3.182 \(\int \frac{1}{1+\sinh ^3(x)} \, dx\)

Optimal. Leaf size=139 \[ -\frac{1}{3} \sqrt{2} \tanh ^{-1}\left (\frac{1-\tanh \left (\frac{x}{2}\right )}{\sqrt{2}}\right )-\frac{1}{3} \sqrt [6]{-1} \log \left (-\sqrt [6]{-1} \tanh \left (\frac{x}{2}\right )+(-1)^{5/6}+1\right )+\frac{1}{3} \sqrt [6]{-1} \log \left (\sqrt [3]{-1} \tanh \left (\frac{x}{2}\right )+\sqrt [6]{-1}+1\right )-\frac{2 \sqrt [6]{-1} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \tanh \left (\frac{x}{2}\right )+i}{\sqrt{1-\sqrt [3]{-1}}}\right )}{3 \sqrt{1-\sqrt [3]{-1}}} \]

[Out]

(-2*(-1)^(1/6)*ArcTan[(I + (-1)^(1/6)*Tanh[x/2])/Sqrt[1 - (-1)^(1/3)]])/(3*Sqrt[1 - (-1)^(1/3)]) - (Sqrt[2]*Ar
cTanh[(1 - Tanh[x/2])/Sqrt[2]])/3 - ((-1)^(1/6)*Log[1 + (-1)^(5/6) - (-1)^(1/6)*Tanh[x/2]])/3 + ((-1)^(1/6)*Lo
g[1 + (-1)^(1/6) + (-1)^(1/3)*Tanh[x/2]])/3

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Rubi [A]  time = 0.187819, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {3213, 2660, 618, 204, 617, 206, 616, 31} \[ -\frac{1}{3} \sqrt{2} \tanh ^{-1}\left (\frac{1-\tanh \left (\frac{x}{2}\right )}{\sqrt{2}}\right )-\frac{1}{3} \sqrt [6]{-1} \log \left (-\sqrt [6]{-1} \tanh \left (\frac{x}{2}\right )+(-1)^{5/6}+1\right )+\frac{1}{3} \sqrt [6]{-1} \log \left (\sqrt [3]{-1} \tanh \left (\frac{x}{2}\right )+\sqrt [6]{-1}+1\right )-\frac{2 \sqrt [6]{-1} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \tanh \left (\frac{x}{2}\right )+i}{\sqrt{1-\sqrt [3]{-1}}}\right )}{3 \sqrt{1-\sqrt [3]{-1}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^3)^(-1),x]

[Out]

(-2*(-1)^(1/6)*ArcTan[(I + (-1)^(1/6)*Tanh[x/2])/Sqrt[1 - (-1)^(1/3)]])/(3*Sqrt[1 - (-1)^(1/3)]) - (Sqrt[2]*Ar
cTanh[(1 - Tanh[x/2])/Sqrt[2]])/3 - ((-1)^(1/6)*Log[1 + (-1)^(5/6) - (-1)^(1/6)*Tanh[x/2]])/3 + ((-1)^(1/6)*Lo
g[1 + (-1)^(1/6) + (-1)^(1/3)*Tanh[x/2]])/3

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

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 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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 616

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 1.39943, size = 156, normalized size = 1.12 \[ \frac{2 \tanh ^{-1}\left (\frac{\tanh \left (\frac{x}{2}\right )-1}{\sqrt{2}}\right )+i \sqrt{-1-i \sqrt{3}} \left (\sqrt{3}+i\right ) \tan ^{-1}\left (\frac{2+\left (1-i \sqrt{3}\right ) \tanh \left (\frac{x}{2}\right )}{\sqrt{-2+2 i \sqrt{3}}}\right )+\left (-1-i \sqrt{3}\right ) \sqrt{-1+i \sqrt{3}} \tan ^{-1}\left (\frac{2+\left (1+i \sqrt{3}\right ) \tanh \left (\frac{x}{2}\right )}{\sqrt{-2-2 i \sqrt{3}}}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^3)^(-1),x]

[Out]

