\(\int (a+b \sqrt [3]{x}+c x^{2/3})^{3/2} \, dx\) [57]

Optimal result

Integrand size = 20, antiderivative size = 211 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{512 c^4}+\frac {\left (7 b^2-4 a c\right ) \left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{64 c^3}-\frac {\left (7 b-10 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{20 c^2}+\frac {3 \left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}\right )}{1024 c^{9/2}} \] Output:

-3/512*(-4*a*c+b^2)*(-4*a*c+7*b^2)*(b+2*c*x^(1/3))*(a+b*x^(1/3)+c*x^(2/3)) 
^(1/2)/c^4+1/64*(-4*a*c+7*b^2)*(b+2*c*x^(1/3))*(a+b*x^(1/3)+c*x^(2/3))^(3/ 
2)/c^3-1/20*(7*b-10*c*x^(1/3))*(a+b*x^(1/3)+c*x^(2/3))^(5/2)/c^2+3/1024*(- 
4*a*c+b^2)^2*(-4*a*c+7*b^2)*arctanh(1/2*(b+2*c*x^(1/3))/c^(1/2)/(a+b*x^(1/ 
3)+c*x^(2/3))^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.04 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\frac {\sqrt {a+b \sqrt [3]{x}+c x^{2/3}} \left (-105 b^5+760 a b^3 c-1296 a^2 b c^2+70 b^4 c \sqrt [3]{x}-432 a b^2 c^2 \sqrt [3]{x}+480 a^2 c^3 \sqrt [3]{x}-56 b^3 c^2 x^{2/3}+288 a b c^3 x^{2/3}+48 b^2 c^3 x+2240 a c^4 x+1664 b c^4 x^{4/3}+1280 c^5 x^{5/3}\right )}{2560 c^4}+\frac {3 \left (-7 b^2+4 a c\right ) \left (-b^2+4 a c\right )^2 \log \left (b-2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}+2 c \sqrt [3]{x}\right )}{1024 c^{9/2}} \] Input:

Integrate[(a + b*x^(1/3) + c*x^(2/3))^(3/2),x]
 

Output:

(Sqrt[a + b*x^(1/3) + c*x^(2/3)]*(-105*b^5 + 760*a*b^3*c - 1296*a^2*b*c^2 
+ 70*b^4*c*x^(1/3) - 432*a*b^2*c^2*x^(1/3) + 480*a^2*c^3*x^(1/3) - 56*b^3* 
c^2*x^(2/3) + 288*a*b*c^3*x^(2/3) + 48*b^2*c^3*x + 2240*a*c^4*x + 1664*b*c 
^4*x^(4/3) + 1280*c^5*x^(5/3)))/(2560*c^4) + (3*(-7*b^2 + 4*a*c)*(-b^2 + 4 
*a*c)^2*Log[b - 2*Sqrt[c]*Sqrt[a + b*x^(1/3) + c*x^(2/3)] + 2*c*x^(1/3)])/ 
(1024*c^(9/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1680, 1166, 27, 1160, 1087, 1087, 1092, 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.

\(\displaystyle \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1680

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

\(\Big \downarrow \) 1166

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

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\int \left (2 a+7 b \sqrt [3]{x}\right ) \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{3/2}d\sqrt [3]{x}}{12 c}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \int \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{3/2}d\sqrt [3]{x}}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {a+c x^{2/3}+b \sqrt [3]{x}}d\sqrt [3]{x}}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {a+c x^{2/3}+b \sqrt [3]{x}}}d\sqrt [3]{x}}{8 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^{2/3}}d\frac {b+2 c \sqrt [3]{x}}{\sqrt {a+c x^{2/3}+b \sqrt [3]{x}}}}{4 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

Input:

Int[(a + b*x^(1/3) + c*x^(2/3))^(3/2),x]
 

Output:

3*(((a + b*x^(1/3) + c*x^(2/3))^(5/2)*x^(1/3))/(6*c) - ((7*b*(a + b*x^(1/3 
) + c*x^(2/3))^(5/2))/(5*c) - ((7*b^2 - 4*a*c)*(((b + 2*c*x^(1/3))*(a + b* 
x^(1/3) + c*x^(2/3))^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*x^(1/3))*S 
qrt[a + b*x^(1/3) + c*x^(2/3)])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^ 
(1/3))/(2*Sqrt[c]*Sqrt[a + b*x^(1/3) + c*x^(2/3)])])/(8*c^(3/2))))/(16*c)) 
)/(2*c))/(12*c))
 

