Integrand size = 12, antiderivative size = 62 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=-\frac {1}{4} \left (1+2 \sqrt {x}\right ) \sqrt {1+\sqrt {x}+x}+\frac {2}{3} \left (1+\sqrt {x}+x\right )^{3/2}-\frac {3}{8} \text {arcsinh}\left (\frac {1+2 \sqrt {x}}{\sqrt {3}}\right ) \] Output:
-1/4*(1+2*x^(1/2))*(1+x^(1/2)+x)^(1/2)+2/3*(1+x^(1/2)+x)^(3/2)-3/8*arcsinh (1/3*(1+2*x^(1/2))*3^(1/2))
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=\frac {1}{12} \sqrt {1+\sqrt {x}+x} \left (5+2 \sqrt {x}+8 x\right )+\frac {3}{8} \log \left (-1-2 \sqrt {x}+2 \sqrt {1+\sqrt {x}+x}\right ) \] Input:
Integrate[Sqrt[1 + Sqrt[x] + x],x]
Output:
(Sqrt[1 + Sqrt[x] + x]*(5 + 2*Sqrt[x] + 8*x))/12 + (3*Log[-1 - 2*Sqrt[x] + 2*Sqrt[1 + Sqrt[x] + x]])/8
Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1680, 1160, 1087, 1090, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {x+\sqrt {x}+1} \, dx\) |
\(\Big \downarrow \) 1680 |
\(\displaystyle 2 \int \sqrt {x} \sqrt {x+\sqrt {x}+1}d\sqrt {x}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle 2 \left (\frac {1}{3} \left (x+\sqrt {x}+1\right )^{3/2}-\frac {1}{2} \int \sqrt {x+\sqrt {x}+1}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {3}{8} \int \frac {1}{\sqrt {x+\sqrt {x}+1}}d\sqrt {x}-\frac {1}{4} \sqrt {x+\sqrt {x}+1} \left (2 \sqrt {x}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x}+1\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {1}{8} \sqrt {3} \int \frac {1}{\sqrt {\frac {x}{3}+1}}d\left (2 \sqrt {x}+1\right )-\frac {1}{4} \sqrt {x+\sqrt {x}+1} \left (2 \sqrt {x}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x}+1\right )^{3/2}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {3}{8} \text {arcsinh}\left (\frac {2 \sqrt {x}+1}{\sqrt {3}}\right )-\frac {1}{4} \sqrt {x+\sqrt {x}+1} \left (2 \sqrt {x}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x}+1\right )^{3/2}\right )\) |
Input:
Int[Sqrt[1 + Sqrt[x] + x],x]
Output:
2*((1 + Sqrt[x] + x)^(3/2)/3 + (-1/4*((1 + 2*Sqrt[x])*Sqrt[1 + Sqrt[x] + x ]) - (3*ArcSinh[(1 + 2*Sqrt[x])/Sqrt[3]])/8)/2)
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]
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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[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]
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] + Simp[(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[p, -1]
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[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] && Fr actionQ[n]
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {2 \left (1+\sqrt {x}+x \right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {x}\right ) \sqrt {1+\sqrt {x}+x}}{4}-\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (\sqrt {x}+\frac {1}{2}\right )}{3}\right )}{8}\) | \(42\) |
default | \(\frac {2 \left (1+\sqrt {x}+x \right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {x}\right ) \sqrt {1+\sqrt {x}+x}}{4}-\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (\sqrt {x}+\frac {1}{2}\right )}{3}\right )}{8}\) | \(42\) |
Input:
int((1+x^(1/2)+x)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*(1+x^(1/2)+x)^(3/2)-1/4*(1+2*x^(1/2))*(1+x^(1/2)+x)^(1/2)-3/8*arcsinh( 2/3*3^(1/2)*(x^(1/2)+1/2))
Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=\frac {1}{12} \, {\left (8 \, x + 2 \, \sqrt {x} + 5\right )} \sqrt {x + \sqrt {x} + 1} + \frac {3}{16} \, \log \left (4 \, \sqrt {x + \sqrt {x} + 1} {\left (2 \, \sqrt {x} + 1\right )} - 8 \, x - 8 \, \sqrt {x} - 5\right ) \] Input:
integrate((1+x^(1/2)+x)^(1/2),x, algorithm="fricas")
Output:
1/12*(8*x + 2*sqrt(x) + 5)*sqrt(x + sqrt(x) + 1) + 3/16*log(4*sqrt(x + sqr t(x) + 1)*(2*sqrt(x) + 1) - 8*x - 8*sqrt(x) - 5)
Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.77 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=2 \left (\frac {\sqrt {x}}{12} + \frac {x}{3} + \frac {5}{24}\right ) \sqrt {\sqrt {x} + x + 1} - \frac {3 \operatorname {asinh}{\left (\frac {2 \sqrt {3} \left (\sqrt {x} + \frac {1}{2}\right )}{3} \right )}}{8} \] Input:
integrate((1+x**(1/2)+x)**(1/2),x)
Output:
2*(sqrt(x)/12 + x/3 + 5/24)*sqrt(sqrt(x) + x + 1) - 3*asinh(2*sqrt(3)*(sqr t(x) + 1/2)/3)/8
\[ \int \sqrt {1+\sqrt {x}+x} \, dx=\int { \sqrt {x + \sqrt {x} + 1} \,d x } \] Input:
integrate((1+x^(1/2)+x)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(x + sqrt(x) + 1), x)
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=\frac {1}{12} \, {\left (2 \, \sqrt {x} {\left (4 \, \sqrt {x} + 1\right )} + 5\right )} \sqrt {x + \sqrt {x} + 1} + \frac {3}{8} \, \log \left (2 \, \sqrt {x + \sqrt {x} + 1} - 2 \, \sqrt {x} - 1\right ) \] Input:
integrate((1+x^(1/2)+x)^(1/2),x, algorithm="giac")
Output:
1/12*(2*sqrt(x)*(4*sqrt(x) + 1) + 5)*sqrt(x + sqrt(x) + 1) + 3/8*log(2*sqr t(x + sqrt(x) + 1) - 2*sqrt(x) - 1)
Timed out. \[ \int \sqrt {1+\sqrt {x}+x} \, dx=\int \sqrt {x+\sqrt {x}+1} \,d x \] Input:
int((x + x^(1/2) + 1)^(1/2),x)
Output:
int((x + x^(1/2) + 1)^(1/2), x)
Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \sqrt {1+\sqrt {x}+x} \, dx=\frac {\sqrt {x}\, \sqrt {\sqrt {x}+x +1}}{6}+\frac {2 \sqrt {\sqrt {x}+x +1}\, x}{3}+\frac {5 \sqrt {\sqrt {x}+x +1}}{12}-\frac {3 \,\mathrm {log}\left (\frac {2 \sqrt {\sqrt {x}+x +1}+2 \sqrt {x}+1}{\sqrt {3}}\right )}{8} \] Input:
int((1+x^(1/2)+x)^(1/2),x)
Output:
(4*sqrt(x)*sqrt(sqrt(x) + x + 1) + 16*sqrt(sqrt(x) + x + 1)*x + 10*sqrt(sq rt(x) + x + 1) - 9*log((2*sqrt(sqrt(x) + x + 1) + 2*sqrt(x) + 1)/sqrt(3))) /24