(I*Sqrt[-1 - I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[(2 + (1 - I*Sqrt[3])*Tanh[x/2])/Sqrt[-2 + (2*I)*Sqrt[3]]] + (-1 -
I*Sqrt[3])*Sqrt[-1 + I*Sqrt[3]]*ArcTan[(2 + (1 + I*Sqrt[3])*Tanh[x/2])/Sqrt[-2 - (2*I)*Sqrt[3]]] + 2*ArcTanh[
(-1 + Tanh[x/2])/Sqrt[2]])/(3*Sqrt[2])

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Maple [C]  time = 0.029, size = 82, normalized size = 0.6 \begin{align*}{\frac{2}{3}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{4}+2\,{{\it \_Z}}^{3}+2\,{{\it \_Z}}^{2}-2\,{\it \_Z}+1 \right ) }{\frac{-{{\it \_R}}^{2}-{\it \_R}+1}{2\,{{\it \_R}}^{3}+3\,{{\it \_R}}^{2}+2\,{\it \_R}-1}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -{\it \_R} \right ) }}+{\frac{\sqrt{2}}{3}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) -2 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sinh(x)^3),x)

[Out]

2/3*sum((-_R^2-_R+1)/(2*_R^3+3*_R^2+2*_R-1)*ln(tanh(1/2*x)-_R),_R=RootOf(_Z^4+2*_Z^3+2*_Z^2-2*_Z+1))+1/3*2^(1/
2)*arctanh(1/4*(2*tanh(1/2*x)-2)*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{x} - 1}{\sqrt{2} + e^{x} + 1}\right ) - \int \frac{2 \,{\left (e^{\left (3 \, x\right )} - 4 \, e^{\left (2 \, x\right )} - e^{x}\right )}}{3 \,{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 2 \, e^{x} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*log(-(sqrt(2) - e^x - 1)/(sqrt(2) + e^x + 1)) - integrate(2/3*(e^(3*x) - 4*e^(2*x) - e^x)/(e^(4*x)
- 2*e^(3*x) + 2*e^(2*x) + 2*e^x + 1), x)

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Fricas [B]  time = 1.93521, size = 591, normalized size = 4.25 \begin{align*} -\frac{1}{6} \, \sqrt{3} \log \left (-4 \,{\left (\sqrt{3} + 1\right )} e^{x} + 4 \, \sqrt{3} + 4 \, e^{\left (2 \, x\right )} + 8\right ) + \frac{1}{6} \, \sqrt{3} \log \left (4 \,{\left (\sqrt{3} - 1\right )} e^{x} - 4 \, \sqrt{3} + 4 \, e^{\left (2 \, x\right )} + 8\right ) + \frac{1}{6} \, \sqrt{2} \log \left (-\frac{2 \,{\left (\sqrt{2} - 1\right )} e^{x} + 2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{e^{\left (2 \, x\right )} + 2 \, e^{x} - 1}\right ) + \frac{2}{3} \, \arctan \left (-{\left (\sqrt{3} + 1\right )} e^{x} + \sqrt{{\left (\sqrt{3} - 1\right )} e^{x} - \sqrt{3} + e^{\left (2 \, x\right )} + 2}{\left (\sqrt{3} + 1\right )} - 1\right ) - \frac{2}{3} \, \arctan \left (-{\left (\sqrt{3} - 1\right )} e^{x} + \frac{1}{2} \, \sqrt{-4 \,{\left (\sqrt{3} + 1\right )} e^{x} + 4 \, \sqrt{3} + 4 \, e^{\left (2 \, x\right )} + 8}{\left (\sqrt{3} - 1\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(-4*(sqrt(3) + 1)*e^x + 4*sqrt(3) + 4*e^(2*x) + 8) + 1/6*sqrt(3)*log(4*(sqrt(3) - 1)*e^x - 4*s
qrt(3) + 4*e^(2*x) + 8) + 1/6*sqrt(2)*log(-(2*(sqrt(2) - 1)*e^x + 2*sqrt(2) - e^(2*x) - 3)/(e^(2*x) + 2*e^x -
1)) + 2/3*arctan(-(sqrt(3) + 1)*e^x + sqrt((sqrt(3) - 1)*e^x - sqrt(3) + e^(2*x) + 2)*(sqrt(3) + 1) - 1) - 2/3
*arctan(-(sqrt(3) - 1)*e^x + 1/2*sqrt(-4*(sqrt(3) + 1)*e^x + 4*sqrt(3) + 4*e^(2*x) + 8)*(sqrt(3) - 1) + 1)

