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

Optimal. Leaf size=202 \[ -(1-x)^{5/6} \sqrt [6]{x+1}-\frac {\log \left (\frac {\sqrt [3]{1-x}}{\sqrt [3]{x+1}}-\frac {\sqrt {3} \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+1\right )}{2 \sqrt {3}}+\frac {\log \left (\frac {\sqrt [3]{1-x}}{\sqrt [3]{x+1}}+\frac {\sqrt {3} \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+1\right )}{2 \sqrt {3}}-\frac {2}{3} \tan ^{-1}\left (\frac {\sqrt [6]{1-x}}{\sqrt [6]{x+1}}\right )+\frac {1}{3} \tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [6]{1-x}}{\sqrt [6]{x+1}}\right )-\frac {1}{3} \tan ^{-1}\left (\frac {2 \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+\sqrt {3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {6125, 50, 63, 331, 295, 634, 618, 204, 628, 203} \[ -(1-x)^{5/6} \sqrt [6]{x+1}-\frac {\log \left (\frac {\sqrt [3]{1-x}}{\sqrt [3]{x+1}}-\frac {\sqrt {3} \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+1\right )}{2 \sqrt {3}}+\frac {\log \left (\frac {\sqrt [3]{1-x}}{\sqrt [3]{x+1}}+\frac {\sqrt {3} \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+1\right )}{2 \sqrt {3}}-\frac {2}{3} \tan ^{-1}\left (\frac {\sqrt [6]{1-x}}{\sqrt [6]{x+1}}\right )+\frac {1}{3} \tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [6]{1-x}}{\sqrt [6]{x+1}}\right )-\frac {1}{3} \tan ^{-1}\left (\frac {2 \sqrt [6]{1-x}}{\sqrt [6]{x+1}}+\sqrt {3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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 63

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 203

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

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 295

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)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
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 331

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 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 6125

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

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 39, normalized size = 0.19 \[ 2 e^{\frac {1}{3} \tanh ^{-1}(x)} \left (\, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-e^{2 \tanh ^{-1}(x)}\right )-\frac {1}{e^{2 \tanh ^{-1}(x)}+1}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 295, normalized size = 1.46 \[ \frac {1}{6} \, \sqrt {3} \log \left (4 \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 4 \, \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 4\right ) - \frac {1}{6} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 4 \, \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 4\right ) + {\left (x - 1\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} - \frac {2}{3} \, \arctan \left (\sqrt {3} + \sqrt {-4 \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 4 \, \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 4} - 2 \, \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}}\right ) - \frac {2}{3} \, \arctan \left (-\sqrt {3} + 2 \, \sqrt {\sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 1} - 2 \, \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}}\right ) + \frac {2}{3} \, \arctan \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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