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

Optimal. Leaf size=75 \[ \frac{2}{3} \left (x+\sqrt{x+1}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x+1}+1\right ) \sqrt{x+\sqrt{x+1}+1}+\frac{1}{4} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}+1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0440683, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {1980, 640, 612, 620, 206} \[ \frac{2}{3} \left (x+\sqrt{x+1}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x+1}+1\right ) \sqrt{x+\sqrt{x+1}+1}+\frac{1}{4} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x + Sqrt[1 + x]],x]

[Out]

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

Rule 1980

Int[(u_)^(p_.)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{c, m, p}, x] &&
GeneralizedBinomialQ[u, x] &&  !GeneralizedBinomialMatchQ[u, x]

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0565714, size = 62, normalized size = 0.83 \[ \frac{1}{12} \sqrt{x+\sqrt{x+1}+1} \left (8 x+2 \sqrt{x+1}+\frac{3 \sinh ^{-1}\left (\sqrt [4]{x+1}\right )}{\sqrt [4]{x+1} \sqrt{\sqrt{x+1}+1}}+5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x + Sqrt[1 + x]],x]

[Out]

(Sqrt[1 + x + Sqrt[1 + x]]*(5 + 8*x + 2*Sqrt[1 + x] + (3*ArcSinh[(1 + x)^(1/4)])/((1 + x)^(1/4)*Sqrt[1 + Sqrt[
1 + x]])))/12

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Maple [A]  time = 0.004, size = 55, normalized size = 0.7 \begin{align*}{\frac{2}{3} \left ( 1+x+\sqrt{1+x} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{4} \left ( 1+2\,\sqrt{1+x} \right ) \sqrt{1+x+\sqrt{1+x}}}+{\frac{1}{8}\ln \left ({\frac{1}{2}}+\sqrt{1+x}+\sqrt{1+x+\sqrt{1+x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(1+x+(1+x)^(1/2))^(3/2)-1/4*(1+2*(1+x)^(1/2))*(1+x+(1+x)^(1/2))^(1/2)+1/8*ln(1/2+(1+x)^(1/2)+(1+x+(1+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} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + sqrt(x + 1) + 1), x)

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Fricas [A]  time = 4.14999, size = 196, normalized size = 2.61 \begin{align*} \frac{1}{12} \,{\left (8 \, x + 2 \, \sqrt{x + 1} + 5\right )} \sqrt{x + \sqrt{x + 1} + 1} + \frac{1}{16} \, \log \left (-4 \, \sqrt{x + \sqrt{x + 1} + 1}{\left (2 \, \sqrt{x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt{x + 1} - 9\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*x + 2*sqrt(x + 1) + 5)*sqrt(x + sqrt(x + 1) + 1) + 1/16*log(-4*sqrt(x + sqrt(x + 1) + 1)*(2*sqrt(x + 1
) + 1) - 8*x - 8*sqrt(x + 1) - 9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x + sqrt(x + 1) + 1), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out