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Sympy [B]  time = 174.633, size = 1423, normalized size = 10.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

167721543*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + sqrt(2))/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*
sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 8216
22480*sqrt(6)*I + 1162008936*sqrt(3)*I) + 118590990*sqrt(2)*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + sqrt(2))/(
711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I
) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) + 193668156*sqrt(6)*
I*log(tanh(x/2) - 1 + sqrt(2))/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545
940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936
*sqrt(3)*I) + 273874160*sqrt(3)*I*log(tanh(x/2) - 1 + sqrt(2))/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt
(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) +
821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) + 118590990*sqrt(6)*I*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + sqrt
(2))/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqr
t(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) + 167721543*sq
rt(3)*I*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + sqrt(2))/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sq
rt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622
480*sqrt(6)*I + 1162008936*sqrt(3)*I) - 167721543*sqrt(3)*I*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - sqrt(2) - 1)/(
711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I
) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) - 118590990*sqrt(6)*
I*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - sqrt(2) - 1)/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 +
sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sq
rt(6)*I + 1162008936*sqrt(3)*I) - 273874160*sqrt(3)*I*log(tanh(x/2) - sqrt(2) - 1)/(711545940*sqrt(1 + sqrt(3)
*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sq
rt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) - 193668156*sqrt(6)*I*log(tanh(x/2) - sqrt(2)
- 1)/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqr
t(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I) - 118590990*sq
rt(2)*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - sqrt(2) - 1)/(711545940*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt
(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 503164629*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 82162248
0*sqrt(6)*I + 1162008936*sqrt(3)*I) - 167721543*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - sqrt(2) - 1)/(711545940*sq
rt(1 + sqrt(3)*I) + 503164629*sqrt(2)*sqrt(1 + sqrt(3)*I) + 711545940*sqrt(3)*I*sqrt(1 + sqrt(3)*I) + 50316462
9*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 821622480*sqrt(6)*I + 1162008936*sqrt(3)*I)

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Giac [A]  time = 1.16328, size = 140, normalized size = 1.01 \begin{align*} \frac{1}{6} \,{\left (\sqrt{3} + i\right )} \log \left (\sqrt{3} + \left (i + 1\right ) \, e^{x} - 1\right ) + \frac{1}{6} \,{\left (\sqrt{3} - i\right )} \log \left (i \, \sqrt{3} + \left (i + 1\right ) \, e^{x} - i\right ) - \frac{1}{6} \,{\left (\sqrt{3} + i\right )} \log \left (-i \, \sqrt{3} + \left (i + 1\right ) \, e^{x} - i\right ) - \frac{1}{6} \,{\left (\sqrt{3} - i\right )} \log \left (-\sqrt{3} + \left (i + 1\right ) \, e^{x} - 1\right ) + \frac{1}{6} \, \sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 2 \, e^{x} + 2 \right |}}{2 \,{\left (\sqrt{2} + e^{x} + 1\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(sqrt(3) + I)*log(sqrt(3) + (I + 1)*e^x - 1) + 1/6*(sqrt(3) - I)*log(I*sqrt(3) + (I + 1)*e^x - I) - 1/6*(s
qrt(3) + I)*log(-I*sqrt(3) + (I + 1)*e^x - I) - 1/6*(sqrt(3) - I)*log(-sqrt(3) + (I + 1)*e^x - 1) + 1/6*sqrt(2
)*log(1/2*abs(-2*sqrt(2) + 2*e^x + 2)/(sqrt(2) + e^x + 1))