\(\int (a+b \sqrt [3]{x}+c x^{2/3})^{5/2} \, dx\) [56]

Optimal result

Integrand size = 20, antiderivative size = 268 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\frac {15 \left (b^2-4 a c\right )^2 \left (9 b^2-4 a c\right ) \left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{16384 c^5}-\frac {5 \left (b^2-4 a c\right ) \left (9 b^2-4 a c\right ) \left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{2048 c^4}+\frac {\left (9 b^2-4 a c\right ) \left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{128 c^3}-\frac {3 \left (9 b-14 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{112 c^2}-\frac {15 \left (b^2-4 a c\right )^3 \left (9 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}\right )}{32768 c^{11/2}} \] Output:

15/16384*(-4*a*c+b^2)^2*(-4*a*c+9*b^2)*(b+2*c*x^(1/3))*(a+b*x^(1/3)+c*x^(2 
/3))^(1/2)/c^5-5/2048*(-4*a*c+b^2)*(-4*a*c+9*b^2)*(b+2*c*x^(1/3))*(a+b*x^( 
1/3)+c*x^(2/3))^(3/2)/c^4+1/128*(-4*a*c+9*b^2)*(b+2*c*x^(1/3))*(a+b*x^(1/3 
)+c*x^(2/3))^(5/2)/c^3-3/112*(9*b-14*c*x^(1/3))*(a+b*x^(1/3)+c*x^(2/3))^(7 
/2)/c^2-15/32768*(-4*a*c+b^2)^3*(-4*a*c+9*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^(11/2)
 

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.16 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\frac {2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}} \left (945 b^7+84 b^5 c \left (-125 a+6 c x^{2/3}\right )-630 b^6 c \sqrt [3]{x}-8 b^4 \left (-791 a c^2 \sqrt [3]{x}+54 c^3 x\right )+16 b^3 c^2 \left (2359 a^2-284 a c x^{2/3}+24 c^2 x^{4/3}\right )+96 b^2 \left (-199 a^2 c^3 \sqrt [3]{x}+36 a c^4 x+648 c^5 x^{5/3}\right )+64 b c^3 \left (-663 a^3+174 a^2 c x^{2/3}+2456 a c^2 x^{4/3}+1584 c^3 x^2\right )+896 \left (15 a^3 c^4 \sqrt [3]{x}+118 a^2 c^5 x+136 a c^6 x^{5/3}+48 c^7 x^{7/3}\right )\right )+105 \left (b^2-4 a c\right )^3 \left (9 b^2-4 a c\right ) \log \left (b-2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}+2 c \sqrt [3]{x}\right )}{229376 c^{11/2}} \] Input:

Integrate[(a + b*x^(1/3) + c*x^(2/3))^(5/2),x]
 

Output:

(2*Sqrt[c]*Sqrt[a + b*x^(1/3) + c*x^(2/3)]*(945*b^7 + 84*b^5*c*(-125*a + 6 
*c*x^(2/3)) - 630*b^6*c*x^(1/3) - 8*b^4*(-791*a*c^2*x^(1/3) + 54*c^3*x) + 
16*b^3*c^2*(2359*a^2 - 284*a*c*x^(2/3) + 24*c^2*x^(4/3)) + 96*b^2*(-199*a^ 
2*c^3*x^(1/3) + 36*a*c^4*x + 648*c^5*x^(5/3)) + 64*b*c^3*(-663*a^3 + 174*a 
^2*c*x^(2/3) + 2456*a*c^2*x^(4/3) + 1584*c^3*x^2) + 896*(15*a^3*c^4*x^(1/3 
) + 118*a^2*c^5*x + 136*a*c^6*x^(5/3) + 48*c^7*x^(7/3))) + 105*(b^2 - 4*a* 
c)^3*(9*b^2 - 4*a*c)*Log[b - 2*Sqrt[c]*Sqrt[a + b*x^(1/3) + c*x^(2/3)] + 2 
*c*x^(1/3)])/(229376*c^(11/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1680, 1166, 27, 1160, 1087, 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.

\(\displaystyle \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx\)

\(\Big \downarrow \) 1680

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

\(\Big \downarrow \) 1166

\(\displaystyle 3 \left (\frac {\int -\frac {1}{2} \left (2 a+9 b \sqrt [3]{x}\right ) \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{5/2}d\sqrt [3]{x}}{8 c}+\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\int \left (2 a+9 b \sqrt [3]{x}\right ) \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{5/2}d\sqrt [3]{x}}{16 c}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \int \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{5/2}d\sqrt [3]{x}}{2 c}}{16 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \int \left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{3/2}d\sqrt [3]{x}}{24 c}\right )}{2 c}}{16 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {a+c x^{2/3}+b \sqrt [3]{x}}d\sqrt [3]{x}}{16 c}\right )}{24 c}\right )}{2 c}}{16 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {a+c x^{2/3}+b \sqrt [3]{x}}}d\sqrt [3]{x}}{8 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}}{16 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^{2/3}}d\frac {b+2 c \sqrt [3]{x}}{\sqrt {a+c x^{2/3}+b \sqrt [3]{x}}}}{4 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}}{16 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{8 c}-\frac {\frac {9 b \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}}{7 c}-\frac {\left (9 b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt [3]{x}\right ) \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{2 \sqrt {c} \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{24 c}\right )}{2 c}}{16 c}\right )\)

Input:

Int[(a + b*x^(1/3) + c*x^(2/3))^(5/2),x]
 

Output:

