### 3.1324 $$\int \frac{1}{\sqrt{1+2 x} (1+x+x^2)} \, dx$$

Optimal. Leaf size=157 $-\frac{\log \left (2 x-\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} 3^{3/4}}+\frac{\log \left (2 x+\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} 3^{3/4}}-\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}\right )}{3^{3/4}}+\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}+1\right )}{3^{3/4}}$

[Out]

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

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Rubi [A]  time = 0.119428, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.444, Rules used = {694, 329, 211, 1165, 628, 1162, 617, 204} $-\frac{\log \left (2 x-\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} 3^{3/4}}+\frac{\log \left (2 x+\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} 3^{3/4}}-\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}\right )}{3^{3/4}}+\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}+1\right )}{3^{3/4}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 694

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 329

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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 1165

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]

Rule 628

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 1162

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.026307, size = 108, normalized size = 0.69 $\frac{-\log \left (2 x-\sqrt [4]{3} \sqrt{4 x+2}+\sqrt{3}+1\right )+\log \left (2 x+\sqrt [4]{3} \sqrt{4 x+2}+\sqrt{3}+1\right )-2 \tan ^{-1}\left (1-\frac{\sqrt{4 x+2}}{\sqrt [4]{3}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{4 x+2}}{\sqrt [4]{3}}+1\right )}{\sqrt{2} 3^{3/4}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 111, normalized size = 0.7 \begin{align*}{\frac{\sqrt [4]{3}\sqrt{2}}{3}\arctan \left ( -1+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\sqrt{1+2\,x}} \right ) }+{\frac{\sqrt [4]{3}\sqrt{2}}{6}\ln \left ({ \left ( 1+2\,x+\sqrt{3}+\sqrt [4]{3}\sqrt{2}\sqrt{1+2\,x} \right ) \left ( 1+2\,x+\sqrt{3}-\sqrt [4]{3}\sqrt{2}\sqrt{1+2\,x} \right ) ^{-1}} \right ) }+{\frac{\sqrt [4]{3}\sqrt{2}}{3}\arctan \left ( 1+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\sqrt{1+2\,x}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.88842, size = 178, normalized size = 1.13 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*arct
an(-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)

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Fricas [A]  time = 1.62952, size = 676, normalized size = 4.31 \begin{align*} -\frac{2}{27} \cdot 27^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{9} \cdot 27^{\frac{1}{4}} \sqrt{2} \sqrt{27^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 18 \, x + 9 \, \sqrt{3} + 9} - \frac{1}{3} \cdot 27^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} - 1\right ) - \frac{2}{27} \cdot 27^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{54} \cdot 27^{\frac{1}{4}} \sqrt{2} \sqrt{-36 \cdot 27^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 648 \, x + 324 \, \sqrt{3} + 324} - \frac{1}{3} \cdot 27^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 1\right ) + \frac{1}{54} \cdot 27^{\frac{3}{4}} \sqrt{2} \log \left (36 \cdot 27^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 648 \, x + 324 \, \sqrt{3} + 324\right ) - \frac{1}{54} \cdot 27^{\frac{3}{4}} \sqrt{2} \log \left (-36 \cdot 27^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 648 \, x + 324 \, \sqrt{3} + 324\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/27*27^(3/4)*sqrt(2)*arctan(1/9*27^(1/4)*sqrt(2)*sqrt(27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 18*x + 9*sqrt(3) + 9)
- 1/3*27^(1/4)*sqrt(2)*sqrt(2*x + 1) - 1) - 2/27*27^(3/4)*sqrt(2)*arctan(1/54*27^(1/4)*sqrt(2)*sqrt(-36*27^(3
/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 324*sqrt(3) + 324) - 1/3*27^(1/4)*sqrt(2)*sqrt(2*x + 1) + 1) + 1/54*27^(3/
4)*sqrt(2)*log(36*27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 324*sqrt(3) + 324) - 1/54*27^(3/4)*sqrt(2)*log(-36*
27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 324*sqrt(3) + 324)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 x + 1} \left (x^{2} + x + 1\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22032, size = 162, normalized size = 1.03 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)