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

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 157, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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 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 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 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 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 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 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 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]

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 = 0.03, size = 108, normalized size = 0.69 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 99, normalized size = 0.63 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\left (\frac {\frac {2 x+1}{\sqrt {2} \sqrt [4]{3}}-\frac {\sqrt [4]{3}}{\sqrt {2}}}{\sqrt {2 x+1}}\right )}{3^{3/4}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} 3^{3/4} \sqrt {2 x+1}}{\sqrt {3} (2 x+1)+3}\right )}{3^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 190, normalized size = 1.21 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 120, normalized size = 0.76 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 111, normalized size = 0.71 \begin {gather*} \frac {3^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )}{3}+\frac {3^{\frac {1}{4}} \sqrt {2}\, \arctan \left (1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )}{3}+\frac {3^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {2 x +1+\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}{2 x +1+\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.99, size = 132, normalized size = 0.84 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 57, normalized size = 0.36 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {2 x + 1} \left (x^{2} + x + 1\right )}\, dx \end {gather*}

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)

________________________________________________________________________________________