\(\int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx\) [147]

Optimal result

Integrand size = 18, antiderivative size = 116 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=-\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}}{1+\sqrt {3}+2 x}\right )}{\sqrt [4]{3}} \] Output:

1/3*2^(1/2)*3^(3/4)*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))+1/3*2^(1/ 
2)*3^(3/4)*arctan(1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))-1/3*2^(1/2)*3^(3/4) 
*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 = 78, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right )\right )}{\sqrt [4]{3}} \] Input:

Integrate[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^(1/4)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.50, 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, 826, 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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 = 1.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
default	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
pseudoelliptic	\(\frac {3^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{6}\)	\(99\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{4} x -6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) x -18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )-108 \sqrt {1+2 x}}{6+x \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}+3 x}\right )}{3}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right ) \ln \left (\frac {x \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{5}+6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{3}-9 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )-18 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )-108 \sqrt {1+2 x}}{-6+x \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}-3 x}\right )}{3}\)	\(197\)

Input:

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

Output:

1/6*3^(3/4)*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 = 97, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{6} \cdot 12^{\frac {3}{4}} \arctan \left (\frac {1}{6} \cdot 12^{\frac {3}{4}} \sqrt {2 \, x + 1} + 1\right ) + \frac {1}{6} \cdot 12^{\frac {3}{4}} \arctan \left (\frac {1}{6} \cdot 12^{\frac {3}{4}} \sqrt {2 \, x + 1} - 1\right ) - \frac {1}{12} \cdot 12^{\frac {3}{4}} \log \left (4 \, x + 2 \cdot 12^{\frac {1}{4}} \sqrt {2 \, x + 1} + 2 \, \sqrt {3} + 2\right ) + \frac {1}{12} \cdot 12^{\frac {3}{4}} \log \left (4 \, x - 2 \cdot 12^{\frac {1}{4}} \sqrt {2 \, x + 1} + 2 \, \sqrt {3} + 2\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(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 = 1.14 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \] Input:

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

Output:

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)*arctan(-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)

Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \] Input:

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

Output:

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 + sqrt( 
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 = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right )+\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right ) \] Input:

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

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {\sqrt {6}\, 3^{\frac {1}{4}} \left (2 \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}-\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}+\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+\mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )-\mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )\right )}{6} \] Input:

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

Output:

(sqrt(6)*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