Integrand size = 17, antiderivative size = 82 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\frac {5}{32} \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}-\frac {5}{12} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}-\frac {5}{32} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {\sqrt {x}+x}}\right ) \] Output:
5/32*(1+2*x^(1/2))*(x^(1/2)+x)^(1/2)-5/12*(x^(1/2)+x)^(3/2)+1/2*x^(1/2)*(x ^(1/2)+x)^(3/2)-5/32*arctanh(x^(1/2)/(x^(1/2)+x)^(1/2))
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\frac {1}{96} \sqrt {\sqrt {x}+x} \left (15-10 \sqrt {x}+8 x+48 x^{3/2}\right )-\frac {5}{32} \text {arctanh}\left (\frac {\sqrt {\sqrt {x}+x}}{\sqrt {x}}\right ) \] Input:
Integrate[Sqrt[x]*Sqrt[Sqrt[x] + x],x]
Output:
(Sqrt[Sqrt[x] + x]*(15 - 10*Sqrt[x] + 8*x + 48*x^(3/2)))/96 - (5*ArcTanh[S qrt[Sqrt[x] + x]/Sqrt[x]])/32
Time = 0.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1924, 1134, 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.
\(\displaystyle \int \sqrt {x} \sqrt {x+\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 1924 |
\(\displaystyle 2 \int x \sqrt {x+\sqrt {x}}d\sqrt {x}\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{8} \int \sqrt {x} \sqrt {x+\sqrt {x}}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{8} \left (\frac {1}{3} \left (x+\sqrt {x}\right )^{3/2}-\frac {1}{2} \int \sqrt {x+\sqrt {x}}d\sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{8} \int \frac {1}{\sqrt {x+\sqrt {x}}}d\sqrt {x}-\frac {1}{4} \left (2 \sqrt {x}+1\right ) \sqrt {x+\sqrt {x}}\right )+\frac {1}{3} \left (x+\sqrt {x}\right )^{3/2}\right )\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}-\frac {1}{4} \left (2 \sqrt {x}+1\right ) \sqrt {x+\sqrt {x}}\right )+\frac {1}{3} \left (x+\sqrt {x}\right )^{3/2}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}\right )-\frac {1}{4} \left (2 \sqrt {x}+1\right ) \sqrt {x+\sqrt {x}}\right )+\frac {1}{3} \left (x+\sqrt {x}\right )^{3/2}\right )\right )\) |
Input:
Int[Sqrt[x]*Sqrt[Sqrt[x] + x],x]
Output:
2*((Sqrt[x]*(Sqrt[x] + x)^(3/2))/4 - (5*((Sqrt[x] + x)^(3/2)/3 + (-1/4*((1 + 2*Sqrt[x])*Sqrt[Sqrt[x] + x]) + ArcTanh[Sqrt[x]/Sqrt[Sqrt[x] + x]]/4)/2 ))/8)
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_))^(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[(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 = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56
method | result | size |
meijerg | \(-\frac {-\frac {\sqrt {\pi }\, x^{\frac {1}{4}} \left (336 x^{\frac {3}{2}}+56 x -70 \sqrt {x}+105\right ) \sqrt {1+\sqrt {x}}}{672}+\frac {5 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {1}{4}}\right )}{32}}{\sqrt {\pi }}\) | \(46\) |
derivativedivides | \(\frac {\sqrt {x}\, \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{2}-\frac {5 \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{12}+\frac {5 \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}}{32}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {\sqrt {x}+x}\right )}{64}\) | \(54\) |
