\(\int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx\) [1323]

Optimal result

Integrand size = 18, antiderivative size = 157 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=-\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}} \]

[Out]

1/6*3^(3/4)*ln(1+2*x+3^(1/2)-3^(1/4)*2^(1/2)*(1+2*x)^(1/2))*2^(1/2)-1/6*3^(3/4)*ln(1+2*x+3^(1/2)+3^(1/4)*2^(1/
2)*(1+2*x)^(1/2))*2^(1/2)+1/3*3^(3/4)*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)+1/3*3^(3/4)*arctan(
1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {708, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=-\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt [4]{3}}+\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2} \sqrt [4]{3}} \]

[In]

Int[Sqrt[1 + 2*x]/(1 + x + x^2),x]

[Out]

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

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

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 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 708

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {x}}{\frac {3}{4}+\frac {x^2}{4}} \, dx,x,1+2 x\right ) \\ & = \text {Subst}\left (\int \frac {x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {3}-x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {3}+x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right ) \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}+2 x}{-\sqrt {3}-\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}-2 x}{-\sqrt {3}+\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}+\text {Subst}\left (\int \frac {1}{\sqrt {3}-\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )+\text {Subst}\left (\int \frac {1}{\sqrt {3}+\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right ) \\ & = \frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}+\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}-\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}} \\ & = -\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right )\right )}{\sqrt [4]{3}} \]

[In]

Integrate[Sqrt[1 + 2*x]/(1 + x + x^2),x]

[Out]

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

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
default	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
pseudoelliptic	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{4} x -6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) x +108 \sqrt {1+2 x}-18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )}{6+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} x +3 x}\right )}{3}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{5} x +6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{3}-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right ) x -18 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )-108 \sqrt {1+2 x}}{-6+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} x -3 x}\right )}{3}\)	\(198\)

[In]

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

[Out]

1/6*3^(3/4)*2^(1/2)*(ln((1+2*x+3^(1/2)-3^(1/4)*2^(1/2)*(1+2*x)^(1/2))/(1+2*x+3^(1/2)+3^(1/4)*2^(1/2)*(1+2*x)^(
1/2)))+2*arctan(1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))+2*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\left (\frac {1}{6} i - \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (\left (i + 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) - \left (\frac {1}{6} i + \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-\left (i - 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) + \left (\frac {1}{6} i + \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (\left (i - 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) - \left (\frac {1}{6} i - \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-\left (i + 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\int \frac {\sqrt {2 x + 1}}{x^{2} + x + 1}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \]

[In]

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

[Out]

1/3*108^(1/4)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1))) + 1/3*108^(1/4)*arctan(-1/6*3^(3
/4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) - 1/6*108^(1/4)*log(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqr
t(3) + 1) + 1/6*108^(1/4)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right )+\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right ) \]

[In]

int((2*x + 1)^(1/2)/(x + x^2 + 1),x)

[Out]

2^(1/2)*3^(3/4)*atan(2^(1/2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 - 1i/6))*(1/3 - 1i/3) + 2^(1/2)*3^(3/4)*atan(2^(1/2)
*3^(3/4)*(2*x + 1)^(1/2)*(1/6 + 1i/6))*(1/3 + 1i/3)