Integrand size = 11, antiderivative size = 62 \[ \int \sqrt {\sqrt {x}+x} \, dx=-\frac {1}{4} \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}+\frac {2}{3} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {\sqrt {x}+x}}\right ) \] Output:
-1/4*(1+2*x^(1/2))*(x^(1/2)+x)^(1/2)+2/3*(x^(1/2)+x)^(3/2)+1/4*arctanh(x^( 1/2)/(x^(1/2)+x)^(1/2))
Time = 0.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \sqrt {\sqrt {x}+x} \, dx=\frac {1}{12} \sqrt {\sqrt {x}+x} \left (-3+2 \sqrt {x}+8 x\right )+\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {\sqrt {x}+x}}{\sqrt {x}}\right ) \] Input:
Integrate[Sqrt[Sqrt[x] + x],x]
Output:
(Sqrt[Sqrt[x] + x]*(-3 + 2*Sqrt[x] + 8*x))/12 + ArcTanh[Sqrt[Sqrt[x] + x]/ Sqrt[x]]/4
Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {1910, 1924, 1134, 1160, 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.
\(\displaystyle \int \sqrt {x+\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 1910 |
\(\displaystyle \frac {1}{6} \int \frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}dx+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
\(\Big \downarrow \) 1924 |
\(\displaystyle \frac {1}{3} \int \frac {x}{\sqrt {x+\sqrt {x}}}d\sqrt {x}+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \sqrt {x} \sqrt {x+\sqrt {x}}-\frac {3}{4} \int \frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}d\sqrt {x}\right )+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \sqrt {x} \sqrt {x+\sqrt {x}}-\frac {3}{4} \left (\sqrt {x+\sqrt {x}}-\frac {1}{2} \int \frac {1}{\sqrt {x+\sqrt {x}}}d\sqrt {x}\right )\right )+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \sqrt {x} \sqrt {x+\sqrt {x}}-\frac {3}{4} \left (\sqrt {x+\sqrt {x}}-\int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}\right )\right )+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \sqrt {x} \sqrt {x+\sqrt {x}}-\frac {3}{4} \left (\sqrt {x+\sqrt {x}}-\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}\right )\right )\right )+\frac {2}{3} \sqrt {x+\sqrt {x}} x\) |
Input:
Int[Sqrt[Sqrt[x] + x],x]
Output:
(2*x*Sqrt[Sqrt[x] + x])/3 + ((Sqrt[x]*Sqrt[Sqrt[x] + x])/2 - (3*(Sqrt[Sqrt [x] + x] - ArcTanh[Sqrt[x]/Sqrt[Sqrt[x] + x]]))/4)/3
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[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_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^ (m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1)) Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && !IntegerQ[p] && LtQ[0, j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp [1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x ], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j ] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1 ]
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.66
method | result | size |
meijerg | \(-\frac {\frac {\sqrt {\pi }\, x^{\frac {1}{4}} \left (-40 x -10 \sqrt {x}+15\right ) \sqrt {1+\sqrt {x}}}{60}-\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {1}{4}}\right )}{4}}{\sqrt {\pi }}\) | \(41\) |
derivativedivides | \(\frac {2 \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {\sqrt {x}+x}\right )}{8}\) | \(42\) |
default | \(\frac {2 \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {\sqrt {x}+x}\right )}{8}\) | \(42\) |
Input:
int((x^(1/2)+x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/Pi^(1/2)*(1/60*Pi^(1/2)*x^(1/4)*(-40*x-10*x^(1/2)+15)*(1+x^(1/2))^(1/2) -1/4*Pi^(1/2)*arcsinh(x^(1/4)))
Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79 \[ \int \sqrt {\sqrt {x}+x} \, dx=\frac {1}{12} \, {\left (8 \, x + 2 \, \sqrt {x} - 3\right )} \sqrt {x + \sqrt {x}} + \frac {1}{16} \, \log \left (4 \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} + 1\right )} + 8 \, x + 8 \, \sqrt {x} + 1\right ) \] Input:
integrate((x^(1/2)+x)^(1/2),x, algorithm="fricas")
Output:
1/12*(8*x + 2*sqrt(x) - 3)*sqrt(x + sqrt(x)) + 1/16*log(4*sqrt(x + sqrt(x) )*(2*sqrt(x) + 1) + 8*x + 8*sqrt(x) + 1)
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.77 \[ \int \sqrt {\sqrt {x}+x} \, dx=2 \sqrt {\sqrt {x} + x} \left (\frac {\sqrt {x}}{12} + \frac {x}{3} - \frac {1}{8}\right ) + \frac {\log {\left (2 \sqrt {x} + 2 \sqrt {\sqrt {x} + x} + 1 \right )}}{8} \] Input:
integrate((x**(1/2)+x)**(1/2),x)
Output:
2*sqrt(sqrt(x) + x)*(sqrt(x)/12 + x/3 - 1/8) + log(2*sqrt(x) + 2*sqrt(sqrt (x) + x) + 1)/8
\[ \int \sqrt {\sqrt {x}+x} \, dx=\int { \sqrt {x + \sqrt {x}} \,d x } \] Input:
integrate((x^(1/2)+x)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(x + sqrt(x)), x)
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int \sqrt {\sqrt {x}+x} \, dx=\frac {1}{12} \, {\left (2 \, \sqrt {x} {\left (4 \, \sqrt {x} + 1\right )} - 3\right )} \sqrt {x + \sqrt {x}} - \frac {1}{8} \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right ) \] Input:
integrate((x^(1/2)+x)^(1/2),x, algorithm="giac")
Output:
1/12*(2*sqrt(x)*(4*sqrt(x) + 1) - 3)*sqrt(x + sqrt(x)) - 1/8*log(-2*sqrt(x + sqrt(x)) + 2*sqrt(x) + 1)
Time = 23.64 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.44 \[ \int \sqrt {\sqrt {x}+x} \, dx=\frac {4\,x\,\sqrt {x+\sqrt {x}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\sqrt {x}\right )}{5\,\sqrt {\sqrt {x}+1}} \] Input:
int((x + x^(1/2))^(1/2),x)
Output:
(4*x*(x + x^(1/2))^(1/2)*hypergeom([-1/2, 5/2], 7/2, -x^(1/2)))/(5*(x^(1/2 ) + 1)^(1/2))
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int \sqrt {\sqrt {x}+x} \, dx=\frac {x^{\frac {3}{4}} \sqrt {\sqrt {x}+1}}{6}+\frac {2 x^{\frac {5}{4}} \sqrt {\sqrt {x}+1}}{3}-\frac {x^{\frac {1}{4}} \sqrt {\sqrt {x}+1}}{4}+\frac {\mathrm {log}\left (\sqrt {\sqrt {x}+1}+x^{\frac {1}{4}}\right )}{4} \] Input:
int((x^(1/2)+x)^(1/2),x)
Output:
(2*x**(3/4)*sqrt(sqrt(x) + 1) + 8*x**(1/4)*sqrt(sqrt(x) + 1)*x - 3*x**(1/4 )*sqrt(sqrt(x) + 1) + 3*log(sqrt(sqrt(x) + 1) + x**(1/4)))/12