3.1323 \(\int \frac{\sqrt{1+2 x}}{1+x+x^2} \, dx\)

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

[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))

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Rubi [A]  time = 0.120514, 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, 297, 1162, 617, 204, 1165, 628} \[ \frac{\log \left (2 x-\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} \sqrt [4]{3}}-\frac{\log \left (2 x+\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2} \sqrt [4]{3}}-\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt [4]{3}}+\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt [4]{3}} \]

Antiderivative was successfully verified.

[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 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 297

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 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])

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{1+2 x}}{1+x+x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\frac{3}{4}+\frac{x^2}{4}} \, dx,x,1+2 x\right )\\ &=\operatorname{Subst}\left (\int \frac{x^2}{\frac{3}{4}+\frac{x^4}{4}} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\left (\frac{1}{2} \operatorname{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} \operatorname{Subst}\left (\int \frac{\sqrt{3}+x^2}{\frac{3}{4}+\frac{x^4}{4}} \, dx,x,\sqrt{1+2 x}\right )\\ &=\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} \sqrt [4]{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} \sqrt [4]{3}}+\operatorname{Subst}\left (\int \frac{1}{\sqrt{3}-\sqrt{2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt{1+2 x}\right )+\operatorname{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} \operatorname{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} \operatorname{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 [C]  time = 0.0058424, size = 32, normalized size = 0.2 \[ \frac{4}{9} (2 x+1)^{3/2} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\frac{1}{3} (2 x+1)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(1 + 2*x)^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -(1 + 2*x)^2/3])/9

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Maple [A]  time = 0.041, size = 111, normalized size = 0.7 \begin{align*}{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\arctan \left ( 1+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\sqrt{1+2\,x}} \right ) }+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\arctan \left ( -1+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{3}\sqrt{1+2\,x}} \right ) }+{\frac{\sqrt{2}{3}^{{\frac{3}{4}}}}{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)+1/6*2^(1/2)*3^(3/4)*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)))

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Maxima [A]  time = 2.18092, size = 178, normalized size = 1.13 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.59657, size = 621, normalized size = 3.96 \begin{align*} -\frac{2}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 2 \, x + \sqrt{3} + 1} - \frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} - 1\right ) - \frac{2}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \arctan \left (\frac{1}{6} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{-4 \cdot 3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 8 \, x + 4 \, \sqrt{3} + 4} - \frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 1\right ) - \frac{1}{6} \cdot 3^{\frac{3}{4}} \sqrt{2} \log \left (4 \cdot 3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 8 \, x + 4 \, \sqrt{3} + 4\right ) + \frac{1}{6} \cdot 3^{\frac{3}{4}} \sqrt{2} \log \left (-4 \cdot 3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 8 \, x + 4 \, \sqrt{3} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*3^(3/4)*sqrt(2)*arctan(1/3*3^(3/4)*sqrt(2)*sqrt(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 1/3*
3^(3/4)*sqrt(2)*sqrt(2*x + 1) - 1) - 2/3*3^(3/4)*sqrt(2)*arctan(1/6*3^(3/4)*sqrt(2)*sqrt(-4*3^(1/4)*sqrt(2)*sq
rt(2*x + 1) + 8*x + 4*sqrt(3) + 4) - 1/3*3^(3/4)*sqrt(2)*sqrt(2*x + 1) + 1) - 1/6*3^(3/4)*sqrt(2)*log(4*3^(1/4
)*sqrt(2)*sqrt(2*x + 1) + 8*x + 4*sqrt(3) + 4) + 1/6*3^(3/4)*sqrt(2)*log(-4*3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 8*
x + 4*sqrt(3) + 4)

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Sympy [A]  time = 2.76604, size = 155, normalized size = 0.99 \begin{align*} \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \log{\left (2 x - \sqrt{2} \sqrt [4]{3} \sqrt{2 x + 1} + 1 + \sqrt{3} \right )}}{6} - \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \log{\left (2 x + \sqrt{2} \sqrt [4]{3} \sqrt{2 x + 1} + 1 + \sqrt{3} \right )}}{6} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \operatorname{atan}{\left (\frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \sqrt{2 x + 1}}{3} - 1 \right )}}{3} + \frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \operatorname{atan}{\left (\frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \sqrt{2 x + 1}}{3} + 1 \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16337, size = 162, normalized size = 1.03 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)