\(\int \frac {1}{\sqrt {1+2 x} (1+x+x^2)} \, dx\) [148]

Optimal result

Integrand size = 18, antiderivative size = 115 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=-\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3^{3/4}}+\frac {\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3^{3/4}}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}}{1+\sqrt {3}+2 x}\right )}{3^{3/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )+\text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right )\right )}{3^{3/4}} \] Input:

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

Output:

(Sqrt[2]*(ArcTan[(-3 + Sqrt[3] + 2*Sqrt[3]*x)/(3^(3/4)*Sqrt[2 + 4*x])] + A 
rcTanh[(3^(3/4)*Sqrt[2 + 4*x])/(3 + Sqrt[3] + 2*Sqrt[3]*x)]))/3^(3/4)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.60, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {1118, 27, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {2 x+1} \left (x^2+x+1\right )} \, dx\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {1}{2} \int \frac {4}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \frac {1}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\)

\(\Big \downarrow \) 266

\(\displaystyle 4 \int \frac {1}{(2 x+1)^2+3}d\sqrt {2 x+1}\)

\(\Big \downarrow \) 755

\(\displaystyle 4 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}+1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle 4 \left (\frac {\frac {1}{2} \int \frac {1}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}+\frac {1}{2} \int \frac {1}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle 4 \left (\frac {\frac {\int \frac {1}{-2 x-2}d\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\int \frac {1}{-2 x-2}d\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 4 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle 4 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 4 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 4 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[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] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
default	\(\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
pseudoelliptic	\(\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{4} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )-12 \sqrt {1+2 x}}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}-x -2}\right )}{3}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{5} x +4 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}-3 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )+12 \sqrt {1+2 x}-6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}+x +2}\right )}{3}\)	\(232\)

Input:

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

Output:

1/6*3^(1/4)*2^(1/2)*(ln((1+2*x+3^(1/4)*(1+2*x)^(1/2)*2^(1/2)+3^(1/2))/(1+2 
*x-3^(1/4)*(1+2*x)^(1/2)*2^(1/2)+3^(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 [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\frac {1}{54} \cdot 108^{\frac {3}{4}} \arctan \left (\frac {1}{3} \cdot 108^{\frac {1}{4}} \sqrt {2 \, x + 1} + 1\right ) + \frac {1}{54} \cdot 108^{\frac {3}{4}} \arctan \left (\frac {1}{3} \cdot 108^{\frac {1}{4}} \sqrt {2 \, x + 1} - 1\right ) + \frac {1}{108} \cdot 108^{\frac {3}{4}} \log \left (108^{\frac {3}{4}} \sqrt {2 \, x + 1} + 36 \, x + 18 \, \sqrt {3} + 18\right ) - \frac {1}{108} \cdot 108^{\frac {3}{4}} \log \left (-108^{\frac {3}{4}} \sqrt {2 \, x + 1} + 36 \, x + 18 \, \sqrt {3} + 18\right ) \] Input:

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

Output:

1/54*108^(3/4)*arctan(1/3*108^(1/4)*sqrt(2*x + 1) + 1) + 1/54*108^(3/4)*ar 
ctan(1/3*108^(1/4)*sqrt(2*x + 1) - 1) + 1/108*108^(3/4)*log(108^(3/4)*sqrt 
(2*x + 1) + 36*x + 18*sqrt(3) + 18) - 1/108*108^(3/4)*log(-108^(3/4)*sqrt( 
2*x + 1) + 36*x + 18*sqrt(3) + 18)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\int \frac {1}{\sqrt {2 x + 1} \left (x^{2} + x + 1\right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\frac {1}{3} \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{6} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {1}{6} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\frac {1}{3} \cdot 12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{6} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {1}{6} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \] Input:

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

Output:

1/3*12^(1/4)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1) 
)) + 1/3*12^(1/4)*arctan(-1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2* 
x + 1))) + 1/6*12^(1/4)*log(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) 
+ 1) - 1/6*12^(1/4)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1 
)
 

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right ) \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx=\frac {\sqrt {2}\, 3^{\frac {1}{4}} \left (2 \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}-\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}+\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )-\mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )+\mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )\right )}{6} \] Input:

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

Output:

(sqrt(2)*3**(1/4)*(2*atan((2*sqrt(2*x + 1) - sqrt(2)*3**(1/4))/(sqrt(2)*3* 
*(1/4))) + 2*atan((2*sqrt(2*x + 1) + sqrt(2)*3**(1/4))/(sqrt(2)*3**(1/4))) 
 - log( - sqrt(2*x + 1)*sqrt(2)*3**(1/4) + sqrt(3) + 2*x + 1) + log(sqrt(2 
*x + 1)*sqrt(2)*3**(1/4) + sqrt(3) + 2*x + 1)))/6