3.8.2 \(\int \sqrt {1+x+\sqrt {1+x}} \, dx\) [702]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {1976, 654, 626, 634, 212} \begin {gather*} \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 ) \end {gather*}

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 212

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

Rule 626

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 634

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 654

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 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rubi steps

\begin {align*} \int \sqrt {1+x+\sqrt {1+x}} \, dx &=2 \text {Subst}\left (\int x \sqrt {x (1+x)} \, dx,x,\sqrt {1+x}\right )\\ &=2 \text {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}-\text {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} \text {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} \text {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 = 0.09, size = 59, normalized size = 0.79 \begin {gather*} \frac {1}{12} \left (\sqrt {1+x+\sqrt {1+x}} \left (5+8 x+2 \sqrt {1+x}\right )+3 \tanh ^{-1}\left (\frac {\sqrt {1+x+\sqrt {1+x}}}{\sqrt {1+x}}\right )\right ) \end {gather*}

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*ArcTanh[Sqrt[1 + x + Sqrt[1 + x]]/Sqrt[1 + x]])/12

________________________________________________________________________________________

Maple [A]
time = 0.02, size = 55, normalized size = 0.73

method	result	size
derivativedivides	\(\frac {2 \left (1+x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {1+x +\sqrt {1+x}}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {1+x +\sqrt {1+x}}\right )}{8}\)	\(55\)
default	\(\frac {2 \left (1+x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {1+x +\sqrt {1+x}}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {1+x +\sqrt {1+x}}\right )}{8}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x+(1+x)^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[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))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.56, size = 61, normalized size = 0.81 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {x + \sqrt {x + 1} + 1}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [A]
time = 2.10, size = 55, normalized size = 0.73 \begin {gather*} \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 ) \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x+\sqrt {x+1}+1} \,d x \end {gather*}

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)

________________________________________________________________________________________