\(\int \sqrt {x+\sqrt {1+x}} \, dx\) [1043]

Optimal result

Integrand size = 13, antiderivative size = 78 \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\frac {1}{6} \sqrt {1+x} \sqrt {x+\sqrt {1+x}}+\frac {1}{12} (-3+8 x) \sqrt {x+\sqrt {1+x}}-\frac {5}{8} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]

[Out]

1/6*(1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2)+1/12*(-3+8*x)*(x+(1+x)^(1/2))^(1/2)-5/8*ln(1+2*(1+x)^(1/2)-2*(x+(1+x)^(1
/2))^(1/2))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {654, 626, 635, 212} \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\frac {5}{8} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )+\frac {2}{3} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {1}{4} \left (2 \sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}} \]

[In]

Int[Sqrt[x + Sqrt[1 + x]],x]

[Out]

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

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 635

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

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]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right ) \\ & = \frac {2}{3} \left (x+\sqrt {1+x}\right )^{3/2}-\text {Subst}\left (\int \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right ) \\ & = \frac {2}{3} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {1}{4} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {5}{8} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = \frac {2}{3} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {1}{4} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {5}{4} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right ) \\ & = \frac {2}{3} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {1}{4} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {5}{8} \text {arctanh}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\frac {1}{12} \sqrt {x+\sqrt {1+x}} \left (-11+2 \sqrt {1+x}+8 (1+x)\right )-\frac {5}{8} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]

[In]

Integrate[Sqrt[x + Sqrt[1 + x]],x]

[Out]

(Sqrt[x + Sqrt[1 + x]]*(-11 + 2*Sqrt[1 + x] + 8*(1 + x)))/12 - (5*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 +
x]]])/8

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{4}+\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{8}\)	\(52\)
default	\(\frac {2 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{4}+\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{8}\)	\(52\)

[In]

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

[Out]

2/3*(x+(1+x)^(1/2))^(3/2)-1/4*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)+5/8*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(
1/2))

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76 \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\frac {1}{12} \, {\left (8 \, x + 2 \, \sqrt {x + 1} - 3\right )} \sqrt {x + \sqrt {x + 1}} + \frac {5}{16} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + 8 \, x + 8 \, \sqrt {x + 1} + 5\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \sqrt {x+\sqrt {1+x}} \, dx=2 \sqrt {x + \sqrt {x + 1}} \left (\frac {x}{3} + \frac {\sqrt {x + 1}}{12} - \frac {1}{8}\right ) + \frac {5 \log {\left (2 \sqrt {x + 1} + 2 \sqrt {x + \sqrt {x + 1}} + 1 \right )}}{8} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sqrt {x+\sqrt {1+x}} \, dx=\int { \sqrt {x + \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(x + sqrt(x + 1)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.68 \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\frac {1}{12} \, {\left (2 \, \sqrt {x + 1} {\left (4 \, \sqrt {x + 1} + 1\right )} - 11\right )} \sqrt {x + \sqrt {x + 1}} - \frac {5}{8} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \]

[In]

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

[Out]

1/12*(2*sqrt(x + 1)*(4*sqrt(x + 1) + 1) - 11)*sqrt(x + sqrt(x + 1)) - 5/8*log(-2*sqrt(x + sqrt(x + 1)) + 2*sqr
t(x + 1) + 1)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x+\sqrt {1+x}} \, dx=\int \sqrt {x+\sqrt {x+1}} \,d x \]

[In]

int((x + (x + 1)^(1/2))^(1/2),x)

[Out]

int((x + (x + 1)^(1/2))^(1/2), x)