right )+\frac {1}{36} \log \left (1-\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}+\frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}}\right )-\frac {1}{36} \log \left (1+\frac {\sqrt [6]{1+x}}{\sqrt [6]{1-x}}+\frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.27 \[ -\frac {(1-x)^{5/6} \left (2 x^2 \, _2F_1\left (\frac {5}{6},1;\frac {11}{6};\frac {1-x}{x+1}\right )+5 \left (4 x^2+7 x+3\right )\right )}{30 x^2 (x+1)^{5/6}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/30*((1 - x)^(5/6)*(5*(3 + 7*x + 4*x^2) + 2*x^2*Hypergeometric2F1[5/6, 1, 11/6, (1 - x)/(1 + x)]))/(x^2*(1 +
x)^(5/6))

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fricas [A]  time = 0.60, size = 253, normalized size = 1.13 \[ -\frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 2 \, \sqrt {3} x^{2} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + x^{2} \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 1\right ) - x^{2} \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 1\right ) + 2 \, x^{2} \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 1\right ) - 2 \, x^{2} \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} - 1\right ) - 6 \, {\left (4 \, x^{2} - x - 3\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}}}{36 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(2*sqrt(3)*x^2*arctan(2/3*sqrt(3)*(-sqrt(-x^2 + 1)/(x - 1))^(1/3) + 1/3*sqrt(3)) + 2*sqrt(3)*x^2*arctan(
2/3*sqrt(3)*(-sqrt(-x^2 + 1)/(x - 1))^(1/3) - 1/3*sqrt(3)) + x^2*log((-sqrt(-x^2 + 1)/(x - 1))^(2/3) + (-sqrt(
-x^2 + 1)/(x - 1))^(1/3) + 1) - x^2*log((-sqrt(-x^2 + 1)/(x - 1))^(2/3) - (-sqrt(-x^2 + 1)/(x - 1))^(1/3) + 1)
+ 2*x^2*log((-sqrt(-x^2 + 1)/(x - 1))^(1/3) + 1) - 2*x^2*log((-sqrt(-x^2 + 1)/(x - 1))^(1/3) - 1) - 6*(4*x^2
- x - 3)*(-sqrt(-x^2 + 1)/(x - 1))^(1/3))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {1+x}{\sqrt {-x^{2}+1}}\right )^{\frac {1}{3}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {x+1}{\sqrt {1-x^2}}\right )}^{1/3}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [3]{\frac {x + 1}{\sqrt {1 - x^{2}}}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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