default | \(\frac {\sqrt {x}\, \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{2}-\frac {5 \left (\sqrt {x}+x \right )^{\frac {3}{2}}}{12}+\frac {5 \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}}{32}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {\sqrt {x}+x}\right )}{64}\) | \(54\) |
Input:
int(x^(1/2)*(x^(1/2)+x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/Pi^(1/2)*(-1/672*Pi^(1/2)*x^(1/4)*(336*x^(3/2)+56*x-70*x^(1/2)+105)*(1+ x^(1/2))^(1/2)+5/32*Pi^(1/2)*arcsinh(x^(1/4)))
Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.66 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\frac {1}{96} \, {\left (2 \, {\left (24 \, x - 5\right )} \sqrt {x} + 8 \, x + 15\right )} \sqrt {x + \sqrt {x}} + \frac {5}{128} \, \log \left (4 \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} + 1\right )} - 8 \, x - 8 \, \sqrt {x} - 1\right ) \] Input:
integrate(x^(1/2)*(x^(1/2)+x)^(1/2),x, algorithm="fricas")
Output:
1/96*(2*(24*x - 5)*sqrt(x) + 8*x + 15)*sqrt(x + sqrt(x)) + 5/128*log(4*sqr t(x + sqrt(x))*(2*sqrt(x) + 1) - 8*x - 8*sqrt(x) - 1)
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=2 \sqrt {\sqrt {x} + x} \left (\frac {x^{\frac {3}{2}}}{4} - \frac {5 \sqrt {x}}{96} + \frac {x}{24} + \frac {5}{64}\right ) - \frac {5 \log {\left (2 \sqrt {x} + 2 \sqrt {\sqrt {x} + x} + 1 \right )}}{64} \] Input:
integrate(x**(1/2)*(x**(1/2)+x)**(1/2),x)
Output:
2*sqrt(sqrt(x) + x)*(x**(3/2)/4 - 5*sqrt(x)/96 + x/24 + 5/64) - 5*log(2*sq rt(x) + 2*sqrt(sqrt(x) + x) + 1)/64
\[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\int { \sqrt {x + \sqrt {x}} \sqrt {x} \,d x } \] Input:
integrate(x^(1/2)*(x^(1/2)+x)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(x + sqrt(x))*sqrt(x), x)
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.61 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\frac {1}{96} \, {\left (2 \, {\left (4 \, \sqrt {x} {\left (6 \, \sqrt {x} + 1\right )} - 5\right )} \sqrt {x} + 15\right )} \sqrt {x + \sqrt {x}} + \frac {5}{64} \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right ) \] Input:
integrate(x^(1/2)*(x^(1/2)+x)^(1/2),x, algorithm="giac")
Output:
1/96*(2*(4*sqrt(x)*(6*sqrt(x) + 1) - 5)*sqrt(x) + 15)*sqrt(x + sqrt(x)) + 5/64*log(-2*sqrt(x + sqrt(x)) + 2*sqrt(x) + 1)
Timed out. \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\int \sqrt {x}\,\sqrt {x+\sqrt {x}} \,d x \] Input:
int(x^(1/2)*(x + x^(1/2))^(1/2),x)
Output:
int(x^(1/2)*(x + x^(1/2))^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx=\frac {x^{\frac {7}{4}} \sqrt {\sqrt {x}+1}}{2}-\frac {5 x^{\frac {3}{4}} \sqrt {\sqrt {x}+1}}{48}+\frac {x^{\frac {5}{4}} \sqrt {\sqrt {x}+1}}{12}+\frac {5 x^{\frac {1}{4}} \sqrt {\sqrt {x}+1}}{32}-\frac {5 \,\mathrm {log}\left (\sqrt {\sqrt {x}+1}+x^{\frac {1}{4}}\right )}{32} \] Input:
int(x^(1/2)*(x^(1/2)+x)^(1/2),x)
Output:
(48*x**(3/4)*sqrt(sqrt(x) + 1)*x - 10*x**(3/4)*sqrt(sqrt(x) + 1) + 8*x**(1 /4)*sqrt(sqrt(x) + 1)*x + 15*x**(1/4)*sqrt(sqrt(x) + 1) - 15*log(sqrt(sqrt (x) + 1) + x**(1/4)))/96