\(\int \sqrt {(1+\sqrt [3]{x}) x} \, dx\) [307]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 126 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\frac {7}{64} \sqrt {\left (1+\sqrt [3]{x}\right ) x}-\frac {21 \sqrt {\left (1+\sqrt [3]{x}\right ) x}}{128 \sqrt [3]{x}}-\frac {7}{80} \sqrt [3]{x} \sqrt {\left (1+\sqrt [3]{x}\right ) x}+\frac {3}{40} x^{2/3} \sqrt {\left (1+\sqrt [3]{x}\right ) x}+\frac {3}{5} x \sqrt {\left (1+\sqrt [3]{x}\right ) x}+\frac {21}{128} \text {arctanh}\left (\frac {x^{2/3}}{\sqrt {\left (1+\sqrt [3]{x}\right ) x}}\right ) \] Output:

21/128*arctanh(x^(2/3)/((1+x^(1/3))*x)^(1/2))+7/64*((1+x^(1/3))*x)^(1/2)-2 
1/128*((1+x^(1/3))*x)^(1/2)/x^(1/3)-7/80*x^(1/3)*((1+x^(1/3))*x)^(1/2)+3/4 
0*x^(2/3)*((1+x^(1/3))*x)^(1/2)+3/5*x*((1+x^(1/3))*x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.55 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\frac {\sqrt {x+x^{4/3}} \left (-105+70 \sqrt [3]{x}-56 x^{2/3}+48 x+384 x^{4/3}\right )}{640 \sqrt [3]{x}}+\frac {21}{128} \text {arctanh}\left (\frac {x^{2/3}}{\sqrt {x+x^{4/3}}}\right ) \] Input:

Integrate[Sqrt[(1 + x^(1/3))*x],x]
 

Output:

(Sqrt[x + x^(4/3)]*(-105 + 70*x^(1/3) - 56*x^(2/3) + 48*x + 384*x^(4/3)))/ 
(640*x^(1/3)) + (21*ArcTanh[x^(2/3)/Sqrt[x + x^(4/3)]])/128
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {2078, 1910, 1930, 1930, 1930, 1916, 1935, 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.

\(\displaystyle \int \sqrt {\left (\sqrt [3]{x}+1\right ) x} \, dx\)

\(\Big \downarrow \) 2078

\(\displaystyle \int \sqrt {x^{4/3}+x}dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{10} \int \frac {x}{\sqrt {x^{4/3}+x}}dx+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \int \frac {x^{2/3}}{\sqrt {x^{4/3}+x}}dx\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \left (\sqrt [3]{x} \sqrt {x^{4/3}+x}-\frac {5}{6} \int \frac {\sqrt [3]{x}}{\sqrt {x^{4/3}+x}}dx\right )\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \left (\sqrt [3]{x} \sqrt {x^{4/3}+x}-\frac {5}{6} \left (\frac {3}{2} \sqrt {x^{4/3}+x}-\frac {3}{4} \int \frac {1}{\sqrt {x^{4/3}+x}}dx\right )\right )\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 1916

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \left (\sqrt [3]{x} \sqrt {x^{4/3}+x}-\frac {5}{6} \left (\frac {3}{2} \sqrt {x^{4/3}+x}-\frac {3}{4} \left (\frac {3 \sqrt {x^{4/3}+x}}{\sqrt [3]{x}}-\frac {1}{2} \int \frac {1}{\sqrt [3]{x} \sqrt {x^{4/3}+x}}dx\right )\right )\right )\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 1935

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \left (\sqrt [3]{x} \sqrt {x^{4/3}+x}-\frac {5}{6} \left (\frac {3}{2} \sqrt {x^{4/3}+x}-\frac {3}{4} \left (\frac {3 \sqrt {x^{4/3}+x}}{\sqrt [3]{x}}-3 \int \frac {1}{1-\frac {x^{4/3}}{x^{4/3}+x}}d\frac {x^{2/3}}{\sqrt {x^{4/3}+x}}\right )\right )\right )\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{10} \left (\frac {3}{4} x^{2/3} \sqrt {x^{4/3}+x}-\frac {7}{8} \left (\sqrt [3]{x} \sqrt {x^{4/3}+x}-\frac {5}{6} \left (\frac {3}{2} \sqrt {x^{4/3}+x}-\frac {3}{4} \left (\frac {3 \sqrt {x^{4/3}+x}}{\sqrt [3]{x}}-3 \text {arctanh}\left (\frac {x^{2/3}}{\sqrt {x^{4/3}+x}}\right )\right )\right )\right )\right )+\frac {3}{5} \sqrt {x^{4/3}+x} x\)

