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]

-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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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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]]

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*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.00 \[ 2 \sqrt {x}-\log \left (x+\sqrt {x}+1\right )-\frac {2 \tan ^{-1}\left (\frac {2 \sqrt {x}+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

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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fricas [A]  time = 0.94, size = 35, normalized size = 0.83 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \sqrt {x} + \frac {1}{3} \, \sqrt {3}\right ) + 2 \, \sqrt {x} - \log \left (x + \sqrt {x} + 1\right ) \]

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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giac [A]  time = 0.39, size = 33, normalized size = 0.79 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \sqrt {x} + 1\right )}\right ) + 2 \, \sqrt {x} - \log \left (x + \sqrt {x} + 1\right ) \]

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)

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maple [A]  time = 0.00, size = 34, normalized size = 0.81 \[ -\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \sqrt {x}+1\right ) \sqrt {3}}{3}\right )}{3}-\ln \left (x +\sqrt {x}+1\right )+2 \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.01, size = 33, normalized size = 0.79 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \sqrt {x} + 1\right )}\right ) + 2 \, \sqrt {x} - \log \left (x + \sqrt {x} + 1\right ) \]

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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mupad [B]  time = 3.45, size = 35, normalized size = 0.83 \[ 2\,\sqrt {x}-\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}+\frac {2\,\sqrt {3}\,\sqrt {x}}{3}\right )}{3}-\ln \left (x+\sqrt {x}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x + x^(1/2) + 1),x)

[Out]

2*x^(1/2) - (2*3^(1/2)*atan(3^(1/2)/3 + (2*3^(1/2)*x^(1/2))/3))/3 - log(x + x^(1/2) + 1)

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sympy [A]  time = 0.25, size = 49, normalized size = 1.17 \[ 2 \sqrt {x} - \log {\left (4 \sqrt {x} + 4 x + 4 \right )} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} \sqrt {x}}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

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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