\(\int \frac {x^2 (a+b x^3)^{4/3}}{c+d x^3} \, dx\) [713]

Optimal result

Integrand size = 24, antiderivative size = 187 \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=-\frac {(b c-a d) \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3}}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}} \] Output:

-(-a*d+b*c)*(b*x^3+a)^(1/3)/d^2+1/4*(b*x^3+a)^(4/3)/d-1/3*(-a*d+b*c)^(4/3) 
*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 
)/d^(7/3)-1/6*(-a*d+b*c)^(4/3)*ln(d*x^3+c)/d^(7/3)+1/2*(-a*d+b*c)^(4/3)*ln 
((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(7/3)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.18 \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (-4 b c+5 a d+b d x^3\right )-4 \sqrt {3} (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+4 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 (b c-a d)^{4/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 d^{7/3}} \] Input:

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

Output:

(3*d^(1/3)*(a + b*x^3)^(1/3)*(-4*b*c + 5*a*d + b*d*x^3) - 4*Sqrt[3]*(b*c - 
 a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/S 
qrt[3]] + 4*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^ 
(1/3)] - 2*(b*c - a*d)^(4/3)*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)])/(12*d^(7/3))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {946, 60, 60, 70, 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.

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

\(\Big \downarrow \) 946

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 70

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

((3*(a + b*x^3)^(4/3))/(4*d) - ((b*c - a*d)*((3*(a + b*x^3)^(1/3))/d - ((b 
*c - a*d)*(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d 
)^(1/3))/Sqrt[3]])/(d^(1/3)*(b*c - a*d)^(2/3))) - Log[c + d*x^3]/(2*d^(1/3 
)*(b*c - a*d)^(2/3)) + (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3 
)])/(2*d^(1/3)*(b*c - a*d)^(2/3))))/d))/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 60

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

rule 70

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) 
, x] + (Simp[3/(2*b*q)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] + Simp[3/(2*b*q^2)   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 = 1.73 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {\frac {15 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d \left (\left (\frac {b \,x^{3}}{5}+a \right ) d -\frac {4 b c}{5}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2}+\left (a d -b c \right )^{2} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{3}}\)	\(205\)

Input:

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

Output:

1/6/((a*d-b*c)/d)^(2/3)*(15/2*((a*d-b*c)/d)^(2/3)*d*((1/5*b*x^3+a)*d-4/5*b 
*c)*(b*x^3+a)^(1/3)+(a*d-b*c)^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))))/d^3
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.32 \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {4 \, \sqrt {3} {\left (b c - a d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 2 \, {\left (b c - a d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 4 \, {\left (b c - a d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (b d x^{3} - 4 \, b c + 5 \, a d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, d^{2}} \] Input:

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

Output:

1/12*(4*sqrt(3)*(b*c - a*d)*(-(b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)* 
(b*x^3 + a)^(1/3)*d*(-(b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a 
*d)) + 2*(b*c - a*d)*(-(b*c - a*d)/d)^(1/3)*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)) - 4*(b*c - a* 
d)*(-(b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) 
+ 3*(b*d*x^3 - 4*b*c + 5*a*d)*(b*x^3 + a)^(1/3))/d^2
 

Sympy [F]

\[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\int \frac {x^{2} \left (a + b x^{3}\right )^{\frac {4}{3}}}{c + d x^{3}}\, dx \] 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. \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] 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 [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.59 \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=-\frac {{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c d^{4} - a d^{5}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{3}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} d^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a d^{3}}{4 \, d^{4}} \] Input:

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

Output:

-1/3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(-(b*c - a*d)/d)^(1/3)*log(abs( 
(b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c*d^4 - a*d^5) + 1/3*sqrt( 
3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*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))/d^3 + 1/6*(-b*c*d^ 
2 + a*d^3)^(1/3)*(b*c - a*d)*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))/d^3 - 1/4*(4*(b*x^3 + a)^(1/ 
3)*b*c*d^2 - (b*x^3 + a)^(4/3)*d^3 - 4*(b*x^3 + a)^(1/3)*a*d^3)/d^4
 

Mupad [B] (verification not implemented)

Time = 3.39 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.63 \[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,d}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )-\frac {{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{7/3}}+\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}{d^2}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )+\frac {\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{7/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a^2\,d^2-6\,a\,b\,c\,d+3\,b^2\,c^2\right )-\frac {\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{d^{7/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}} \] Input:

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

Output:

(a + b*x^3)^(4/3)/(4*d) + (log((a + b*x^3)^(1/3)*(3*a^2*d^2 + 3*b^2*c^2 - 
6*a*b*c*d) - ((a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*(a*d - 
 b*c)^(4/3))/(3*d^(7/3)) + ((a + b*x^3)^(1/3)*(a*d - b*c))/d^2 - (log((a + 
 b*x^3)^(1/3)*(3*a^2*d^2 + 3*b^2*c^2 - 6*a*b*c*d) + (((3^(1/2)*1i)/2 + 1/2 
)*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*((3^(1/2)*1i)/2 + 
1/2)*(a*d - b*c)^(4/3))/(3*d^(7/3)) + (log((a + b*x^3)^(1/3)*(3*a^2*d^2 + 
3*b^2*c^2 - 6*a*b*c*d) - (((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3)*(9*a*d^ 
3 - 9*b*c*d^2))/d^(7/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3))/d^(7/3)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {x^2 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d -3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c \,x^{3}-4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b \,d^{2}+8 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{2} c d -4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{3} c^{2}}{4 b c d} \] Input:

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

Output:

(4*(a + b*x**3)**(1/3)*a**2*d - 3*(a + b*x**3)**(1/3)*a*b*c + (a + b*x**3) 
**(1/3)*b**2*c*x**3 - 4*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d*x**3 + b 
*c*x**3 + b*d*x**6),x)*a**2*b*d**2 + 8*int(((a + b*x**3)**(1/3)*x**5)/(a*c 
 + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b**2*c*d - 4*int(((a + b*x**3)**(1 
/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b**3*c**2)/(4*b*c*d)