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

Optimal. Leaf size=62 \[ \frac{2}{3} \left (x+\sqrt{x}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x}+1\right ) \sqrt{x+\sqrt{x}+1}-\frac{3}{8} \sinh ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right ) \]

[Out]

-((1 + 2*Sqrt[x])*Sqrt[1 + Sqrt[x] + x])/4 + (2*(1 + Sqrt[x] + x)^(3/2))/3 - (3*ArcSinh[(1 + 2*Sqrt[x])/Sqrt[3
]])/8

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Rubi [A]  time = 0.0253673, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1341, 640, 612, 619, 215} \[ \frac{2}{3} \left (x+\sqrt{x}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x}+1\right ) \sqrt{x+\sqrt{x}+1}-\frac{3}{8} \sinh ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Sqrt[x] + x],x]

[Out]

-((1 + 2*Sqrt[x])*Sqrt[1 + Sqrt[x] + x])/4 + (2*(1 + Sqrt[x] + x)^(3/2))/3 - (3*ArcSinh[(1 + 2*Sqrt[x])/Sqrt[3
]])/8

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 640

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

Rule 612

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \sqrt{1+\sqrt{x}+x} \, dx &=2 \operatorname{Subst}\left (\int x \sqrt{1+x+x^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} \left (1+\sqrt{x}+x\right )^{3/2}-\operatorname{Subst}\left (\int \sqrt{1+x+x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1+\sqrt{x}+x}+\frac{2}{3} \left (1+\sqrt{x}+x\right )^{3/2}-\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x+x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1+\sqrt{x}+x}+\frac{2}{3} \left (1+\sqrt{x}+x\right )^{3/2}-\frac{1}{8} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 \sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1+\sqrt{x}+x}+\frac{2}{3} \left (1+\sqrt{x}+x\right )^{3/2}-\frac{3}{8} \sinh ^{-1}\left (\frac{1+2 \sqrt{x}}{\sqrt{3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0186094, size = 49, normalized size = 0.79 \[ \frac{1}{24} \left (2 \sqrt{x+\sqrt{x}+1} \left (8 x+2 \sqrt{x}+5\right )-9 \sinh ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Sqrt[x] + x],x]

[Out]

(2*Sqrt[1 + Sqrt[x] + x]*(5 + 2*Sqrt[x] + 8*x) - 9*ArcSinh[(1 + 2*Sqrt[x])/Sqrt[3]])/24

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Maple [A]  time = 0.006, size = 42, normalized size = 0.7 \begin{align*}{\frac{2}{3} \left ( 1+x+\sqrt{x} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{4} \left ( 1+2\,\sqrt{x} \right ) \sqrt{1+x+\sqrt{x}}}-{\frac{3}{8}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( \sqrt{x}+{\frac{1}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(1+x+x^(1/2))^(3/2)-1/4*(1+2*x^(1/2))*(1+x+x^(1/2))^(1/2)-3/8*arcsinh(2/3*3^(1/2)*(x^(1/2)+1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x + \sqrt{x} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + sqrt(x) + 1), x)

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Fricas [A]  time = 4.14163, size = 167, normalized size = 2.69 \begin{align*} \frac{1}{12} \,{\left (8 \, x + 2 \, \sqrt{x} + 5\right )} \sqrt{x + \sqrt{x} + 1} + \frac{3}{16} \, \log \left (4 \, \sqrt{x + \sqrt{x} + 1}{\left (2 \, \sqrt{x} + 1\right )} - 8 \, x - 8 \, \sqrt{x} - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*x + 2*sqrt(x) + 5)*sqrt(x + sqrt(x) + 1) + 3/16*log(4*sqrt(x + sqrt(x) + 1)*(2*sqrt(x) + 1) - 8*x - 8*
sqrt(x) - 5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sqrt{x} + x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sqrt(x) + x + 1), x)

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Giac [A]  time = 1.14488, size = 61, normalized size = 0.98 \begin{align*} \frac{1}{12} \,{\left (2 \, \sqrt{x}{\left (4 \, \sqrt{x} + 1\right )} + 5\right )} \sqrt{x + \sqrt{x} + 1} + \frac{3}{8} \, \log \left (2 \, \sqrt{x + \sqrt{x} + 1} - 2 \, \sqrt{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*sqrt(x)*(4*sqrt(x) + 1) + 5)*sqrt(x + sqrt(x) + 1) + 3/8*log(2*sqrt(x + sqrt(x) + 1) - 2*sqrt(x) - 1)