3*(((a + b*x^(1/3) + c*x^(2/3))^(7/2)*x^(1/3))/(8*c) - ((9*b*(a + b*x^(1/3 
) + c*x^(2/3))^(7/2))/(7*c) - ((9*b^2 - 4*a*c)*(((b + 2*c*x^(1/3))*(a + b* 
x^(1/3) + c*x^(2/3))^(5/2))/(12*c) - (5*(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))*Sqrt[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)))/(24*c)))/(2*c))/(16*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 = 0.02 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {3 x^{\frac {1}{3}} \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}{8 c}-\frac {27 b \left (\frac {\left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}{7 c}-\frac {b \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )}{16 c}-\frac {3 a \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{8 c}\)	\(378\)
default	\(\frac {3 x^{\frac {1}{3}} \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}{8 c}-\frac {27 b \left (\frac {\left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}{7 c}-\frac {b \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )}{16 c}-\frac {3 a \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \,x^{\frac {1}{3}}\right ) \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{\frac {1}{3}}}{\sqrt {c}}+\sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{8 c}\)	\(378\)

Input:

int((a+b*x^(1/3)+c*x^(2/3))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

3/8*x^(1/3)*(a+b*x^(1/3)+c*x^(2/3))^(7/2)/c-27/16*b/c*(1/7*(a+b*x^(1/3)+c* 
x^(2/3))^(7/2)/c-1/2*b/c*(1/12*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^( 
5/2)+5/24*(4*a*c-b^2)/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))))))-3/8*a/c*(1/12*(b+2*c*x^(1/3))/c*(a+b*x^(1/3)+c*x^(2/3))^(5/ 
2)+5/24*(4*a*c-b^2)/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. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+b*x^(1/3)+c*x^(2/3))^(5/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 1.76 (sec) , antiderivative size = 1744, normalized size of antiderivative = 6.51 \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\text {Too large to display} \] Input:

integrate((a+b*x**(1/3)+c*x**(2/3))**(5/2),x)
 

Output:

3*Piecewise(((-a*(a**3 - 3*a*(3*a**2*c + 3*a*b**2 - 5*a*(17*a*c**2/8 + 243 
*b**2*c/224)/(6*c) - 9*b*(237*a*b*c/56 + b**3 - 11*b*(17*a*c**2/8 + 243*b* 
*2*c/224)/(12*c))/(10*c))/(4*c) - 5*b*(3*a**2*b - 4*a*(237*a*b*c/56 + b**3 
 - 11*b*(17*a*c**2/8 + 243*b**2*c/224)/(12*c))/(5*c) - 7*b*(3*a**2*c + 3*a 
*b**2 - 5*a*(17*a*c**2/8 + 243*b**2*c/224)/(6*c) - 9*b*(237*a*b*c/56 + b** 
3 - 11*b*(17*a*c**2/8 + 243*b**2*c/224)/(12*c))/(10*c))/(8*c))/(6*c))/(2*c 
) - b*(-2*a*(3*a**2*b - 4*a*(237*a*b*c/56 + b**3 - 11*b*(17*a*c**2/8 + 243 
*b**2*c/224)/(12*c))/(5*c) - 7*b*(3*a**2*c + 3*a*b**2 - 5*a*(17*a*c**2/8 + 
 243*b**2*c/224)/(6*c) - 9*b*(237*a*b*c/56 + b**3 - 11*b*(17*a*c**2/8 + 24 
3*b**2*c/224)/(12*c))/(10*c))/(8*c))/(3*c) - 3*b*(a**3 - 3*a*(3*a**2*c + 3 
*a*b**2 - 5*a*(17*a*c**2/8 + 243*b**2*c/224)/(6*c) - 9*b*(237*a*b*c/56 + b 
**3 - 11*b*(17*a*c**2/8 + 243*b**2*c/224)/(12*c))/(10*c))/(4*c) - 5*b*(3*a 
**2*b - 4*a*(237*a*b*c/56 + b**3 - 11*b*(17*a*c**2/8 + 243*b**2*c/224)/(12 
*c))/(5*c) - 7*b*(3*a**2*c + 3*a*b**2 - 5*a*(17*a*c**2/8 + 243*b**2*c/224) 
/(6*c) - 9*b*(237*a*b*c/56 + b**3 - 11*b*(17*a*c**2/8 + 243*b**2*c/224)/(1 
2*c))/(10*c))/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqr 
t(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))*(33*b*c*x**2/112 + c 
**2*x**(7/3)/8 + x**(5/3)*(17*a*c**2/8 + 243*b**2*c/224)/(6*c) + x**(4/...
 

Maxima [F(-2)]

Exception generated. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*x^(1/3)+c*x^(2/3))^(5/2),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. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*x^(1/3)+c*x^(2/3))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage3:=type(sage2):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m 
& i,const vecteur & l) Error: Bad Argument ValueDone
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\int {\left (a+b\,x^{1/3}+c\,x^{2/3}\right )}^{5/2} \,d x \] Input:

int((a + b*x^(1/3) + c*x^(2/3))^(5/2),x)
 

Output:

int((a + b*x^(1/3) + c*x^(2/3))^(5/2), x)
 

Reduce [F]

\[ \int \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2} \, dx=\int \left (x^{\frac {2}{3}} c +x^{\frac {1}{3}} b +a \right )^{\frac {5}{2}}d x \] Input:

int((a+b*x^(1/3)+c*x^(2/3))^(5/2),x)
 

Output:

int((a+b*x^(1/3)+c*x^(2/3))^(5/2),x)