Input:

Int[Sqrt[(1 + x^(1/3))*x],x]
 

Output:

(3*x*Sqrt[x + x^(4/3)])/5 + ((3*x^(2/3)*Sqrt[x + x^(4/3)])/4 - (7*(x^(1/3) 
*Sqrt[x + x^(4/3)] - (5*((3*Sqrt[x + x^(4/3)])/2 - (3*((3*Sqrt[x + x^(4/3) 
])/x^(1/3) - 3*ArcTanh[x^(2/3)/Sqrt[x + x^(4/3)]]))/4))/6))/8)/10
 

Defintions of rubi rules used

rule 219
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])
 

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 1916
Int[1/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[-2*(Sqrt 
[a*x^j + b*x^n]/(b*(n - 2)*x^(n - 1))), x] - Simp[a*((2*n - j - 2)/(b*(n - 
2)))   Int[1/(x^(n - j)*Sqrt[a*x^j + b*x^n]), x], x] /; FreeQ[{a, b}, x] && 
 LtQ[2*(n - 1), j, n]
 

rule 1930
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1935
Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 

rule 2078
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && G 
eneralizedBinomialQ[u, x] &&  !GeneralizedBinomialMatchQ[u, x]
 
Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40

method result size
meijerg \(-\frac {3 \left (\frac {\sqrt {\pi }\, x^{\frac {1}{6}} \left (-1152 x^{\frac {4}{3}}-144 x +168 x^{\frac {2}{3}}-210 x^{\frac {1}{3}}+315\right ) \sqrt {x^{\frac {1}{3}}+1}}{2880}-\frac {7 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {1}{6}}\right )}{64}\right )}{2 \sqrt {\pi }}\) \(51\)
derivativedivides \(\frac {\sqrt {\left (x^{\frac {1}{3}}+1\right ) x}\, \left (768 x^{\frac {2}{3}} \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}-672 x^{\frac {1}{3}} \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}+560 \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}-420 \sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}\, x^{\frac {1}{3}}-210 \sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}+105 \ln \left (\frac {1}{2}+x^{\frac {1}{3}}+\sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}\right )\right )}{1280 x^{\frac {1}{3}} \sqrt {\left (x^{\frac {1}{3}}+1\right ) x^{\frac {1}{3}}}}\) \(108\)
default \(\frac {\sqrt {\left (x^{\frac {1}{3}}+1\right ) x}\, \left (768 x^{\frac {2}{3}} \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}-672 x^{\frac {1}{3}} \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}+560 \left (x^{\frac {2}{3}}+x^{\frac {1}{3}}\right )^{\frac {3}{2}}-420 \sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}\, x^{\frac {1}{3}}-210 \sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}+105 \ln \left (\frac {1}{2}+x^{\frac {1}{3}}+\sqrt {x^{\frac {2}{3}}+x^{\frac {1}{3}}}\right )\right )}{1280 x^{\frac {1}{3}} \sqrt {\left (x^{\frac {1}{3}}+1\right ) x^{\frac {1}{3}}}}\) \(108\)

Input:

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

Output:

-3/2/Pi^(1/2)*(1/2880*Pi^(1/2)*x^(1/6)*(-1152*x^(4/3)-144*x+168*x^(2/3)-21 
0*x^(1/3)+315)*(x^(1/3)+1)^(1/2)-7/64*Pi^(1/2)*arcsinh(x^(1/6)))
 

Fricas [A] (verification not implemented)

