\(\int (\sqrt {x}+x)^{2/3} \, dx\) [325]

Optimal result

Integrand size = 11, antiderivative size = 769 \[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=-\frac {3 \left (1+2 \sqrt {x}\right ) \sqrt [3]{-\sqrt {x}-x}}{7 \sqrt [3]{2} \left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right ) \sqrt [3]{\sqrt {x}+x}}-\frac {3}{14} \left (1+2 \sqrt {x}\right ) \left (\sqrt {x}+x\right )^{2/3}+\frac {3}{5} \left (\sqrt {x}+x\right )^{5/3}-\frac {3 \sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (1-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}+2 \sqrt [3]{2} \left (-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right )^2}} \sqrt [3]{-\sqrt {x}-x} E\left (\arcsin \left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}\right )|-7+4 \sqrt {3}\right )}{14 \sqrt [3]{2} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right )^2}} \left (1+2 \sqrt {x}\right ) \sqrt [3]{\sqrt {x}+x}}+\frac {\sqrt [6]{2} 3^{3/4} \left (1-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}+2 \sqrt [3]{2} \left (-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right )^2}} \sqrt [3]{-\sqrt {x}-x} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}\right ),-7+4 \sqrt {3}\right )}{7 \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\left (\left (1+\sqrt {x}\right ) \sqrt {x}\right )}\right )^2}} \left (1+2 \sqrt {x}\right ) \sqrt [3]{\sqrt {x}+x}} \] Output:

-3/14*(1+2*x^(1/2))*(-x^(1/2)-x)^(1/3)*2^(2/3)/(1-3^(1/2)-2^(2/3)*(-(1+x^( 
1/2))*x^(1/2))^(1/3))/(x^(1/2)+x)^(1/3)-3/14*(1+2*x^(1/2))*(x^(1/2)+x)^(2/ 
3)+3/5*(x^(1/2)+x)^(5/3)-3/28*3^(1/4)*(1/2*6^(1/2)+1/2*2^(1/2))*(1-2^(2/3) 
*(-(1+x^(1/2))*x^(1/2))^(1/3))*((1+2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3)+2* 
2^(1/3)*(-(1+x^(1/2))*x^(1/2))^(2/3))/(1-3^(1/2)-2^(2/3)*(-(1+x^(1/2))*x^( 
1/2))^(1/3))^2)^(1/2)*(-x^(1/2)-x)^(1/3)*EllipticE((1+3^(1/2)-2^(2/3)*(-(1 
+x^(1/2))*x^(1/2))^(1/3))/(1-3^(1/2)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3)) 
,2*I-I*3^(1/2))*2^(2/3)/(-(1-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))/(1-3^(1 
/2)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))^2)^(1/2)/(1+2*x^(1/2))/(x^(1/2)+ 
x)^(1/3)+1/7*2^(1/6)*3^(3/4)*(1-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))*((1+ 
2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3)+2*2^(1/3)*(-(1+x^(1/2))*x^(1/2))^(2/3 
))/(1-3^(1/2)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))^2)^(1/2)*(-x^(1/2)-x)^ 
(1/3)*EllipticF((1+3^(1/2)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))/(1-3^(1/2 
)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3)),2*I-I*3^(1/2))/(-(1-2^(2/3)*(-(1+x 
^(1/2))*x^(1/2))^(1/3))/(1-3^(1/2)-2^(2/3)*(-(1+x^(1/2))*x^(1/2))^(1/3))^2 
)^(1/2)/(1+2*x^(1/2))/(x^(1/2)+x)^(1/3)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.06 \[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\frac {3 x \left (\sqrt {x}+x\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {8}{3},\frac {11}{3},-\sqrt {x}\right )}{4 \left (1+\sqrt {x}\right )^{2/3}} \] Input:

Integrate[(Sqrt[x] + x)^(2/3),x]
 

Output:

(3*x*(Sqrt[x] + x)^(2/3)*Hypergeometric2F1[-2/3, 8/3, 11/3, -Sqrt[x]])/(4* 
(1 + Sqrt[x])^(2/3))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1910, 1924, 1137, 74}

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.

\(\displaystyle \int \left (x+\sqrt {x}\right )^{2/3} \, dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{5} \int \frac {\sqrt {x}}{\sqrt [3]{x+\sqrt {x}}}dx+\frac {3}{5} \left (x+\sqrt {x}\right )^{2/3} x\)

\(\Big \downarrow \) 1924

\(\displaystyle \frac {2}{5} \int \frac {x}{\sqrt [3]{x+\sqrt {x}}}d\sqrt {x}+\frac {3}{5} \left (x+\sqrt {x}\right )^{2/3} x\)

