\(\int \frac {x^2}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\) [766]

Optimal result

Integrand size = 24, antiderivative size = 167 \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}+\frac {\sqrt [3]{d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}} \] Output:

-1/(-a*d+b*c)/(b*x^3+a)^(1/3)+1/3*d^(1/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a 
)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))*3^(1/2)/(-a*d+b*c)^(4/3)-1/6*d^(1/3)*ln 
(d*x^3+c)/(-a*d+b*c)^(4/3)+1/2*d^(1/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+ 
a)^(1/3))/(-a*d+b*c)^(4/3)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.32 \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}+\frac {\sqrt [3]{d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 (b c-a d)^{4/3}}-\frac {\sqrt [3]{d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 (b c-a d)^{4/3}} \] Input:

Integrate[x^2/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

-(1/((b*c - a*d)*(a + b*x^3)^(1/3))) + (d^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a 
+ b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*(b*c - a*d)^(4/3)) + 
 (d^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*(b*c - a* 
d)^(4/3)) - (d^(1/3)*Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a 
+ b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(6*(b*c - a*d)^(4/3))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {946, 61, 68, 16, 1082, 217}

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.

\(\displaystyle \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 946

\(\displaystyle \frac {1}{3} \int \frac {1}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx^3\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{3} \left (-\frac {d \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3}{b c-a d}-\frac {3}{\sqrt [3]{a+b x^3} (b c-a d)}\right )\)

\(\Big \downarrow \) 68

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[x^2/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

(-3/((b*c - a*d)*(a + b*x^3)^(1/3)) - (d*(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3 
)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(d^(2/3)*(b*c - a*d)^(1/ 
3))) + Log[c + d*x^3]/(2*d^(2/3)*(b*c - a*d)^(1/3)) - (3*Log[(b*c - a*d)^( 
1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(2/3)*(b*c - a*d)^(1/3))))/(b*c - 
a*d))/3
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 61

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

rule 68

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 946

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^n], 
x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m - n 
+ 1, 0]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(\frac {6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}+\left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (6 a d -6 b c \right )}\)	\(197\)

Input:

int(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

2/((a*d-b*c)/d)^(1/3)*(3*((a*d-b*c)/d)^(1/3)+1/2*(2*arctan(1/3*3^(1/2)*(2* 
(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(1/3))/((a*d-b*c)/d)^(1/3))*3^(1/2)+2*ln((b* 
x^3+a)^(1/3)-((a*d-b*c)/d)^(1/3))-ln((b*x^3+a)^(2/3)+((a*d-b*c)/d)^(1/3)*( 
b*x^3+a)^(1/3)+((a*d-b*c)/d)^(2/3)))*(b*x^3+a)^(1/3))/(b*x^3+a)^(1/3)/(6*a 
*d-6*b*c)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.57 \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \log \left (-{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} d - {\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}}\right ) + 2 \, {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \log \left ({\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} d\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )}} \] Input:

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-1/6*(2*sqrt(3)*(b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*arctan(2/3*sqrt(3)*(b*x 
^3 + a)^(1/3)*(-d/(b*c - a*d))^(1/3) + 1/3*sqrt(3)) - (b*x^3 + a)*(-d/(b*c 
 - a*d))^(1/3)*log(-(b*x^3 + a)^(1/3)*(b*c - a*d)*(-d/(b*c - a*d))^(2/3) + 
 (b*x^3 + a)^(2/3)*d - (b*c - a*d)*(-d/(b*c - a*d))^(1/3)) + 2*(b*x^3 + a) 
*(-d/(b*c - a*d))^(1/3)*log((b*c - a*d)*(-d/(b*c - a*d))^(2/3) + (b*x^3 + 
a)^(1/3)*d) + 6*(b*x^3 + a)^(2/3))/((b^2*c - a*b*d)*x^3 + a*b*c - a^2*d)
 

Sympy [F]

\[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{2}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**2/(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(x**2/((a + b*x**3)**(4/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (135) = 270\).

Time = 0.14 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.71 \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d - 2 \, \sqrt {3} a b c d^{2} + \sqrt {3} a^{2} d^{3}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}} \] Input:

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

1/3*d*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^ 
(1/3)))/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) + (-b*c*d^2 + a*d^3)^(2/3)*arctan( 
1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d 
)^(1/3))/(sqrt(3)*b^2*c^2*d - 2*sqrt(3)*a*b*c*d^2 + sqrt(3)*a^2*d^3) - 1/6 
*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c 
 - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d - 2*a*b*c*d^2 + a^2* 
d^3) - 1/((b*x^3 + a)^(1/3)*(b*c - a*d))
 

Mupad [B] (verification not implemented)

Time = 3.65 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.33 \[ \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {1}{{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}+\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{{\left (a\,d-b\,c\right )}^{4/3}} \] Input:

int(x^2/((a + b*x^3)^(4/3)*(c + d*x^3)),x)
 

Output:

1/((a + b*x^3)^(1/3)*(a*d - b*c)) + (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 
- b*c*d^3) - (d^(2/3)*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a 
^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(9*(a*d - b*c)^(8/3))))/(3*(a*d - b*c)^( 
4/3)) - (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 - b*c*d^3) - (d^(2/3)*((3^(1 
/2)*1i)/2 + 1/2)^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2* 
b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(9*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/2 + 1/ 
2))/(3*(a*d - b*c)^(4/3)) + (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 - b*c*d^ 
3) - (d^(2/3)*((3^(1/2)*1i)/6 - 1/6)^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b 
^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(a*d - b*c)^(8/3))*((3^ 
(1/2)*1i)/6 - 1/6))/(a*d - b*c)^(4/3)
 

Reduce [F]

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

int(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(x**2/((a + b*x**3)**(1/3)*a*c + (a + b*x**3)**(1/3)*a*d*x**3 + (a + b* 
x**3)**(1/3)*b*c*x**3 + (a + b*x**3)**(1/3)*b*d*x**6),x)