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]

1/4*arctanh((1+x)^(1/2)/(1+x+(1+x)^(1/2))^(1/2))+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)

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.87, size = 61, normalized size = 0.81 \[ \frac {1}{12} \, {\left (8 \, x + 2 \, \sqrt {x + 1} + 5\right )} \sqrt {x + \sqrt {x + 1} + 1} + \frac {1}{16} \, \log \left (-4 \, \sqrt {x + \sqrt {x + 1} + 1} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 9\right ) \]

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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giac [A]  time = 0.41, size = 55, normalized size = 0.73 \[ \frac {1}{12} \, {\left (2 \, \sqrt {x + 1} {\left (4 \, \sqrt {x + 1} + 1\right )} - 3\right )} \sqrt {x + \sqrt {x + 1} + 1} - \frac {1}{8} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1} + 1} + 2 \, \sqrt {x + 1} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.73 \[ \frac {\ln \left (\sqrt {x +1}+\frac {1}{2}+\sqrt {x +1+\sqrt {x +1}}\right )}{8}+\frac {2 \left (x +1+\sqrt {x +1}\right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {x +1}\right ) \sqrt {x +1+\sqrt {x +1}}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x + \sqrt {x + 1} + 1}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {x+\sqrt {x+1}+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + (x + 1)^(1/2) + 1)^(1/2),x)

[Out]

int((x + (x + 1)^(1/2) + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x + \sqrt {x + 1} + 1}\, dx \]

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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