Integrand size = 14, antiderivative size = 75 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\frac {2}{3} \left (1+x+\sqrt {1+x}\right )^{3/2}-\frac {1}{4} \sqrt {1+x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {1+x}}{\sqrt {1+x+\sqrt {1+x}}}\right ) \] Output:
2/3*(1+x+(1+x)^(1/2))^(3/2)-1/4*(1+x+(1+x)^(1/2))^(1/2)*(1+2*(1+x)^(1/2))+ 1/4*arctanh((1+x)^(1/2)/(1+x+(1+x)^(1/2))^(1/2))
Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\frac {1}{12} \left (\sqrt {1+x+\sqrt {1+x}} \left (5+8 x+2 \sqrt {1+x}\right )+3 \text {arctanh}\left (\frac {\sqrt {1+x+\sqrt {1+x}}}{\sqrt {1+x}}\right )\right ) \] Input:
Integrate[Sqrt[1 + x + Sqrt[1 + x]],x]
Output:
(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
Time = 0.41 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {7267, 2048, 1160, 1087, 1091, 219}
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.
\(\displaystyle \int \sqrt {x+\sqrt {x+1}+1} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int \sqrt {x+1} \sqrt {\sqrt {x+1} \left (\sqrt {x+1}+1\right )}d\sqrt {x+1}\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle 2 \int \sqrt {x+1} \sqrt {x+\sqrt {x+1}+1}d\sqrt {x+1}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle 2 \left (\frac {1}{3} \left (x+\sqrt {x+1}+1\right )^{3/2}-\frac {1}{2} \int \sqrt {x+\sqrt {x+1}+1}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{8} \int \frac {1}{\sqrt {x+\sqrt {x+1}+1}}d\sqrt {x+1}-\frac {1}{4} \sqrt {x+\sqrt {x+1}+1} \left (2 \sqrt {x+1}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x+1}+1\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{4} \int -\frac {1}{x}d\frac {\sqrt {x+1}}{\sqrt {x+\sqrt {x+1}+1}}-\frac {1}{4} \sqrt {x+\sqrt {x+1}+1} \left (2 \sqrt {x+1}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x+1}+1\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {x+1}}{\sqrt {x+\sqrt {x+1}+1}}\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}+1} \left (2 \sqrt {x+1}+1\right )\right )+\frac {1}{3} \left (x+\sqrt {x+1}+1\right )^{3/2}\right )\) |
Input:
Int[Sqrt[1 + x + Sqrt[1 + x]],x]
Output:
2*((1 + x + Sqrt[1 + x])^(3/2)/3 + (-1/4*(Sqrt[1 + x + Sqrt[1 + x]]*(1 + 2 *Sqrt[1 + x])) + ArcTanh[Sqrt[1 + x]/Sqrt[1 + x + Sqrt[1 + x]]]/4)/2)
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] && (Gt Q[a, 0] || LtQ[b, 0])
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[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
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[(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] /; F reeQ[{a, b, c, d, e, n, p}, x]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {2 \left (1+x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {1+x +\sqrt {1+x}}\, \left (1+2 \sqrt {1+x}\right )}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {1+x +\sqrt {1+x}}\right )}{8}\) | \(55\) |
default | \(\frac {2 \left (1+x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {1+x +\sqrt {1+x}}\, \left (1+2 \sqrt {1+x}\right )}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {1+x +\sqrt {1+x}}\right )}{8}\) | \(55\) |
Input:
int((1+x+(1+x)^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*(1+x+(1+x)^(1/2))^(3/2)-1/4*(1+x+(1+x)^(1/2))^(1/2)*(1+2*(1+x)^(1/2))+ 1/8*ln(1/2+(1+x)^(1/2)+(1+x+(1+x)^(1/2))^(1/2))
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\frac {1}{12} \, {\left (8 \, x + 2 \, \sqrt {x + 1} + 5\right )} \sqrt {x + \sqrt {x + 1} + 1} + \frac {1}{16} \, \log \left (-4 \, \sqrt {x + \sqrt {x + 1} + 1} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 9\right ) \] Input:
integrate((1+x+(1+x)^(1/2))^(1/2),x, algorithm="fricas")
Output:
1/12*(8*x + 2*sqrt(x + 1) + 5)*sqrt(x + sqrt(x + 1) + 1) + 1/16*log(-4*sqr t(x + sqrt(x + 1) + 1)*(2*sqrt(x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 9)
Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=2 \left (\frac {x}{3} + \frac {\sqrt {x + 1}}{12} + \frac {5}{24}\right ) \sqrt {x + \sqrt {x + 1} + 1} + \frac {\log {\left (2 \sqrt {x + 1} + 2 \sqrt {x + \sqrt {x + 1} + 1} + 1 \right )}}{8} \] Input:
integrate((1+x+(1+x)**(1/2))**(1/2),x)
Output:
2*(x/3 + sqrt(x + 1)/12 + 5/24)*sqrt(x + sqrt(x + 1) + 1) + log(2*sqrt(x + 1) + 2*sqrt(x + sqrt(x + 1) + 1) + 1)/8
\[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\int { \sqrt {x + \sqrt {x + 1} + 1} \,d x } \] Input:
integrate((1+x+(1+x)^(1/2))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(x + sqrt(x + 1) + 1), x)
Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\frac {1}{12} \, {\left (2 \, \sqrt {x + 1} {\left (4 \, \sqrt {x + 1} + 1\right )} - 3\right )} \sqrt {x + \sqrt {x + 1} + 1} - \frac {1}{8} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1} + 1} + 2 \, \sqrt {x + 1} + 1\right ) \] Input:
integrate((1+x+(1+x)^(1/2))^(1/2),x, algorithm="giac")
Output:
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)
Timed out. \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\int \sqrt {x+\sqrt {x+1}+1} \,d x \] Input:
int((x + (x + 1)^(1/2) + 1)^(1/2),x)
Output:
int((x + (x + 1)^(1/2) + 1)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \sqrt {1+x+\sqrt {1+x}} \, dx=\frac {\left (x +1\right )^{\frac {3}{4}} \sqrt {\sqrt {x +1}+1}}{6}+\frac {2 \left (x +1\right )^{\frac {1}{4}} \sqrt {\sqrt {x +1}+1}\, x}{3}+\frac {5 \left (x +1\right )^{\frac {1}{4}} \sqrt {\sqrt {x +1}+1}}{12}+\frac {\mathrm {log}\left (\sqrt {\sqrt {x +1}+1}+\left (x +1\right )^{\frac {1}{4}}\right )}{4} \] Input:
int((1+x+(1+x)^(1/2))^(1/2),x)
Output:
(2*(x + 1)**(3/4)*sqrt(sqrt(x + 1) + 1) + 8*(x + 1)**(1/4)*sqrt(sqrt(x + 1 ) + 1)*x + 5*(x + 1)**(1/4)*sqrt(sqrt(x + 1) + 1) + 3*log(sqrt(sqrt(x + 1) + 1) + (x + 1)**(1/4)))/12