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 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 1087

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])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1160

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]
 

rule 1166

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[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1680

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k 
 = Denominator[n]}, Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* 
n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr 
actionQ[n]
 

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {x^{\frac {1}{3}} \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{2 c}-\frac {7 b \left (\frac {\left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{4 c}-\frac {a \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\)	\(292\)
default	\(\frac {x^{\frac {1}{3}} \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{2 c}-\frac {7 b \left (\frac {\left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{4 c}-\frac {a \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\)	\(292\)

Input:

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

Output:

1/2*x^(1/3)*(a+b*x^(1/3)+c*x^(2/3))^(5/2)/c-7/4*b/c*(1/5*(a+b*x^(1/3)+c*x^ 
(2/3))^(5/2)/c-1/2*b/c*(1/8*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^(3/2 
)+3/16*(4*a*c-b^2)/c*(1/4*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^(1/2)+ 
1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x^(1/3))/c^(1/2)+(a+b*x^(1/3)+c*x^(2/3 
))^(1/2)))))-1/2*a/c*(1/8*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^(3/2)+ 
3/16*(4*a*c-b^2)/c*(1/4*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^(1/2)+1/ 
8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x^(1/3))/c^(1/2)+(a+b*x^(1/3)+c*x^(2/3)) 
^(1/2))))
 

Fricas [F(-1)]

Timed out. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.65 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

3*Piecewise(((-a*(a**2 - 3*a*(7*a*c/6 + b**2/40)/(4*c) - 5*b*(17*a*b/15 - 
7*b*(7*a*c/6 + b**2/40)/(8*c))/(6*c))/(2*c) - b*(-2*a*(17*a*b/15 - 7*b*(7* 
a*c/6 + b**2/40)/(8*c))/(3*c) - 3*b*(a**2 - 3*a*(7*a*c/6 + b**2/40)/(4*c) 
- 5*b*(17*a*b/15 - 7*b*(7*a*c/6 + b**2/40)/(8*c))/(6*c))/(4*c))/(2*c))*Pie 
cewise((log(b + 2*sqrt(c)*sqrt(a + b*x**(1/3) + c*x**(2/3)) + 2*c*x**(1/3) 
)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x**(1/3))*log(b/(2*c) + x** 
(1/3))/sqrt(c*(b/(2*c) + x**(1/3))**2), True)) + sqrt(a + b*x**(1/3) + c*x 
**(2/3))*(13*b*x**(4/3)/60 + c*x**(5/3)/6 + x**(2/3)*(17*a*b/15 - 7*b*(7*a 
*c/6 + b**2/40)/(8*c))/(3*c) + x**(1/3)*(a**2 - 3*a*(7*a*c/6 + b**2/40)/(4 
*c) - 5*b*(17*a*b/15 - 7*b*(7*a*c/6 + b**2/40)/(8*c))/(6*c))/(2*c) + x*(7* 
a*c/6 + b**2/40)/(4*c) + (-2*a*(17*a*b/15 - 7*b*(7*a*c/6 + b**2/40)/(8*c)) 
/(3*c) - 3*b*(a**2 - 3*a*(7*a*c/6 + b**2/40)/(4*c) - 5*b*(17*a*b/15 - 7*b* 
(7*a*c/6 + b**2/40)/(8*c))/(6*c))/(4*c))/c), Ne(c, 0)), (2*(a**2*(a + b*x* 
*(1/3))**(5/2)/5 - 2*a*(a + b*x**(1/3))**(7/2)/7 + (a + b*x**(1/3))**(9/2) 
/9)/b**3, Ne(b, 0)), (a**(3/2)*x/3, True))
 

Maxima [F(-2)]

Exception generated. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value3*sageVARc*(2*(((((7680*sageVARc^5*1/9 
2160/sageVAR
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\int {\left (a+b\,x^{1/3}+c\,x^{2/3}\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2} \, dx=\int \left (x^{\frac {2}{3}} c +x^{\frac {1}{3}} b +a \right )^{\frac {3}{2}}d x \] Input:

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

Output:

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