\(\Big \downarrow \) 1137

\(\displaystyle \frac {2 \sqrt [3]{\sqrt {x}+1} \sqrt [6]{x} \int \frac {x^{5/6}}{\sqrt [3]{\sqrt {x}+1}}d\sqrt {x}}{5 \sqrt [3]{x+\sqrt {x}}}+\frac {3}{5} \left (x+\sqrt {x}\right )^{2/3} x\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {3 \sqrt [3]{\sqrt {x}+1} x^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {11}{3},-\sqrt {x}\right )}{20 \sqrt [3]{x+\sqrt {x}}}+\frac {3}{5} \left (x+\sqrt {x}\right )^{2/3} x\)

Input:

Int[(Sqrt[x] + x)^(2/3),x]
 

Output:

(3*x*(Sqrt[x] + x)^(2/3))/5 + (3*(1 + Sqrt[x])^(1/3)*x^(3/2)*Hypergeometri 
c2F1[1/3, 8/3, 11/3, -Sqrt[x]])/(20*(Sqrt[x] + x)^(1/3))
 

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

rule 1910

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]
 

rule 1924

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 
]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.02

method	result	size
meijerg	\(\frac {3 x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {8}{3}\right ], \left [\frac {11}{3}\right ], -\sqrt {x}\right )}{4}\)	\(17\)

Input:

int((x^(1/2)+x)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

3/4*x^(4/3)*hypergeom([-2/3,8/3],[11/3],-x^(1/2))
 

Fricas [F]

\[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\int { {\left (x + \sqrt {x}\right )}^{\frac {2}{3}} \,d x } \] Input:

integrate((x^(1/2)+x)^(2/3),x, algorithm="fricas")
 

Output:

integral((x + sqrt(x))^(2/3), x)
 

Sympy [F]

\[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\int \left (\sqrt {x} + x\right )^{\frac {2}{3}}\, dx \] Input:

integrate((x**(1/2)+x)**(2/3),x)
 

Output:

Integral((sqrt(x) + x)**(2/3), x)
 

Maxima [F]

\[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\int { {\left (x + \sqrt {x}\right )}^{\frac {2}{3}} \,d x } \] Input:

integrate((x^(1/2)+x)^(2/3),x, algorithm="maxima")
 

Output:

integrate((x + sqrt(x))^(2/3), x)
 

Giac [F]

\[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\int { {\left (x + \sqrt {x}\right )}^{\frac {2}{3}} \,d x } \] Input:

integrate((x^(1/2)+x)^(2/3),x, algorithm="giac")
 

Output:

integrate((x + sqrt(x))^(2/3), x)
 

Mupad [B] (verification not implemented)

Time = 23.66 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.04 \[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\frac {3\,x\,{\left (x+\sqrt {x}\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},\frac {8}{3};\ \frac {11}{3};\ -\sqrt {x}\right )}{4\,{\left (\sqrt {x}+1\right )}^{2/3}} \] Input:

int((x + x^(1/2))^(2/3),x)
 

Output:

(3*x*(x + x^(1/2))^(2/3)*hypergeom([-2/3, 8/3], 11/3, -x^(1/2)))/(4*(x^(1/ 
2) + 1)^(2/3))
 

Reduce [F]

\[ \int \left (\sqrt {x}+x\right )^{2/3} \, dx=\frac {42 \sqrt {x}\, \left (\sqrt {x}+1\right )^{\frac {2}{3}} x -15 \sqrt {x}\, \left (\sqrt {x}+1\right )^{\frac {2}{3}}+12 \left (\sqrt {x}+1\right )^{\frac {2}{3}} x +30 \left (\sqrt {x}+1\right )^{\frac {2}{3}}+5 x^{\frac {1}{6}} \left (\int \frac {\left (\sqrt {x}+1\right )^{\frac {2}{3}}}{2 x^{\frac {7}{6}}+x^{\frac {5}{3}}+x^{\frac {2}{3}}}d x \right )+5 x^{\frac {1}{6}} \left (\int \frac {x^{\frac {2}{3}} \left (\sqrt {x}+1\right )^{\frac {2}{3}}}{2 x^{\frac {7}{3}}+x^{\frac {17}{6}}+x^{\frac {11}{6}}}d x \right )}{70 x^{\frac {1}{6}}} \] Input:

int((x^(1/2)+x)^(2/3),x)
 

Output:

(42*sqrt(x)*(sqrt(x) + 1)**(2/3)*x - 15*sqrt(x)*(sqrt(x) + 1)**(2/3) + 12* 
(sqrt(x) + 1)**(2/3)*x + 30*(sqrt(x) + 1)**(2/3) + 5*x**(1/6)*int((sqrt(x) 
 + 1)**(2/3)/(2*x**(7/6) + x**(2/3)*x + x**(2/3)),x) + 5*x**(1/6)*int((x** 
(2/3)*(sqrt(x) + 1)**(2/3))/(2*x**(4/3)*x + x**(5/6)*x**2 + x**(5/6)*x),x) 
)/(70*x**(1/6))