\(\int \sqrt {b \sqrt [3]{x}+a x} \, dx\) [113]

Optimal result

Integrand size = 15, antiderivative size = 323 \[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=-\frac {4 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{3/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}+\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}+\frac {4 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}}-\frac {2 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

-4/5*b^2*(b+a*x^(2/3))*x^(1/3)/a^(3/2)/(b^(1/2)+a^(1/2)*x^(1/3))/(b*x^(1/3 
)+a*x)^(1/2)+4/15*b*x^(1/3)*(b*x^(1/3)+a*x)^(1/2)/a+2/3*x*(b*x^(1/3)+a*x)^ 
(1/2)+4/5*b^(9/4)*(b^(1/2)+a^(1/2)*x^(1/3))*((b+a*x^(2/3))/(b^(1/2)+a^(1/2 
)*x^(1/3))^2)^(1/2)*x^(1/6)*EllipticE(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4) 
)),1/2*2^(1/2))/a^(7/4)/(b*x^(1/3)+a*x)^(1/2)-2/5*b^(9/4)*(b^(1/2)+a^(1/2) 
*x^(1/3))*((b+a*x^(2/3))/(b^(1/2)+a^(1/2)*x^(1/3))^2)^(1/2)*x^(1/6)*Invers 
eJacobiAM(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)),1/2*2^(1/2))/a^(7/4)/(b*x^(1/3 
)+a*x)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.29 \[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\frac {2 \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x} \left (\left (b+a x^{2/3}\right ) \sqrt {1+\frac {a x^{2/3}}{b}}-b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {a x^{2/3}}{b}\right )\right )}{3 a \sqrt {1+\frac {a x^{2/3}}{b}}} \] Input:

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

Output:

(2*x^(1/3)*Sqrt[b*x^(1/3) + a*x]*((b + a*x^(2/3))*Sqrt[1 + (a*x^(2/3))/b] 
- b*Hypergeometric2F1[-1/2, 3/4, 7/4, -((a*x^(2/3))/b)]))/(3*a*Sqrt[1 + (a 
*x^(2/3))/b])
 

Rubi [A] (warning: unable to verify)

Time = 0.76 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {1910, 1924, 1930, 1938, 266, 834, 27, 761, 1510}

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.

\(\displaystyle \int \sqrt {a x+b \sqrt [3]{x}} \, dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {2}{9} b \int \frac {\sqrt [3]{x}}{\sqrt {\sqrt [3]{x} b+a x}}dx+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}}\)

\(\Big \downarrow \) 1924

\(\displaystyle \frac {2}{3} b \int \frac {x}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {2}{3} b \left (\frac {2 \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{5 a}-\frac {3 b \int \frac {\sqrt [3]{x}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{5 a}\right )+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}}\)

\(\Big \downarrow \) 1938

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{3} b \left (\frac {2 \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{5 a}-\frac {6 b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \left (\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a x^{4/3}+b}}-\frac {\int \frac {\sqrt {b}-\sqrt {a} x^{2/3}}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{\sqrt {a}}\right )}{5 a \sqrt {a x+b \sqrt [3]{x}}}\right )+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{3} b \left (\frac {2 \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{5 a}-\frac {6 b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \left (\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a x^{4/3}+b}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{a} \sqrt {a x^{4/3}+b}}-\frac {\sqrt [6]{x} \sqrt {a x^{4/3}+b}}{\sqrt {a} x^{2/3}+\sqrt {b}}}{\sqrt {a}}\right )}{5 a \sqrt {a x+b \sqrt [3]{x}}}\right )+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}}\)

Input:

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

Output:

(2*x*Sqrt[b*x^(1/3) + a*x])/3 + (2*b*((2*x^(1/3)*Sqrt[b*x^(1/3) + a*x])/(5 
*a) - (6*b*Sqrt[b + a*x^(2/3)]*x^(1/6)*(-((-((x^(1/6)*Sqrt[b + a*x^(4/3)]) 
/(Sqrt[b] + Sqrt[a]*x^(2/3))) + (b^(1/4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[ 
(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^(2/3))^2]*EllipticE[2*ArcTan[(a^(1/4) 
*x^(1/6))/b^(1/4)], 1/2])/(a^(1/4)*Sqrt[b + a*x^(4/3)]))/Sqrt[a]) + (b^(1/ 
4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^( 
2/3))^2]*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(2*a^(3/4)*S 
qrt[b + a*x^(4/3)])))/(5*a*Sqrt[b*x^(1/3) + a*x])))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

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 
]
 

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 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {2 x \sqrt {b \,x^{\frac {1}{3}}+a x}}{3}+\frac {4 b \,x^{\frac {1}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{15 a}-\frac {2 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{5 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(207\)
default	\(\frac {2 x \sqrt {b \,x^{\frac {1}{3}}+a x}}{3}+\frac {4 b \,x^{\frac {1}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{15 a}-\frac {2 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{5 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(207\)

Input:

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

Output:

2/3*x*(b*x^(1/3)+a*x)^(1/2)+4/15*b*x^(1/3)*(b*x^(1/3)+a*x)^(1/2)/a-2/5/a^2 
*b^2*(-a*b)^(1/2)*((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-2*(x 
^(1/3)-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x^(1/3)/(-a*b)^(1/2)*a)^( 
1/2)/(b*x^(1/3)+a*x)^(1/2)*(-2/a*(-a*b)^(1/2)*EllipticE(((x^(1/3)+1/a*(-a* 
b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+1/a*(-a*b)^(1/2)*EllipticF((( 
x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))
 

Fricas [F]

\[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\int { \sqrt {a x + b x^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

integral(sqrt(a*x + b*x^(1/3)), x)
 

Sympy [F]

\[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\int \sqrt {a x + b \sqrt [3]{x}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\int { \sqrt {a x + b x^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\int { \sqrt {a x + b x^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.12 \[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\frac {6\,x\,\sqrt {a\,x+b\,x^{1/3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{7\,\sqrt {\frac {a\,x^{2/3}}{b}+1}} \] Input:

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

Output:

(6*x*(a*x + b*x^(1/3))^(1/2)*hypergeom([-1/2, 7/4], 11/4, -(a*x^(2/3))/b)) 
/(7*((a*x^(2/3))/b + 1)^(1/2))
 

Reduce [F]

\[ \int \sqrt {b \sqrt [3]{x}+a x} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {x^{\frac {2}{3}} a +b}\, b}{15}+\frac {2 x^{\frac {7}{6}} \sqrt {x^{\frac {2}{3}} a +b}\, a}{3}-\frac {2 \left (\int \frac {x^{\frac {1}{6}} \sqrt {x^{\frac {2}{3}} a +b}}{x^{\frac {2}{3}} b +x^{\frac {4}{3}} a}d x \right ) b^{2}}{15}}{a} \] Input:

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

Output:

(2*(2*sqrt(x)*sqrt(x**(2/3)*a + b)*b + 5*x**(1/6)*sqrt(x**(2/3)*a + b)*a*x 
 - int((x**(1/6)*sqrt(x**(2/3)*a + b))/(x**(2/3)*b + x**(1/3)*a*x),x)*b**2 
))/(15*a)