Time = 39.89 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\frac {35 \, x \log \left (\frac {32 \, x^{2} + 48 \, x^{\frac {5}{3}} + 2 \, {\left (16 \, x^{\frac {4}{3}} + 16 \, x + 3 \, x^{\frac {2}{3}}\right )} \sqrt {x^{\frac {4}{3}} + x} + 18 \, x^{\frac {4}{3}} + x}{x}\right ) + 2 \, {\left (384 \, x^{2} + 3 \, {\left (16 \, x - 35\right )} x^{\frac {2}{3}} - 56 \, x^{\frac {4}{3}} + 70 \, x\right )} \sqrt {x^{\frac {4}{3}} + x}}{1280 \, x} \] Input:

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

Output:

1/1280*(35*x*log((32*x^2 + 48*x^(5/3) + 2*(16*x^(4/3) + 16*x + 3*x^(2/3))* 
sqrt(x^(4/3) + x) + 18*x^(4/3) + x)/x) + 2*(384*x^2 + 3*(16*x - 35)*x^(2/3 
) - 56*x^(4/3) + 70*x)*sqrt(x^(4/3) + x))/x
 

Sympy [F]

\[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\int \sqrt {x \left (\sqrt [3]{x} + 1\right )}\, dx \] Input:

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

Output:

Integral(sqrt(x*(x**(1/3) + 1)), x)
 

Maxima [F]

\[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\int { \sqrt {x {\left (x^{\frac {1}{3}} + 1\right )}} \,d x } \] Input:

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

Output:

integrate(sqrt(x*(x^(1/3) + 1)), x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\frac {1}{1280} \, {\left (2 \, {\left (2 \, {\left (4 \, {\left (6 \, x^{\frac {1}{3}} {\left (8 \, x^{\frac {1}{3}} + 1\right )} - 7\right )} x^{\frac {1}{3}} + 35\right )} x^{\frac {1}{3}} - 105\right )} \sqrt {x^{\frac {2}{3}} + x^{\frac {1}{3}}} - 105 \, \log \left ({\left | 2 \, \sqrt {x^{\frac {2}{3}} + x^{\frac {1}{3}}} - 2 \, x^{\frac {1}{3}} - 1 \right |}\right )\right )} \mathrm {sgn}\left (x\right ) \] Input:

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

Output:

1/1280*(2*(2*(4*(6*x^(1/3)*(8*x^(1/3) + 1) - 7)*x^(1/3) + 35)*x^(1/3) - 10 
5)*sqrt(x^(2/3) + x^(1/3)) - 105*log(abs(2*sqrt(x^(2/3) + x^(1/3)) - 2*x^( 
1/3) - 1)))*sgn(x)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.21 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=\frac {2\,x\,\sqrt {x+x^{4/3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {9}{2};\ \frac {11}{2};\ -x^{1/3}\right )}{3\,\sqrt {x^{1/3}+1}} \] Input:

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

Output:

(2*x*(x + x^(4/3))^(1/2)*hypergeom([-1/2, 9/2], 11/2, -x^(1/3)))/(3*(x^(1/ 
3) + 1)^(1/2))
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \sqrt {\left (1+\sqrt [3]{x}\right ) x} \, dx=-\frac {7 x^{\frac {5}{6}} \sqrt {x^{\frac {1}{3}}+1}}{80}+\frac {3 \sqrt {x}\, \sqrt {x^{\frac {1}{3}}+1}\, x}{5}+\frac {7 \sqrt {x}\, \sqrt {x^{\frac {1}{3}}+1}}{64}+\frac {3 x^{\frac {7}{6}} \sqrt {x^{\frac {1}{3}}+1}}{40}-\frac {21 x^{\frac {1}{6}} \sqrt {x^{\frac {1}{3}}+1}}{128}+\frac {21 \,\mathrm {log}\left (\sqrt {x^{\frac {1}{3}}+1}+x^{\frac {1}{6}}\right )}{128} \] Input:

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

Output:

( - 56*x**(5/6)*sqrt(x**(1/3) + 1) + 384*sqrt(x)*sqrt(x**(1/3) + 1)*x + 70 
*sqrt(x)*sqrt(x**(1/3) + 1) + 48*x**(1/6)*sqrt(x**(1/3) + 1)*x - 105*x**(1 
/6)*sqrt(x**(1/3) + 1) + 105*log(sqrt(x**(1/3) + 1) + x**(1/6)))/640