3.936 \(\int \frac{\sqrt{x}}{1+\sqrt{x}+x} \, dx\)

Optimal. Leaf size=42 \[ 2 \sqrt{x}-\log \left (x+\sqrt{x}+1\right )-\frac{2 \tan ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0287223, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1357, 703, 634, 618, 204, 628} \[ 2 \sqrt{x}-\log \left (x+\sqrt{x}+1\right )-\frac{2 \tan ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(1 + Sqrt[x] + x),x]

[Out]

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

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*
e, 0] && GtQ[m, 1]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 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{x}}{1+\sqrt{x}+x} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^2}{1+x+x^2} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x}+2 \operatorname{Subst}\left (\int \frac{-1-x}{1+x+x^2} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x}-\operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\sqrt{x}\right )-\operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x}-\log \left (1+\sqrt{x}+x\right )+2 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt{x}\right )\\ &=2 \sqrt{x}-\frac{2 \tan ^{-1}\left (\frac{1+2 \sqrt{x}}{\sqrt{3}}\right )}{\sqrt{3}}-\log \left (1+\sqrt{x}+x\right )\\ \end{align*}

Mathematica [A]  time = 0.0141231, size = 42, normalized size = 1. \[ 2 \sqrt{x}-\log \left (x+\sqrt{x}+1\right )-\frac{2 \tan ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(1 + Sqrt[x] + x),x]

[Out]

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

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Maple [A]  time = 0.005, size = 34, normalized size = 0.8 \begin{align*} -\ln \left ( 1+x+\sqrt{x} \right ) -{\frac{2\,\sqrt{3}}{3}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 1+2\,\sqrt{x} \right ) } \right ) }+2\,\sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.69182, size = 45, normalized size = 1.07 \begin{align*} -\frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, \sqrt{x} + 1\right )}\right ) + 2 \, \sqrt{x} - \log \left (x + \sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*sqrt(x) + 1)) + 2*sqrt(x) - log(x + sqrt(x) + 1)

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Fricas [A]  time = 1.45813, size = 123, normalized size = 2.93 \begin{align*} -\frac{2}{3} \, \sqrt{3} \arctan \left (\frac{2}{3} \, \sqrt{3} \sqrt{x} + \frac{1}{3} \, \sqrt{3}\right ) + 2 \, \sqrt{x} - \log \left (x + \sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(3)*arctan(2/3*sqrt(3)*sqrt(x) + 1/3*sqrt(3)) + 2*sqrt(x) - log(x + sqrt(x) + 1)

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Sympy [A]  time = 0.247189, size = 49, normalized size = 1.17 \begin{align*} 2 \sqrt{x} - \log{\left (4 \sqrt{x} + 4 x + 4 \right )} - \frac{2 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} \sqrt{x}}{3} + \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) - log(4*sqrt(x) + 4*x + 4) - 2*sqrt(3)*atan(2*sqrt(3)*sqrt(x)/3 + sqrt(3)/3)/3

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Giac [A]  time = 1.12951, size = 45, normalized size = 1.07 \begin{align*} -\frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, \sqrt{x} + 1\right )}\right ) + 2 \, \sqrt{x} - \log \left (x + \sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*sqrt(x) + 1)) + 2*sqrt(x) - log(x + sqrt(x) + 1)