Integrand size = 18, antiderivative size = 157 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=-\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2} \sqrt [4]{3}} \]
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)
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right )\right )}{\sqrt [4]{3}} \]
(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)
Time = 0.36 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.11, 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 |
\(\displaystyle 4 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}\right )+\frac {1}{2} \left (\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}\right )\right )\) |
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)
3.14.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]]))
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]
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]
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]
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]
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]
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 | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{4} x -6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right ) x +108 \sqrt {1+2 x}-18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2}\right )}{6+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} x +3 x}\right )}{3}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{5} x +6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{3}-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right ) x -18 \operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )-108 \sqrt {1+2 x}}{-6+\operatorname {RootOf}\left (\textit {\_Z}^{4}+27\right )^{2} x -3 x}\right )}{3}\) | \(198\) |
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)))
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\left (\frac {1}{6} i - \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (\left (i + 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) - \left (\frac {1}{6} i + \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-\left (i - 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) + \left (\frac {1}{6} i + \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (\left (i - 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) - \left (\frac {1}{6} i - \frac {1}{6}\right ) \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-\left (i + 1\right ) \cdot 3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right ) \]
(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)) + (1/6*I + 1/6)*3^(3/4)*sqrt(2)*log((I - 1)*3^(1/4)*sqrt(2) + 2*s qrt(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))
\[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\int \frac {\sqrt {2 x + 1}}{x^{2} + x + 1}\, dx \]
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \]
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)
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{3} \cdot 108^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{6} \cdot 108^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) \]
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)
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx=\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right )+\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right ) \]